Selected Solutions to Munkres’s Topology, 2nd Ed.
Takumi Murayama
December 20, 2014
These solutions are the result of taking MAT365 Topology in the Fall of 2012 at Princeton University. This is not a complete set of solutions; see the List of Solved Exercises at the end. Please e-mail [email protected] with any corrections.
Contents
I
General Topology
3
1 Set Theory and Logic 3
7 Countable and Uncountable Sets . . . 3
2 Topological Spaces and Continuous Functions 3 13 Basis for a Topology . . . 3
16 The Subspace Topology . . . 5
17 Closed Sets and Limit Points . . . 6
18 Continuous Functions . . . 8
19 The Product Topology . . . 9
20 The Metric Topology . . . 10
21 The Metric Topology (continued) . . . 15
22 The Quotient Topology . . . 18
3 Connectedness and Compactness 20 23 Connected Spaces . . . 20
24 Connected Subspaces of the Real Line . . . 21
25 Components and Local Connectedness . . . 26
27 Compact Subspaces of the Real Line . . . 27
4 Countability and Separation Axioms 28
30 The Countability Axioms . . . 28
31 The Separation Axioms . . . 30
32 Normal Spaces . . . 31
33 The Urysohn Lemma . . . 32
34 The Urysohn Metrization Theorem . . . 32
36 Imbeddings of Manifolds . . . 33
II
Algebraic Topology
35
9 The Fundamental Group 35 51 Homotopy of Paths . . . 3552 The Fundamental Group . . . 36
53 Covering Spaces . . . 37
54 The Fundamental Group of the Circle . . . 38
58 Deformation Retracts and Homotopy Type . . . 39
59 The Fundamental Group of Sn . . . . 43
60 Fundamental Groups of Some Surfaces . . . 44
11 The Seifert-van Kampen Theorem 45 67 Direct Sums of Abelian Groups . . . 45
68 Free Products of Groups . . . 45
71 The Fundamental Group of a Wedge of Circles . . . 46
73 The Fundamental Groups of the Torus and the Dunce Cap . . . 48
12 Classification of Surfaces 49 74 Fundamental Groups of Surfaces . . . 49
Part I
General Topology
1
Set Theory and Logic
7
Countable and Uncountable Sets
Exercise 7.5. Determine, for each of the following sets, whether or not it is count-able. Justify your answers.
(j) The set J of all finite subsets of Z+.
Solution for (j). We claim J is countable. Consider I =S∞
n=0In where In is the set
of sequences with n elements. Each In is countable by Theorem 7.6 so I is countable
by Theorem 7.5. Identifying each finite subset in J with the finite sequence with the same elements in increasing order, we see that J ⊆ I, and so J is countable by Corollary 7.3.
2
Topological Spaces and Continuous Functions
13
Basis for a Topology
Exercise 13.3. Show that the collection Tc given in Example 4 of §12 is a topology
on the set X. Is the collection
T∞ = {U | X \ U is infinite or empty or all of X}
a topology on X?
Proof. Recall Example 12.4: Let X be a set; let Tc be the collection of all subsets
U of X such that X \ U either is countable or is all of X. We claim this forms a topology on X; we will follow the numbering for the definition of a topology on p. 76. (1) X \ ∅ = X and X \ X = ∅ is countable; (2) X \S
α∈AUα =
T
α∈A(X \ Uα) is
countable since it is an intersection of countable sets, unless every Uα = ∅, in which
case X \S
α∈AUα = X; (3) X \
T
α∈A finiteUα =
S
α∈A finite(X \ Uα) is countable
since it is the finite union of countable sets, unless every Uα = ∅, in which case
X \T
α∈A finiteUα = X.
Now consider T∞. It is not a topology, for if we let X = [−1, 1] ⊆ R, and let
U1 = [−1, 0) and U2 = (0, 1], we see that both U1, U2 ∈ T∞, but X \ (U1∪ U2) = {0},
Exercise 13.5. Show that if A is a basis for a topology on X, then the topology generated by A equals the intersection of all topologies on X that contain A. Prove the same if A is a subbasis.
Proof. Let TA be the topology generated by A, and TI be the intersection of all
topologies that contain A.
TI ⊆ TA. This follows from the fact that TA ⊇ A, and so is one of the topologies
that is intersected over in the construction of TI.
TA ⊆ TI. Let U ∈ TA; by Lemma 13.1, U =SαAα for some collection {Aα}α ⊆
A. But U =S
αAα ∈ TI since each Aα ∈ TI.
Now let A be a subbasis. The proof that TI ⊆ TA is identical; it remains to
show TA ⊆ TI. Let U ∈ TA; by definition of the topology generated by A, U is the
union of a finite intersection of elements {Aα}α ⊆ A. But then U ∈ TI since each
Aα ∈ TI.
Exercise 13.6. Show that the topologies of R` and RK are not comparable.
Proof. R` 6⊆ RK. For [a, b) ∈ R`, there is no basis element U ∈ RK such that
a ∈ U, U ⊆ [a, b).
RK 6⊆ R`. For (−1, 1) \ K ∈ RK which contains 0, there is no basis element
[a, b) ∈ R` such that 0 ∈ [a, b), [a, b) ⊆ (−1, 1) \ K by the Archimedean property,
that is, for all > 0, there exists N ∈ N such that 1/N < . Exercise 13.7. Consider the following topologies on R:
T1 = the standard topology,
T2 = the topology of RK,
T3 = the finite complement topology,
T4 = the upper limit topology, having all sets (a, b] as basis,
T5 = the topology having all sets (−∞, a) = {x | x < a} as basis.
Determine, for each of these topologies, which of the others it contains. Proof. We claim we have the following Hasse diagram:
T1
T3 T5
T2
T3 ( T1. Inclusion is true since U ∈ T3 =⇒ Uc finite, and so if we let Uc =
{xi}ni=1 with xi in increasing order, U =
Sn
i=0(xi, xi+1), where x0 = −∞, xn+1 = ∞.
Inequality follows since for (a, b) such that −∞ < a, b < ∞, R \ (a, b) is not finite. T5 ( T1. Inclusion is clear since (−∞, a) is of the form (b, c). Inequality follows
since for (b, c) ∈ T1 and x ∈ (b, c), there is no basis element (−∞, a) ∈ T5 such that
x ∈ (−∞, a), (−∞, a) ⊆ (b, c) (if b > −∞).
T3 and T5 are not comparable. T3 6⊆ T5 since R \ {0} ∈ T3, but if we take x > 0,
which is in this set, there is no basis element (−∞, a) ∈ T5 that contains x but is
contained in R \ {0}. T5 6⊆ T3 since (−∞, 0)c is not finite.
T1 ( T2 by Lemma 13.4.
T2 ( T4. For (a, b) ∈ T2 and x ∈ (a, b), (a, x] ∈ T4 and (a, x] ⊆ (a, b). For
(a, b) \ K ∈ T2 and x ∈ (a, b) \ K, we note that x ∈ (1/(n + 1), c] where x < c < 1/n,
x ∈ (a, 0], or x ∈ (1, d], where x < d < b; in all three cases, these sets are subsets of (a, b) \ K and are members of T4. Inequality follows since for (a, b] ∈ T4, there is no
basis element U ∈ T2 such that b ∈ U, U ⊂ (a, b].
16
The Subspace Topology
Exercise 16.8. If L is a straight line in the plane, describe the topology L inherits as a subspace of R` × R and as a subspace of R`× R`. In each case it is a familiar
topology.
Solution. Note that the basis for R`×R consists of elements of the form [a, b)×(c, d).
If L = {(x, y) | x = x0}, then L∩[a, b)×(c, d) = ∅ or {x0}×(c, d), and so defining the
map ϕ : L ∩ (R`× R) → R, {x0} × (c, d) 7→ (c, d), it is bijective, open, and continuous,
and so the topology L inherits is homeomorphic to R with the standard topology. If L has finite slope, we first note that L ∩ (R`× R) = {(x, mx + b) ∈ R2 | x ∈ R},
and that the basis for our topology are the sets of the form ∅, [(a, ma + b), (c, mc + b)), ((a, ma + b), (c, mc + b)) for a, c ∈ R and a < c, by Lemma 16.1. We then define
ϕ : L ∩ (R`× R) → R`, (a, ma + b) 7→ a.
This implies
((a, ma + b), (c, mc + b)) 7→ (a, c), [(a, ma + b), (c, mc + b)) 7→ [a, c).
We claim this defines a homeomorphism with R`. Clearly, it is continuous, for the
basis elements of R` have preimages that are basis elements in the topology on L.
Lemma 13.4. Finally this is a bijection since there exists an inverse just by reversing the arrows above.
For R`× R`, by following the same steps as above if L = {(x, y) | x = x0}, then
L ∩ (R`× R`) is homeomorphic to R`. For L with |m| < ∞, we must split it up into
two cases. When m ≥ 0, we have a similar situation as above, except we only have to consider basis elements of the form [a, b); thus, L ∩ (R`× R`) is homeomorphic
to R`. When m < 0, since for every point (x, y) ∈ L, we can find a basis element
[x, a) × [y, b) ∈ (R`× R`) such that L ∩ [x, a) × [y, b) = {(x, y)}, and these form the
open sets of our new topology by Lemma 16.1. We see then that the topology on L is homeomorphic to the discrete topology on R.
Exercise 16.9. Show that the dictionary order topology on the set R × R is the same as the product topology Rd×R, where Rd denotes R in the discrete topology. Compare
this topology with the standard topology on R2.
Proof. We see that the basis elements of (R × R)lex consist of intervals of the form
(a × b, c × d) for a < c, and for a = c and b < d, as in Example 14.2. These basis elements are open in Rd× R since
(a × b, c × d) = (a, c) × R ∪ {a} × (c, ∞) ∪ {b} × (−∞, d) ∈ TRd×R.
For the reverse situation, consider the basis elements for Rd × R; these consist of
all {a} × (b, c) since {a | a ∈ R} forms a basis for Rd by Example 13.3. But then,
{a}×(b, c) are open in R×R with the order topology since it is of the form (a×b, c×d) for a = c.
We now compare this to the standard topology on R2. Since (a, b)×(c, d) ∈ Rd×R,
we see that R2 ⊆ Rd× R. Moreover, since {a} × (b, c) ∈ (Rd× R) \ R2, we see that
R2 ( Rd× R.
17
Closed Sets and Limit Points
Exercise 17.2. Show that if A is closed in Y and Y is closed in X, then A is closed in X.
Proof. A is closed in Y iff there exists B ⊆ X closed in X such that A = Y ∩ B by Theorem 17.2. But then, A is the intersection of closed sets, and so is closed. Exercise 17.3. Show that if A is closed in X and B is closed in Y , then A × B is closed in X × Y .
Proof. We see that X \A, Y \B are open in X, Y respectively by definition of a closed set. By definition of the product topology, (X \A)×Y, X ×(Y \B) are open in X ×Y . We see that (X \ A) × Y = (X × Y ) \ (A × Y ), X × (Y \ B) = (X × Y ) \ (X × B), and so A × Y, X × B are closed in X × Y . Finally, A × B = (A × Y ) ∩ (X × B), and so is the intersection of closed sets, i.e., closed.
Exercise 17.5. Let X be an ordered set in the order topology. Show that (a, b) ⊂ [a, b]. Under what conditions does equality hold?
Proof. Since (a, b) ⊆ [a, b] closed, and by the definition of closure, (a, b) = \
K⊇(a,b) closed
K ⊆ [a, b].
(a, b) = [a, b] ⇐⇒ a, b ∈ (a, b) ⇐⇒ any basis elements A 3 a, B 3 b intersect (a, b) by Theorem 17.5(b). We claim that this is equivalent to the fact that there is no immediate successor α of a and no immediate predecessor β of b. If either are the case, say for a, then choosing A with upper bound α would not intersect (a, b), and so equality doesn’t hold since a /∈ (a, b); in the other direction, if neither are the case, we see that, say for a, the upper bound of A, α would be such that (a, α) is non-empty, and so A ∩ (a, b) 6= ∅, satisfying the condition for Theorem 17.5(b). The same argument applies when considering b and β, and so our claim holds.
Exercise 17.13. Show that X is Hausdorff if and only if the diagonal ∆ = {x × x | x ∈ X} is closed in X × X.
Proof. Suppose ∆ is closed in X × X, i.e., the complement ∆c is open. This is
equivalent to for all (x, y) ∈ X × X such that x 6= y, there exists a basis element U × V of X × X for U, V open in X such that (x, y) ∈ U × V but (U × V ) ∩ ∆ = ∅. But then, by definition of ∆, this is equivalent to saying for all x, y ∈ X such that x 6= y, there exist open neighborhoods U 3 x and V 3 y such that U ∩ V = ∅, and so X is Hausdorff.
Exercise 17.16. Consider the five topologies on R given in Exercise 7 of §13. (a) Determine the closure of the set K = {1/n | n ∈ Z+} under each of these
topologies.
(b) Which of these topologies satisfy the Hausdorff axiom? the T1 axiom?
Solution for (a). For T3, A closed ⇐⇒ A finite or all of R. Since no finite set
For T5, we claim K = [0, ∞). For x ∈ [0, ∞), the basis elements that contain
x are of the form (−∞, a) for a > x. Since (−∞, a) ∩ K 6= ∅ by the Archimedean property, that is, ∀ > 0∃N ∈ N such that 1/N < , K = [0, ∞) by Theorem 17.5.
For T1, K0 = {0} by Example 17.8, and so K = K ∪ {0}.
For T2, K is closed since R \ K = (−∞, ∞) \ K is a basis element, and so K = K.
For T4, K = K since T4 is finer than T2, and so R \ K is still open.
Solution for (b). T3 is T1 since any finite point set is closed by definition of the finite
complement topology. It is not Hausdorff, for if we choose U 3 x, V 3 y both open, (U ∩ V )c= Uc∪ Vc is finite, where the equality follows from De Morgan’s Laws, and
so U ∩ V is infinite.
T5 is not Hausdorff and not even T1, for R \ {x0} is not a union of basis elements,
and so {x0} is not closed.
T1 is Hausdorff, for if we have x, y ∈ R and 0 < < |x − y|/2, then (x − , x +
) ∩ (y − , y + ) = ∅. Since Hausdorff =⇒ T1, we see that T1 is also T1.
Since T2, T4 are both finer than T1, we see that the open sets constructed above
are still open and separate x, y, and so T2, T4 are still both Hausdorff and thus T1.
18
Continuous Functions
Exercise 18.1. Prove that for functions f : R → R, the -δ definition of continuity implies the open set definition.
Remark. Recall that f is continuous if for every > 0 and x0 ∈ R, there exists a
δ > 0 such that |f (x) − f (x0)| < for all x ∈ R such that |x − x0| < δ.
Proof. Consider x0 ∈ R, and a corresponding neighborhood V of f(x0); we then
have V ⊇ (f (x0) − , f (x0) + ) for some > 0 since V is open. Then, by hypothesis
there exists a δ > 0 such that f (x) ∈ (f (x0) − , f (x0) + ) for all x ∈ R such that
x ∈ (x0− δ, x0+ δ) = U , which is open. Thus, f (U ) ⊆ V , and so f is continuous by
Theorem 18.1.
Exercise 18.12. Let F : R × R → R be defined by the equation F (x × y) =
(
xy/(x2+ y2) if x × y 6= 0 × 0.
0 if x × y = 0 × 0.
(a) Show that F is continuous in each variable separately.
(b) Compute the function g : R → R defined by g(x) = F (x × x). (c) Show that F is not continuous.
Proof of (a). Since F is symmetric in interchanging x ↔ y, we only have to prove ∀y0 ∈ Y , h(x) = F (x × y0) is continuous as a function R → R. For y0 = 0, this is
trivially true for the image of h is (0, 0) with preimage R. Now suppose y0 6= 0; then
we have h(x) = xy0/(x2 + y02). This is continuous since xy0 and x2 + y02 are both
continuous, and so their quotient is also continuous (since also x2+ y02 6= 0), using the -δ definition of continuity (see Theorem 21.5).
Proof of (b). Since F (x × x) for x 6= 0 equals x2/(x2+ x2) = x2/2x2 = 1/2, we have
g(x) = (
1/2 if x 6= 0. 0 if x = 0.
Proof of (c). We claim F (x×y) is not continuous along the line L = {x = y} at (0, 0), i.e., F |L is not continuous at (0, 0); this suffices by Theorem 18.2(d). Note that the
line L in the subspace topology is homeomorphic to R, where the homeomorphism is given by either of the coordinate projection maps. Now the preimage of the closed set {1/2} ⊆ R is L \ {(0, 0)}, which is not closed since R \ {0} is not closed, hence F |L is not continuous, and neither is F .
19
The Product Topology
Exercise 19.6. Let x1, x2, . . . be a sequence of the points of the product space Q Xα.
Show that this sequence converges to the point x if and only if the sequence πα(x1),
πα(x2), . . . converges to πα(x) for each α. Is this fact true if one uses the box topology
instead of the product topology?
Proof. Suppose {xi} → x, and fix some index γ. Then, for any neighborhood Uγ 3
πγ(x), letting U =Q Uα where Uα = Xα for all α 6= γ, there exists N ∈ N such that
xi ∈ U for all i ≥ N , and so πγ(xi) ∈ πγ(U ) = Uγ for all i ≥ N , i.e., {πγ(xi)} →
πγ(x). Note that this direction does not depend on the topology being the product
or box topology.
In the other direction, suppose {πα(xi)} → πα(x) for all α. We take an arbitrary
neighborhood V of x ∈ Q Xα; it then contains a basis element of Q Xα containing
x, which is a product of open setsQ Uα. In the case of the product topology, there
then exist only finite Uα ( Xα, and for these open sets there exist Nα ∈ N such that
πα(xi) ∈ Uα for all i ≥ Nα for each α. Nα = 1 works for all other α. Thus, we can
take N = max(Nα); then, xi ∈Q Uα ⊆ V for all i ≥ N .
We construct a counterexample for this direction in the case of the box topology. Let RN be the box topology on the product of copies of R indexed by N, and let
Then, for each α ∈ N, {πα(xi)} → (0, 0, 0, . . .) =: x, but this sequence does not
converge in the box topology, for the open set Y
i∈N
(−1i,1i) = (−1, 1) × (−12,12) × (−13,13) × · · ·
in the box topology contains x = (0, 0, 0, . . .), but does not contain any xi.
20
The Metric Topology
Exercise 20.4. Consider the product, uniform, and box topologies on Rω. (a) In which topologies are the following functions from R to Rω continuous?
f (t) = (t, 2t, 3t, . . .), g(t) = (t, t, t, . . .), h(t) = (t,12t,13t, . . .).
(b) In which topologies do the following sequences converge? w1 = (1, 1, 1, 1, . . .), x1 = (1, 1, 1, 1, . . .), w2 = (0, 2, 2, 2, . . .), x2 = (0,12,12,12, . . .), w3 = (0, 0, 3, 3, . . .), x3 = (0, 0,13,13, . . .), . . . . y1 = (1, 0, 0, 0, . . .), z1 = (1, 1, 0, 0, . . .), y2 = (12,12, 0, 0, . . .), z2 = (12,12, 0, 0, . . .), y3 = (13,13,31, 0, . . .), z3 = (13,13, 0, 0, . . .), . . . .
Solution for (a). For the product topology, by Theorem 19.6, f, g, h are all continu-ous since each coordinate function is continucontinu-ous. This is because if an open set in the image of a coordinate function is (a, b), its preimage would still be in the form (a0, b0) ⊆ R where a0, b0 are determined by the linear equations defining f, g, h above. Now consider the uniform topology. Note by Theorem 21.1 we can use the familiar -δ definition for continuity since our spaces both are metric spaces. We claim f is not continuous. For, suppose it is continuous. Then, given > 0 and x ∈ R, there exists δ > 0 such that |x−y| < δ =⇒ |f (x)−f (y)| = supn[min(n|x−y|, 1)] < . But, this is a contradiction since for n large, min(n|x − y|, 1) = 1, and so is always greater than . Now consider g. g is continuous since given > 0 and x ∈ R, we let δ < min(, 1) and
therefore have |x−y| < δ =⇒ |f (x)−f (y)| = supn[min(|x−y|, 1)] = min(|x−y|, 1) < min(, 1) ≤ . h is also continuous since given > 0 and x ∈ R, we let δ < min(, 1) and therefore have |x − y| < δ =⇒ |f (x) − f (y)| = supn[min(|x − y|/n, 1)] ≤ min(|x − y|, 1) < min(, 1) ≤ .
For the box topology, since the box topology is finer than the uniform topology by Theorem 20.4, we see that f is not continuous. For, if V open in the uniform topology has preimage that is not open in R, V is still open in the box topology and still has the same non-open preimage. Next, by Example 19.2, we see that g is not continuous. Last, for h, we choose
B = (−1, 1) × (−212, 1 22) × (− 1 32, 1 32) × · · · ,
and suppose its preimage h−1(B) is open. This implies h((−δ, δ)) ⊆ B, and so applying πn gives
hn((−δ, δ)) = (−nδ,nδ) ⊆ (−n12,
1 n2)
for all n, a contradiction.
Solution for (b). We note that since the product topology is Hausdorff by Theorem 19.4 and both the uniform and box topologies are finer than the product topology by Theorems 19.1 and 20.4, if a sequence converges to a point p in one topology, it must converge to the same point in the finer topologies. For, if the sequence converges to q in the finer topology, then it also converges to q in the coarser topology, and by the Hausdorff property p = q.
Consider wn. For the product topology, we recall that any basic open set U =
Q Uα 3 0 is the product of finitely many open subsets of R with infinitely many
copies of R. Letting N be the largest α such that Uα ( R, we see that wn ∈ U
for all n > N since the first N components are zero, and the rest are trivially in the remaining copies of R of U . Thus, wn → 0 in the product topology. Now we
only have to check if the sequence converges to zero in the other topologies by the above. In the uniform topology, ρ(wn, 0) = 1 for all n, and so the sequence does not
converge. For, if we choose any ball U = B(0, r) ⊆ Rω for r < 1, w
n ∈ U for all n./
Finally, since the box topology is finer than the uniform topology by Theorem 20.4, we see that this same open set U is such that wn∈ U for all n, and so w/ n does not
converge in the box topology, either.
Consider xnand yn. We claim they both converge to zero in the uniform topology.
For any open set 0 ∈ U ⊆ Rω in the uniform topology, we can find B(0, ) such that B(0, ) ⊆ U ; then, we can find N such that 1/N < . We then see that xn, yn ∈
B(0, ) ⊆ U for all n ≥ N , and so xn, yn → 0 in the uniform topology. Moreover,
xn, yn → 0 in the product topology as well. For the box topology, though, we see that
neither sequence converges. For, we can construct the set 0 ∈ U =Q∞
n=1(−1/n, 1/n)
(where we only consider sets containing zero by the above), which does not contain xn, yn for any n.
For zn, we see that for any open set 0 ∈ U = Q Uα ⊆ Rω in the box topology,
for N large enough 1/n ∈ U1, U2 for n ≥ N , and so zn ∈ U for all n ≥ N , since by
hypothesis 0 ∈ U , the third component onwards of zn are always in their respective
Uα. Thus, zn → 0 in the box topology; since the box topology is finer than both the
uniform and product topologies, we see that this implies zn converges in the other
two topologies as well.
Exercise 20.5. Let R∞ be the subset of Rω consisting of all sequences that are
eventually zero. What is the closure of R∞ in Rω in the uniform topology? Justify
your answer.
Solution. We claim that A = R∞ is the set of all sequences that converge to zero; we
denote this latter set by X. It suffices to show by Theorem 17.5 that x ∈ X if and only if every basis element U 3 x intersects A. First suppose x ∈ X and let U 3 x be a basis element in the uniform topology; we then see that we can find an open ball B(x, ) ⊆ U . We know we can find N ∈ N such that |xn| < for all n ≥ N by
the definition of convergence. Then, define y such that yn = xn for all n < N , and
zero otherwise; this means y ∈ A. Then, ρ(x, y) < , and so y ∈ B(x, ) ∩ A.
Now suppose x /∈ X; it suffices to find a basis element containing x that does not intersect A. Since x /∈ X, there exists a ball B(0, ) ⊆ R such that {xn}n≥N 6⊆ B(0, )
for any N . The ball B(x, /2), then, does not intersect A, since for any y ∈ B(x, /2), it is not the case that yn = 0 for all n ≥ N for some N .
Exercise 20.6. Let ρ be the uniform metric on Rω. Given x = (x1, x2, . . .) ∈ Rω
and given 0 < < 1, let
U (x, ) = (x1− , x1+ ) × · · · × (xn− , xn+ ) × · · · .
(a) Show that U (x, ) is not equal to the -ball Bρ(x, ).
(b) Show that U (x, ) is not even open in the uniform topology. (c) Show that
Bρ(x, ) =
[
δ<
U (x, δ). Proof of (a). Consider the point
y = (yn)n∈N, yn= xn+ n X k=1 1 2k = xn+ 1 − 1 2n .
yn ∈ (xn− , xn+ ) for all n implies y ∈ U (x, ), while y /∈ Bρ(x, ) since
ρ(x, y) = sup
n
d(xn, yn) = .
Proof of (b). U (x, ) is not open since the point y in (a) has no neighborhood con-tained in Bρ(x, ). For, suppose Bρ(y, δ) ⊆ U (x, ). We can find N such that
δ 2 > ∞ X k=N +1 1 2k,
since P 1/2k converges, and so its tail becomes infinitesimally small. We see that
then, defining y0 such that yn0 = yn for all n 6= N and y0N = yN + δ/2, y0 ∈ Bρ(y, δ)
but y0 ∈ U (x, ) since y/ 0 n = yn+δ/2 > yn+ P∞ k=N +1 1 2k = xn+, a contradiction.
Proof of (c). The ⊇ direction is clear, since each U (x, δ) ⊆ Bρ(x, ) by the fact that
δ < . Now suppose z ∈ Bρ(x, ); if ρ(x, z) = ξ, then we can find δ ∈ (ξ, ) so that
z ∈ U (x, ξ), i.e., Bρ(x, ) ⊆Sδ<U (x, δ).
Exercise 20.8. Let X be the subset of Rω consisting of all sequences x such that
P x2
i converges. Then the formula
d(x, y) = " ∞ X i=1 (xi− yi)2 #1/2
defines a metric on X. On X we have the three topologies it inherits from the box, uniform, and product topologies on Rω. We have also the topology given by the metric
d, which we call the `2-topology.
(a) Show that on X, we have the inclusions
box topology ⊃ `2-topology ⊃ uniform topology.
(b) The set R∞ of all sequences that are eventually zero is contained in X. Show that the four topologies that R∞ inherits as a subspace of X are all distinct. (c) The set
H = Y
n∈Z+
[0, 1/n]
is contained in X; it is called the Hilbert cube. Compare the four topologies that H inherits as a subspace of X.
Proof of (a). box topology ⊃ `2-topology. Let U 3 x be a basis element in the `2
-topology. Then, there exists a basis element U = Bd(x, ) ⊆ U of the `2-topology.
We claim that V = X ∩Q(xi − /2i/2, xi + /2i/2) 3 x basis element of the box
topology is contained in U . Suppose y ∈ V ; then d(x, y)2 = ∞ X i=1 (xi − yi)2 < ∞ X i=1 2 2i = 2 =⇒ y ∈ U,
i.e., V ⊆ U ⊆ U , and box topology ⊃ `2-topology by Lemma 13.3.
`2-topology ⊃ uniform topology. Let U 3 x be a basis element in the uniform
topology. Then, there exists a basis element U = X ∩ Bρ(x, ) ⊆ U of the uniform
topology. We claim V = Bd(x, ) basis element of the `2-topology is contained in U .
If > 1, then trivially V ⊆ U = X, so we assume ≤ 1. Suppose y ∈ V ; then ρ(x, y) = sup |xi− yi| ≤ d(x, y) < =⇒ y ∈ U,
i.e., V ⊆ U ⊆ U , and `2-topology ⊃ uniform topology by Lemma 13.3.
Proof of (b). By (a) and Theorem 20.4, we have the inclusions
box topology ⊃ `2-topology ⊃ uniform topology ⊃ product topology,
since if T1 ⊃ T2 in the ambient space, we would also have T10 ⊃ T20, where Ti0 are
the subspace topologies induced by Ti on X, by the fact that the open sets in the
subspace X are the open sets of the ambient space intersected with X. We claim these are strict inclusions.
box topology ) `2-topology. Consider the open set U = R∞∩Q(−1/i, 1/i) 3 0
in the box topology. Consider any neighborhood V 3 0; then, there exists some V = R∞ ∩ Bd(0, ) for > 0 contained in V. We can find N ∈ N such that
1/N < , and so x such that xi = 0 for all i except xN = 1/N is contained in V
since d(0, x) = 1/N < and therefore V, but not in U . Hence, U is open in the box topology but not in the `2-topology by p. 78, and so box topology ) `2-topology.
`2-topology ) uniform topology. Consider the open set U = R∞ ∩ B
d(0, 1) in
the `2-topology. Consider any neighborhood V 3 0; then, there exists some V =
R∞∩ Bρ(0, ) for > 0 contained in V. We can find N ∈ N such that N2 > 4, and
so x such that xi = /2 for all 1 ≤ i ≤ N and 0 otherwise is contained in V and
therefore V, yet d(x, 0)2 = ∞ X i=1 x2i = N 2 4 > 1 =⇒ d(x, 0) > 1,
and so x /∈ U . Hence, U is open in the `2-topology but not in the uniform topology
by p. 78, and so `2-topology ) uniform topology.
uniform topology ) product topology. Consider the open set U = R∞∩ Bρ(0, 1)
in the uniform topology. Consider any V = R∞∩Q Uα 3 0 where Uα = R for all but
finitely many α. Let N be such that UN = R; then, x such that xi = 0 for all i except
|xN| ≥ 1 is in V but not in U . Hence, U is open in the uniform topology but not in
the product topology by p. 78, and so uniform topology ) product topology. Solution for (c). We claim that
box topology ) (`2-topology = uniform topology = product topology),
i.e., the box topology is strictly finer than the other topologies, which are equal. To show the equality, it suffices to show product topology ⊃ `2-topology, for then we have `2-topology ⊃ uniform topology ⊃ product topology ⊃ `2-topology by
the same argument as in (b), and so we must have equality throughout. So, consider U 3 x open in the `2-topology; then, we can find a basis element U = H∩B
d(x, ) ⊆ U
of the `2-topology for some > 0. Let δ = /[ζ(2) + 1]1/2, and choose N such that
P∞
i=N1/i2 < δ2, which exists since |ζ(2)| < ∞. We claim that
V = H ∩ "N −1 Y i=1 xi− δ i, xi+ δ i × ∞ Y i=N R # 3 x,
basis element of the product topology is contained in U . Suppose y ∈ V ; then
d(x, y)2 = ∞ X i=1 (xi− yi)2 < δ2 N X i=1 1 i2 + ∞ X i=N 1 i2 < δ 2 [ζ(2) + 1] = 2 =⇒ y ∈ U,
i.e., V ⊆ U ⊆ U , and product topology ⊃ `2-topology by Lemma 13.3. Thus, we have the equality desired.
It remains to show the box topology is strictly finer than the other topologies; since the other topologies are equal it suffices to show box topology ) `2-topology. But the open set U = H ∩Q(−1/i, 1/i) 3 0 is open in the box topology but not the `2-topology by the same argument as the proof that box topology ) `2-topology in
(b), and so we are done.
21
The Metric Topology (continued)
Exercise 21.1. Let A ⊆ X. If d is a metric for the topology of X, show that d|A×A
Proof. Clearly d|A×A is a metric since it inherits all the properties for a metric from
the metric d for X; it therefore suffices to show that every basis element for the subspace topology on A contains some open ball defined by d|A×A, and vice versa,
by Lemma 13.3.
So, suppose B is a basis element for the metric topology on A; B = Bd|A×A(x, r)
for some x ∈ A and r ∈ R. But then, B = Bd(x, r) ∩ A by definition, and so the
subspace topology is finer than the metric topology.
Conversely, suppose B is a basis element for the subspace topology on A; it equals Bd(x, r) ∩ A for some basis element Bd(x, r) of X. Let y ∈ Bd(x, r) ∩ A; we see that
the set Bd|A×A(y, r − d(x, y)) ⊆ A ∩ Bd(x, r) is a basis element for the metric topology
contained in A ∩ Bd(x, r), since z ∈ Bd|A×A(y, r − d(x, y)) is such that
d(z, x) ≤ d(z, y) + d(x, y) = d|A×A(z, y) + d(x, y) < r − d(x, y) + d(x, y) = r.
Thus, the metric topology is finer than the subspace topology. Combining the two inclusions, we see the topologies are equal.
Exercise 21.2. Let X and Y be metric spaces with metrics dX and dY, respectively.
Let f : X → Y have the property that for every pair of points x1, x2 of X,
dY(f (x1), f (x2)) = dX(x1, x2).
Show that f is an imbedding. It is called an isometric imbedding of X in Y . Proof. We first show f is injective:
f (x1) = f (x2) =⇒ dY(f (x1), f (x2)) = 0 =⇒ dX(x1, x2) = 0 =⇒ x1 = x2
by properties of metrics, and so we have an injective map.
Now we consider the map f0: X → im(X) ⊆ Y , which is a bijection; it suffices to show that f0, f0−1 are continuous to show f is an imbedding. Let x ∈ X and > 0 be given; then, letting δ = , we have
dX(x, y) < δ =⇒ dY(f0(x), f0(y)) < ,
and so f is continuous. Given y ∈ Y and > 0, letting δ = gives
dY(x, y) = dY(f (a), f (b)) < δ =⇒ dX(f0−1(x), f0−1(y)) = dX(a, b) < ,
where a, b exist by the bijectivity of f , and so f0−1 is continuous.
(a) Show that
ρ(x, y) = max{d1(x1, y1), . . . , dn(xn, yn)}
is a metric for the product space X1× · · · × Xn.
(b) Let di = min{di, 1}. Show that
D(x, y) = sup{di(xi, yi)/i}
is a metric for the product space Q Xi.
Proof of (a). ρ satisfies properties (1) and (2) on p. 119 since the components do; it then suffices to show the triangle inequality. We first have di(xi, zi) ≤ d(xi, yi) +
d(yi, zi) for all i. Then, by definition of ρ, di(xi, zi) ≤ ρ(x, y) + ρ(y, z) for all i. But
since this is true for all i, we have that ρ(x, z) ≤ ρ(x, y) + ρ(y, z).
We now show that this defines a metric for the product space. First let B =Q Ui
be a basis element of Q Xi, and let x ∈ B. For each i, there is an i such that
Bdi(xi, i) ⊆ Ui. Choosing = min{1, . . . , n}, we see that Bρ(x, ) ⊆ B, since if
y ∈ Bρ(x, ), di(xi, yi) ≤ ρ(x, y) < ≤ i, and so y ∈ Q Ui as desired. Thus the
metric topology is finer than the product topology.
Conversely, let Bρ(x, ) be a basis element in the metric topology; since it is the
product Bdi(xi, ), we see that the product topology is finer than the metric topology.
These two facts imply the two topologies are equal.
Proof of (b). D satisfies properties (1) and (2) on p. 119 since the components do; it then suffices to show the triangle inequality. We first have
di(xi, zi) i ≤ di(xi, yi) i + di(yi, zi) i ≤ D(x, y) + D(y, z) for all i. But since this is true for all i, we have that
D(x, z) = sup di(xi, zi) i
≤ D(x, y) + D(y, z).
We now show that this defines a metric for the product space. Let U be open in the metric topology and let x ∈ U ; choose an open ball BD(x, ) ⊆ U . Choose N
such that 1/N < , and let V = Bd
1(x1, ) × · · · × BdN(xN, ) × XN +1× XN +2× · · · .
We claim V ⊆ BD(x, ) ⊆ U . Given y ∈ Q Xi, di(xi, yi)/i ≤ 1/N for i ≥ N .
Therefore, D(x, y) ≤ max d1(x1, z1) 1 , · · · , dN(xN, zN) N , 1 N .
If y ∈ V , this expression is less than , so V ⊆ BD(x, ) as desired, and the product
topology is finer than the metric topology.
Conversely, let U = Q Ui where Ui is open in Xi for α1, . . . , αn and Ui = Xi
otherwise. Let x ∈ U be given, and choose Bd
i(xi, i) ⊆ Xi for i = α1, . . . , αn,
where each i ≤ 1. Then, defining = min{i/i | i = α1, . . . , αn}, we claim that
x ∈ BD(x, ) ⊆ U . Let y be a point of BD(x, ). Then, for all i,
di(xi, yi)
i ≤ D(x, y) < .
Now if i = α1, . . . , αn, then ≤ i/i, so that di(xi, yi) < i ≤ 1; it follows that
|xi− yi| < i, and so y ∈Q Ui as desired. We thus have that the metric topology is
finer than the product topology; combined with the above this implies the topologies are equal.
22
The Quotient Topology
Exercise 22.2.
(a) Let p : X → Y be a continuous map. If there is a continuous map f : Y → X such that p ◦ f equals the identity map of Y , then p is a quotient map.
(b) If A ⊂ X, a retraction of X onto A is a continuous map r : X → A such that r(a) = a for each a ∈ A. Show that a retraction is a quotient map. Proof of (a). If V ⊆ Y with U = p−1(V ) ⊆ X open, f−1(U ) = f−1(p−1(V )) = (p ◦ f )−1(V ) = V is open. Thus, p is a quotient map.
Proof of (b). Let ι : A → X be the inclusion map; then, r ◦ ι is the identity on A, hence r is a quotient map by (a).
Exercise 22.4.
(a) Define an equivalence relation on the plane X = R2 as follows: x0× y0 ∼ x1× y1 if x0 + y20 = x1+ y12.
Let X∗ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it?
(b) Repeat (a) for the equivalence relation
x0× y0 ∼ x1× y1 if x20 + y 2 0 = x 2 1+ y 2 1.
Solution for (a). Set g(x × y) = x + y2 ∈ R. We see it is a surjection onto R
since R × {0} 7→ R. It is continuous since for x0 × y0 ∈ X, given > 0, letting
δ = min(1, /2(|y0| + 1)), ρ(x0× y0, x × y) < δ implies
|g(x0× y0) − g(x × y)| = |(x0+ y20) − (x + y 2 )| ≤ |x0− x| + |y02− y2| ≤ |x0− x| + |y0+ y||y0 − y| ≤ |x0− x| + |y − y0|2+ 2|y0||y0− y| < 2(|y0| + 1)δ < .
If we define f : R → X by x 7→ x × 0, which is continuous since (a, b) × (c, d) maps back to (a, b) which is open in R, we see g◦f is the identity on R, and so g is a quotient map by the lemma above. Since x0 × y0 ∼ x1× y1 ⇐⇒ g(x0 × y0) = g(x1× y1),
by Corollary 22.3, this induces a bijective continuous map g0: X∗ → R, which is a homeomorphism since g was a quotient map.
Solution for (b). Set g(x × y) = x2+ y2 ∈ R. We see it is a surjection onto R
≥0, since
R × {0} 7→ R≥0, and it does not map to anything else since x2+ y2 ≥ 0 for all x, y. It
is continuous since for x0×y0 ∈ X, given > 0, letting δ = min(1, /2(|y0|+|x0|+1)),
ρ(x0× y0, x × y) < δ implies |g(x0× y0) − g(x × y)| = |(x20+ y 2 0) − (x 2 + y2)| ≤ |x2 0− x2| + |y02− y2| ≤ |x0− x||x − x0+ 2x0| + |y0− y||y − y0+ 2y0| ≤ |x0− x|(1 + 2|x0|) + |y0− y|(1 + 2|y0|) ≤ 2δ(|x0| + |y0| + 1) < . We define f : R≥0 → X by x 7→ √
x × 0, which is continuous since the preimage of (a, b) × (c, d), if (c, d) 3 0, is the open set R≥0 ∩ (a0, b0), where a0 = a2 if a ≥ 0,
and −1 otherwise, and similarly for b0 (we chose −1 out of convenience; we really only have to make sure the preimage is a half-open set [0, b0) or the empty set in these cases); if (c, d) 63 0, then the preimage would be empty. We then see g ◦ f is the identity on R≥0, and so g is a quotient map by the lemma above. Since
x0× y0 ∼ x1× y1 ⇐⇒ g(x0× y0) = g(x1 × y1), by Corollary 22.3, this induces a
bijective continuous map g0: X∗ → R≥0, which is a homeomorphism since g was a
quotient map.
Exercise 22.6. Recall that RK denotes the real line in the K-topology. Let Y be the
quotient space obtained by RK by collapsing the set K to a point; let p : RK → Y be
(a) Show that Y satisfies the T1 axiom, but is not Hausdorff.
(b) Show that p × p : RK× RK → Y × Y is not a quotient map.
Proof of (a). Recall by p. 141 that it suffices to show every element in the partition, i.e., one-point sets {x} for x /∈ K and K itself, are closed in RK. The former are
closed since RK is T1 since it is Hausdorff by Example 31.1, and the latter is closed
since it is the complement of R \ K. Thus, Y is T1.
We now show Y is not Hausdorff. We claim that p(0), p(K) are not separable; note p(0) 6= p(K) since they are in different equivalence classes. Suppose Y is Hausdorff, and let V1 3 p(0), V2 3 p(K) be a separation in Y ; they have open preimages
U1 = p−1(V1) 3 0, U2 = p−1(V2) ⊇ K by definition of a quotient map. There then
exists (a, b) \ K 3 0 contained in U1, and choosing n ∈ N such that 1/n < b, there
exists (c, d) 3 1/n contained in U2, where we can assume 1/(n + 1) ≤ c, since if not,
we can take the intersection with (1/(n + 1), d). Then, (c, 1/n) ⊆ U1∩ U2, and so
p((c, 1/n)) ⊆ V1 ∩ V2, which is a contradiction, and so Y is not Hausdorff.
Proof of (b). By Exercise 17.13, we see that since Y is not Hausdorff by (a), the diagonal ∆Y ⊆ Y × Y is not closed. (p−1 × p−1)(∆Y) = ∆K ∪ (K × K), where
∆K ⊆ RK × RK is the diagonal in RK. However, ∆K is closed by Exercise 17.13
since RK is Hausdorff by Example 31.1, and so ∆K ∪ (K × K) is closed since is is
the finite union of closed sets. Thus, the inverse image of the non-closed set ∆Y is
closed, and so p × p is not a quotient map.
3
Connectedness and Compactness
23
Connected Spaces
Exercise 23.8. Determine whether or not Rω is connected in the uniform topology.
Solution. Let Rω = A ∪ B, where A is the set of bounded sequences and B is the set of unbounded sequences of reals. A, B are disjoint, and so it remains to show they are open. Suppose a = (a1, a2, . . .) ∈ A and b = (b1, b2, . . .) ∈ B. Since |ai| < N for
all i for some N , and since |bi| > N + 1 for all i larger than some I, we have that
d(ai, bi) = 1 for all i ≥ I. Thus, ρ(a, b) = 1 for any a ∈ A, b ∈ B, and so the open
balls with radius 1/2 around a, b are fully contained in A, B respectively.
Exercise 23.11. Let p : X → Y be a quotient map. Show that if each set p−1({y}) is connected, and if Y is connected, then X is connected.
Proof. Suppose not. Then, X = A ∪ B for A, B open, disjoint sets. Consider C = {y ∈ Y | p−1({y}) ⊆ A}, D = {y ∈ Y | p−1({y}) ⊆ B}; we see that these sets are such that C ∪ D = Y since p−1({y}) connected implies it is in either A or B by Lemma 23.2. C, D are then disjoint by definition and p−1(C) = A, p−1(D) = B by the fact that p is surjective. p quotient map implies that C, D are then open, and so Y = C ∪ D is a separation, a contradiction.
24
Connected Subspaces of the Real Line
Exercise 24.7.
(a) Let X and Y be ordered sets in the order topology. Show that if f : X → Y is order preserving and surjective, then f is a homeomorphism.
(b) Let X = Y = R+. Given a positive integer n, show that the function f (x) = xn
is order preserving and surjective. Conclude that its inverse, the nth root function, is continuous.
(c) Let X be the subspace (−∞, −1)∪[0, ∞) of R. Show that the function f : X → R defined by setting f (x) = x + 1 if x < −1, and f (x) = x if x ≥ 0, is order preserving and surjective. Is f a homeomorphism? Compare with (a).
Proof of (a). f is injective since if f (a) = f (b) but a 6= b, then (with possible swapping) a < b, and so f (a) < f (b), a contradiction. We thus must show f and f−1 are continuous. But f is continuous since f−1((a, b)) = (f−1(a), f−1(b)) is open (apply the same argument to the intervals of the form [a0, b), (a, b0] for a0, b0
minimal and maximal respectively); the same argument applies for f−1 as well.
Proof of (b). f (x) = xn is order preserving since a < b =⇒ a/b < 1 =⇒ an/bn <
1 =⇒ an < bn =⇒ f (a) < f (b). f is continuous since it is the product of n
copies of the identity function, which is continuous. We want to show f is surjective. Letting N = {xn | x ∈ Z
≥0}, we see that every real number y ∈ Y is between two
consecutive members of N , or it is already an nth power of an integer, in which case it is trivially mapped to by its nth root. In the case y ∈ Y is not an nth power, we have f (n) < y < f (n + 1), and so by the Intermediate value theorem (Theorem 24.3), we see that there exists a point c ∈ X such that f (c) = r, i.e., f is surjective. Since f is order preserving and surjective, by (a) it is then a homeomorphism, and so f−1, the nth root function, is also continuous.
Proof of (c). f is order-preserving on (−∞, −1) since a < b =⇒ f (a) = a + 1 < b + 1 = f (b), and on [0, ∞) since it is the identity. We check that it is order preserving around the boundary. So, suppose a < −1 and b ≥ 0. Then, a < b but
also a + 1 < 0 ≤ b, and so f is order-preserving. f is surjective since if x ∈ R, if x ≥ 0 its preimage is itself, and if x < 0, its preimage is x − 1. f is not a homeomorphism by Theorem 23.6 since R is connected but X is not, by considering f−1(R).
This does not contradict (a) since X is not in the order topology. Even if R is in the order topology, the subspace topology induced on X is not the order topology. For, (−1/2, 1) is open in R, and so (−1/2, 1) ∩ X = [0, 1) is open in X, but not open in the order topology on X.
Exercise 24.8.
(a) Is a product of path-connected spaces necessarily path connected? (b) If A ⊂ X and A is path connected, is A necessarily path connected?
(c) If f : X → Y is continuous and X is path connected, is f (X) necessarily path connected?
(d) If {Aα} is a collection of path-connected subspaces of X and if T Aα 6= ∅, is
S Aα necessarily path connected?
Solution for (a). Yes. Let X =Q Xα, x, y ∈ X. Since each Xαis path connected, we
have fα: [0, 1] → Xα continuous such that fα(0) = xα, fα(1) = yα, where we assume
the closed interval is [0, 1] after composition with multiplication and addition, which are continuous operations. Thus we have the function f = (fα), which is continuous
by Theorem 19.6, with f (0) = x, f (1) = y, and so X is path-connected.
Solution for (b). No, since S in Example 24.7 is not path-connected while S is: it is the image of the continuous map x 7→ (x, sin(1/x)) from R>0 to R2.
Solution for (c). Yes. For, let x, y ∈ f (X), and choose x0 ∈ f−1(x), y0 ∈ f−1(y).
Then, there exists continuous g : [a, b] → X such that g(a) = x0, g(b) = y0, and so
its composition f ◦ g : [a, b] → Y is continuous with (f ◦ g)(a) = x, (f ◦ g)(b) = y. Solution for (d). Yes. Let x, y ∈ S Aα and p ∈ T Aα. Then, there exists a
contin-uous map f : [a, b] → S Aα with f (a) = x, f (b) = p, and similarly g : [b, c] → S Aα
with f (b) = p, f (c) = y, since a, p ∈ Aα for some α and similarly for y (we are free
to have Dom g = [b, c] by composition with multiplication and addition, which are continuous). Then, by the pasting lemma (Theorem 18.3) since f, g are continuous and f (b) = g(b), we see that h = f on [a, b] and h = g on [b, c] is a continuous map such that h(a) = x, h(c) = y.
Exercise 24.12. Recall that SΩ denotes the minimal uncountable well-ordered set.
Let L denote the ordered set SΩ × [0, 1) in the dictionary order, with its smallest
Theorem. The long line is path connected and locally homeomorphic to R, but it cannot be imbedded in R.
(a) Let X be an ordered set; let a < b < c be points of X. Show that [a, c) has the order type of [0, 1) if and only if both [a, b) and [b, c) have the order type of [0, 1).
(b) Let X be an ordered set. Let x0 < x1 < · · · be an increasing sequence of points
of X; suppose b = sup{xi}. Show that [x0, b) has the order type of [0, 1) if and
only if each interval [xi, xi+1) has the order type of [0, 1).
(c) Let a0 denote the smallest element of SΩ. For each element a of SΩ different
from a0, show that the interval [a0× 0, a × 1) of SΩ× [0, 1) has the order type
of [0, 1).
(d) Show that L is path connected.
(e) Show that every point of L has a neighborhood homeomorphic with an open interval in R.
(f ) Show that L cannot be imbedded in R, or indeed in Rn for any n.
Proof of (a). We first note order-preserving maps are injective. Letting f : A → B be such a map, if a1, a2 ∈ A, then one is larger than the other by the comparability
property of order relations, so one of f (a1), f (a2) is larger than the other, hence
unequal.
Now suppose [a, c) has order type [0, 1), and let f : [0, 1) → [a, c) be the order isomorphism. We claim
g(x) = f {[f−1(b)]x}, h(x) = f {f−1(b) + [1 − f−1(b)]x}
define order isomorphisms g : [0, 1) → [a, b) and h : [0, 1) → [b, c). They are order-preserving since if x, y ∈ [0, 1),
x < y =⇒ [f−1(b)]x < [f−1(b)]y =⇒ g(x) = f {[f−1(b)]x} < f {[f−1(b)]y} = g(y) x < y =⇒ f−1(b) + [1 − f−1(b)]x < f−1(b) + [1 − f−1(b)]y
=⇒ h(x) = f {f−1(b) + [1 − f−1(b)]x} < f {f−1(b) + [1 − f−1(b)]y} = h(y)
where the first implications are due to our linear transformations being strictly mono-tonic increasing, and the second since f is order-preserving. This also implies injec-tivity by the above. It remains to show surjecinjec-tivity. Let z ∈ [a, b), z0 ∈ [b, c). Then, 0 ≤ f−1(z) < f−1(b) and f−1(b) ≤ f−1(z0) < 1, and so g f −1(z) f−1(b) = z, h f −1(z0) − f−1(b) 1 − f−1(b) = z0.
Conversely, suppose [a, b) and [b, c) have order type [0, 1), and let g : [0, 1) → [a, b), h : [0, 1) → [b, c) be the order isomorphisms. We claim
f (x) = (
g(2x) if 0 ≤ x < 1/2 h(2x − 1) if 1/2 ≤ x < 1
is an order isomorphism. It preserves orders since g, h preserve orders on their respective domains, and since if x < 1/2 ≤ y, applying f gives
f (x) = g(2x) < b ≤ h(2y − 1) = f (y).
This also shows injectivity by the above. f is surjective since if z ∈ [a, c), z < b =⇒ f [g−1(z)/2] = z, z ≥ b =⇒ f {[h−1(z) + 1]/2} = z.
Proof of (b). Suppose [x0, b) has order type [0, 1). For any i ∈ Z+, by (a), [xi, b) has
order type [0, 1); applying (a) again gives that [xi, xi+1) has order type [0, 1).
Now suppose every [xi, xi+1) has order type [0, 1). If fi: [0, 1) → [xi, xi+1) are
order isomorphisms, first define
f : [0, ∞) → [x0, b), x 7→ fi(x − i) if x ∈ [i, i + 1),
which is well-defined since any x ∈ [0, ∞) is in some set of the form [i, i + 1). We claim f is an order isomorphism. If x, y ∈ [0, ∞), then x ∈ [i, i + 1), y ∈ [j, j + 1) for some i, j. Suppose x < y. Then,
i 6= j =⇒ f (x) = fi(x − i) < xi+1≤ xj ≤ fj(y − j) = f (y),
i = j =⇒ f (x) = fi(x − i) < fi(y − i) = f (y),
since the fi are order-preserving; this also implies injectivity by the above. To show
surjectivity, we first know f maps onto S
i[xi, xi+1) by definition. So let z ∈ [x0, b).
Since b is the least upper bound of the {xi}, z is not an upper bound, so there exists i
such that z ∈ [xi, xi+1). But then, since the fi are bijective as well, f (fi−1(z) +i) = z.
Now let g : [0, 1) → [0, ∞) be defined as x 7→ x/(1 − x); this is an order isomor-phism since it has inverse x/(1 + x), and since it is strictly monotonic increasing. Thus, f ◦ g : [0, 1) → [x0, b) is a bijection, and preserves orientation since f, g do.
Proof of (c). Let a > a0. We proceed by transfinite induction. SΩ is a well-ordered
set, and so if we let J be the set of a ∈ SΩ such that the claim holds, it suffices to
We first show that either a has an immediate predecessor or there exists a se-quence {ai} ⊆ Sa such that a = sup{ai}. Suppose a does not have an immediate
predecessor. Then, we have the section Sa = {bi}, which is countable by definition
of SΩ. (b1, a] 6= ∅ since a has no immediate predecessor, and so let a1 ∈ (b1, a]. We
construct the ai inductively as follows: if we have an, let an+1 ∈ (sup{an, bn+1}, a],
which is nonempty as above. We then get a sequence of elements a1 < a2 < · · · < a.
But since an > bn for all n by construction, we see that a ≥ sup{ai}. Moreover, if
a > sup{ai}, then sup{ai} = bk for some k, for Sa contains all elements less than a,
and hence sup{ai} < ak, contradicting that sup{ai} is an upper bound.
Now suppose Sa ⊆ J. If a has an immediate predecessor a−1, then [a0×0, a×1) =
[a0× 0, (a − 1) × 1) ∪ [a × 0, a × 1) has order type [0, 1) by (a), for we have the order
isomorphism [a × 0, a × 1) → [0, 1) defined by a × x 7→ x, which is trivially bijective and order-preserving since SΩ× [0, 1) was constructed with the dictionary order. On
the other hand, if a does not have an immediate predecessor, then there exists a sequence {ai} ⊆ Sa such that a = sup{ai}, and so the claim follows by (b).
Proof of (d). Let a × b, a0× b0 be two points in L; suppose without loss of generality
that a × b < a0× b0. By (c), [a
0× 0, a × 1) and [a0× 0, a0× 1) have order type [0, 1); by
9a), this implies [a0×0, a×b) and [a0×0, a0×b0) have order type [0, 1). Hence, by (a),
Y = [a × b, a0× b0) has order type [0, 1). Let f : [0, 1) → Y be the order isomorphism.
We claim f is continuous. First, since Y is an interval, it is convex, and so by Theorem 16.4 the order topology on Y is the same as the subspace topology on Y inherited from L. Then, for any basis set A = (c × d, c0× d0) ⊆ Y , f−1(A) = (f−1(c × d), f−1(c0× d0))
since f is an order isomorphism, and moreover this preimage is open. Also, for any basis set B = [a × b, c0× d0) ⊆ Y , f−1(B) = [f−1(a × b), f−1(c0× d0)), which is again
open. Thus, f is continuous. Finally, if we define F (x) =
(
f (x) if x ∈ [0, 1) a0× b0 if x = 1
we have a continuous path F : [0, 1] → [a × b, a0× b0] by the pasting lemma (Theorem
18.3), and so L is path connected.
Proof of (e). Let a × b be a point in L. Since SΩ× [0, 1) does not have a maximal
element, there is some a0× b0 > a × b. Now by (c), there exists an order isomorphism
f : [0, 1) → [a0× 0, a0× b0). Restricting f to (0, 1), we get another order isomorphism
f : (0, 1) → (a0× 0, a0 × b0). The set [a0 × 0, a0× b0) is open in SΩ × [0, 1), and so
(a0× 0, a0× b0) is open in L, and is a neighborhood of a × b.
We claim (a0× 0, a0× b0) is homeomorphic to (0, 1). We already have a bijection
open as well. But if (x, y) ⊆ (0, 1) is a basis set, then f (x, y) = (f (x), f (y)) since f is an order isomorphism, and moreover open since the topology on (a0× 0, a0× b0) is
the order topology.
Proof of (f ). Suppose L could be imbedded in Rn; then, every subspace of Rn has a countable basis by Example 30.1, and since L is homeomorphic with such a subspace, it also has a countable basis. Now, since X = (SΩ× {0}) \ {a0× 0} is a convex subset
of L, the subspace topology on X is the same as the order topology by Theorem 16.4. Thus, the intersection of the countable basis for L with X forms a countable basis by Theorem 30.2. This implies that there is a countable subset Y of X that is dense in X by Theorem 30.3. By Theorem 10.3, though, this subset Y has an upper bound x in X. Thus, ∅ 6= (x, Ω) ⊆ X \ Y , and so the closure of Y is not all of X, a contradiction.
25
Components and Local Connectedness
Exercise 25.2.
(a) What are the components and path components of Rω(in the product topology)? (b) Consider Rω in the uniform topology. Show that x and y lie in the same
component of Rω if and only if the sequence
x − y = (x1− y1, x2− y2, . . .)
is bounded.
(c) Give Rω the box topology. Show that x and y lie in the same component of Rω
if and only if the sequence x − y is “eventually zero.”
Solution for (a). By Exercise 24.8(a), Rω is path connected, for Theorem 19.6 is not
limited to finite product topologies. Thus, Rω is the only path component, and so
Rω is the only component as well since path connected =⇒ connected.
Proof of (b). We first define ϕ : x 7→ x − y. We recall that since ρ(ϕ(x), ϕ(z)) = ρ(x, z), by Exercise 21.2 ϕ is an isometric imbedding that is moreover surjective (the preimage of any z is z + y), ϕ is a homeomorphism. Thus, x − y is in the same component as 0 if and only if x is in the same component as y, for ϕ, ϕ−1 do not modify the topology of Rω.
It therefore suffices to check the case y = 0. Suppose x is bounded; then, we define f : [0, 1] → Rω where f (t) = (x1t, x2t, . . .). This is continuous since given
of x. Thus, f connects 0 and x, i.e., they are in the same path component, and therefore the same component by Theorem 25.5.
Conversely, recall by Exercise 23.8 that we have the separation Rω = A ∪ B,
where A is the set of bounded sequences and B is the set of unbounded sequences of reals. If x is unbounded it is in B and so is not in the same component as 0.
Proof of (c). x “eventually zero” here means that xi = 0 for all i ≥ N for some N .
Note by the same argument as in (b), it suffices to consider the case y = 0.
Suppose first that x is not eventually zero. Define the function f = (fn), where
fn(a) = na/|xn| if xn6= 0, and a otherwise. f is continuous since each fnis continuous
since it is linear, and so if fn−1(Un) = Vn, we have f−1(Q Un) = Q Vn. Note that
this is a bijection since each component has an inverse fn−1(a) = |xn|a/n if xn 6= 0,
and a otherwise, and moreover since the inverse is continuous since each component is linear, we have a homeomorphism f : Rω → Rω. Since there are infinitely many n
such that xn 6= 0, and so infinitely many n such that fn(xn) = n, we have that f (x) is
unbounded, and thus, by the separation of Rω in the box topology in Example 23.6,
we have that f (x) and 0 are in different components. Since f is a homeomorphism, this implies x and 0 are in different components as well.
Conversely, suppose x is eventually zero. Then, xn = 0 for all n ≥ N for some
N , and so x ∈ RN × {0} × {0} × · · · ⊆ Rω; this subspace is homeomorphic to
RN. Since RN is connected by Theorem 23.6, we see that x and 0 are in the same component.
27
Compact Subspaces of the Real Line
Exercise 27.4. Show that a connected metric space having more than one point is uncountable.
Proof. Let X be a connected metric space with the metric d, and let x0, x1 ∈ X
be distinct. Let d(x0, x1) = r, and define f (x) = d(x0, x). f is continuous by the
discussion on p. 175. We see that f (x0) = 0, f (x) = r, and so by the intermediate
value theorem (Theorem 24.3), f (X) ⊇ [0, r], i.e., f maps onto [0, r].
Now suppose X is countable. Then, by Theorem 7.1 there exists a surjective func-tion g : Z+ → X, and so f ◦ g : Z+ → f (X) maps onto [0, r], which is a contradiction
since [0, r] is uncountable by Corollary 27.8.
29
Local Compactness
Proof. Suppose X = [0, 1]ω is locally compact, and in particular at 0. Then, there
exists C compact that contains a neighborhood U 3 0. There exists X ∩Bρ(0, ) ⊆ U ;
we see that {0, /3}ω ⊆ X ∩ B
ρ(0, ). {0, /3}ω is closed since
{0, /3}ω =Y
{0, /3} =Y{0, /3} =Y{0, /3} = {0, /3}ω
in the product topology by Theorem 19.5, which is finer than the uniform topology, and so it is compact by Theorem 26.2 since it is a closed subset of C compact, i.e., limit point compact by Theorem 28.2.
We claim this is a contradiction. Consider x ∈ X, and the ball X ∩ Bρ(x, /9).
Note that the distance between any two distinct points of {0, /3}ω is /3, and so since the diameter of X ∩ Bρ(x, /9) is at most 2/9, X ∩ Bρ(x, /9) contains at most
one point of {0, /3}ω. Thus, {0, /3}ω contains no limit points, and so is not limit
point compact, a contradiction.
Exercise 29.8. Show that the one-point compactification of Z+ is homeomorphic to
the subspace {0} ∪ {1/n | n ∈ Z+} of R.
Proof. Let K = {1/n | n ∈ Z+}. Let f : R+ → R+ such that f (x) = 1/x; this is a
homeomorphism since it is continuous and is its own inverse. By Theorem 18.2(d) and 18.2(e), f : Z+ → f (Z+) = K is continuous, and again is a homeomorphism
since it is its own inverse. Now consider Y = {0} ∪ K, which is closed and bounded and therefore compact by Theorem 27.3, and Hausdorff by Theorem 17.11. Since K0 = {0} by Example 17.8, we know Y is the one-point compactification of K. If X = {p} ∪ Z+ is the one-point compactification of Z+, and letting g : p 7→ 0 ∈ Y ,
which is clearly continuous, the function h : X → Y defined by the pasting lemma (Theorem 18.3) applied to f, g is also continuous, and has continuous inverse defined by the pasting lemma applied to f−1, g−1, and so is a homeomorphism X ↔ Y .
4
Countability and Separation Axioms
30
The Countability Axioms
Exercise 30.4. Every compact metrizable space X has a countable basis.
Proof. For given n ∈ Z+, we have an open cover of X by B(x, 1/n) for each x ∈ X;
since X is compact, let An be the finite subcover.
S
nAn is countable since it is the
countable union of finite sets; we claim it is a basis for X. Let U ⊆ X open and x ∈ U . Since X is metrizable, there exists B(x, δ) ⊆ U for some δ > 0. Let N such
that 2/N < δ. Since AN covers X, there exists B(y, 1/N ) 3 x. B(y, 1/N ) ⊆ B(x, δ),
for if we choose z ∈ B(y, 1/N ), d(x, z) ≤ d(x, y) + d(y, z) ≤ 1/N + 1/N < δ. Thus, x ∈ B(y, 1/N ) ⊆ U , and soS
nAn is a countable basis by Lemma 13.2.
Exercise 30.5.
(a) Show that every metrizable space with a countable dense subset has a countable basis.
(b) Show that every metrizable Lindel¨of space has a countable basis.
Proof of (a). Let X be a metrizable space and A a countable dense subset. We claim that the set of open balls in X below is a basis for X:
B := {B(a, 1/n) ⊆ X | a ∈ A, n ∈ N}.
Note B is countable since is in bijection with A × N. So let x be contained in an open subset U ⊆ X; since X is metrizable, x ∈ B(x, ) ⊆ U for some small . Let n be such that 1/n < /2. Then, since A is dense, some a ∈ A is contained in B(x, 1/n), and conversely x ∈ B(a, 1/n). By the triangle inequality, x ∈ B(a, 1/n) ⊆ U , so by Lemma 13.2 we are done.
Proof of (b). Let X be a metrizable space. Then, the set of open balls e
Bn:= {B(x, 1/n) ⊆ X | x ∈ X}
is an open cover of X for each n ∈ N; since X is Lindel¨of, it has a countable subcover Bn. We claim B :=
S
n∈NBn is a basis for X; note it is countable since it is
a countable union of countable sets. So let x ∈ B(x, ) ⊆ U as before, and let n such that 1/n < /2. Then, there is some x0 ∈ X such that B(x0, 1/n) ∈ B
n contains x.
By the triangle inequality, x ∈ B(x0, 1/n) ⊆ U , so by Lemma 13.2 we are done.
Exercise 30.8. Which of our four countability axioms does Rω in the uniform topol-ogy satisfy?
Solution. Rω is first countable since it is metrizable (see p. 130 and Example 30.2), but is not second countable by Example 30.2. By Exercise 30.5, we then see that Rω does not have a countable dense subset, and is also not Lindel¨of.
Exercise 30.9. Let A be a closed subspace of X. Show that if X is Lindel¨of, then A is Lindel¨of. Show by example that if X has a countable dense subset, A need not have a countable dense subset.
Proof. X is Lindel¨of if and only if a collection of closed subsets of X with empty intersection has a countable subcollection with empty intersection by taking comple-ments in Theorem 30.3(a). Now suppose we have a collection C of closed subsets of A with empty intersection; it is then also a collection of closed subsets of X with empty intersection by Theorem 17.3 since A is closed, and so has a countable subcollection with empty intersection since X is Lindel¨of. Thus, A is also Lindel¨of.
Now let X = R2
`. We see Q2 is countable, and is dense in X since if we take
x ∈ X and a neighborhood U 3 x, there exists a basis element [a, b) × [c, d) ⊆ U containing x, which intersects Q2 by the fact that (a, b) × (c, d) ∩ Q2 6= ∅ since Q is
dense in R. Thus, X has a countable dense subset; we claim that L = {x × (−x) | x ∈ R`} is a closed subspace of X that does not have a countable dense subset.
L is closed since if (x1, x2) ∈ X \ L, then letting d = x1 + x2, the basis element
[x1− d/3, x1+ d/3), [x2 − d/3, x2+ d/3) does not intersect L. But then, L has the
discrete topology since {(x, −x)} = L ∩ [x, b) × [−x, d) is open in L. Thus, if A ⊆ L, A = A by discreteness, and so A = L is true if and only if A = L. Thus, L has no countable dense subset.
Exercise 30.17. Give Rω the box topology. Let Q∞ denote the subspace consisting of sequences of rationals that end in an infinite string of 0’s. Which of our four countability axioms does this space satisfy?
Proof. We claim Q∞is not first countable, and therefore not second countable. Sup-pose we have a countable basis {Ui} at 0 = (0, 0, 0, . . .) ∈ Qω. Let Vj ( πj(Uj) open
in Q with the subspace topology induced by R. Then, the neighborhood Q
jVj 3 0
does not contain any Ui, so {Ui} is not a basis and Q∞is not first or second countable.
We now show Q∞ has a countable dense subset. For, Qn = Qn× {0} × {0} are
countable since they are finite products of countable sets, and so their countable union Q∞ = S Qn
is also countable. Thus, Q∞ is countable and so is a countable dense subset of itself.
We now show Q∞ is Lindel¨of. Suppose V is an open covering of Q∞. Then, since Q∞ is countable, choosing for every x ∈ Q∞ one element V ∈ V such that x ∈ V , we get a countable subcover of Q∞.
31
The Separation Axioms
Exercise 31.3. Show that every order topology is regular.
Proof. Let X be an ordered set with the order topology. X is Hausdorff and therefore T1 by Theorem 17.11. It therefore suffices to show the condition in Lemma 31.1(a).
So let x ∈ X and let U be an open neighborhood of x; we will construct a basis element V of the order topology such that x ∈ V and V ⊆ U .
Suppose x is neither the smallest nor largest element of X. Then, x ∈ (a, b) ⊆ U for some basis element (a, b) of the order topology. If (a, x), (x, b) are nonempty, then let V = (u, v) where u ∈ (a, x) and v ∈ (x, b). If (a, x) = ∅, let u = a, so that V = (u, v) = [x, v); if (x, b) = ∅, let v = b, so that V = (u, v) = (u, x]. Then, x ∈ (u, v) and (u, v) ⊆ (a, b) ⊆ U .
Now suppose x is the smallest (resp. largest) element of X. Then, x ∈ [x, b) ⊆ U (resp. x ∈ (a, x] ⊆ U ) for some basis element [x, b) (resp. (a, x]) of the order topology. Now if (x, b) (resp. (a, x)) is nonempty, then let V = [x, v) (resp. (u, x]) where v ∈ (x, b) (resp. u ∈ (a, x)). On the other hand, if (x, b) (resp. (a, x)) is empty, then let v = b (resp. u = a), so that V = {x}. We then have x ∈ V and V ⊆ U .
If x is the smallest and largest element of X, then X = {x} is trivially regular.
32
Normal Spaces
Exercise 32.1. Show that a closed subspace of a normal space is normal.
Proof. Suppose Y is our closed subspace of our normal space X, and A, B ⊆ Y disjoint and closed. By Theorem 17.3, A, B are closed in X. Let U, V be a separation of A, B in X. Then, Y ∩ U, Y ∩ V separate A, B in Y , and so Y is normal.
Exercise 32.3. Show that every locally compact Hausdorff space is regular.
Proof. Let X be a locally compact Hausdorff space; in particular it is T1. Let x ∈ X
with neighborhood U 3 x. By Theorem 29.2, there exists a neighborhood V 3 x such that V ⊆ U . By Lemma 31.1(a), X is then regular.
Exercise 32.4. Show that every regular Lindel¨of space is normal.
Proof. Let A, B be disjoint closed subsets of X regular and Lindel¨of. For all x ∈ A, there exists a neighborhood U 3 x disjoint from B. By regularity, there exists a neighborhood U ⊆ U ⊆ U containing x; since these U cover A, and A is Lindel¨of by Exercise §30.9, there exists a countable subcover {Ui} such that Ui∩ B = ∅ for all i.
Similarly, we can construct a countable subcover {Vi} of B such that Vi∩ A = ∅ for
all i. By the exact same argument as in the proof of Theorem 32.1, then, the sets U0 = [ n∈Z+ Un\ n [ i=1 Vi , V0 = [ n∈Z+ Vn\ n [ i=1 Ui
Exercise 32.5. Is Rω normal in the product topology? In the uniform topology?
Proof. Since Rω is metrizable in the product topology by Theorem 20.5 and in the
uniform topology by definition on p. 124, both are normal by Theorem 32.2.
33
The Urysohn Lemma
Exercise 33.1. Examine the proof of the Urysohn lemma, and show that for given r, f−1(r) = \ p>r Up− [ q<r Uq, p, q rational.
Proof. ⊆. Suppose x ∈ f−1(r), i.e., x ∈ Ur by definition, and x /∈ Uq for all q < r by
definition in Step 3. Then, x ∈ Ur ⊆ Up for all p > r by construction in Steps 1, 2.
⊇. Suppose x ∈ T
p>rUp −
S
q<rUq. This implies x ∈ Up ⊆ Up for all p > r, and
so f (x) ≤ r by Step 4(1), and also x /∈ Uq for all q < r, and so f (x) ≥ r by Step
4(2). Thus, f (x) = r.
34
The Urysohn Metrization Theorem
Exercise 34.3. Let X be a compact Hausdorff space. Show that X is metrizable if and only if X has a countable basis.
Proof. ⇒. X is compact and metrizable, hence second countable by Exercise 30.4. ⇐. X is compact and Hausdorff, and so X is regular by Theorem 32.3. X is second countable as well, and so by the Urysohn metrization theorem (Theorem 34.1), X is metrizable.
Exercise 34.5. Let X be a locally compact Hausdorff space. Let Y be the one-point compactification of X. Is it true that if X has a countable basis, then Y is metrizable? Is it true that if Y is metrizable, then X has a countable basis?
Solution. If Y is metrizable, then it is second countable by Exercise 30.4. X is then second countable by Theorem 30.2.
Now suppose X has a countable basis B = {Bi}. We claim that B with sets of
the form Y \S Bi, where the union is finite and the closure is taken in Y , form a
basis of Y ; call this larger basis B+. The B
i ∈ B are open by Lemma 16.2, and
the Y \S Bi by Theorem 26.3. B+ is countable since B is countable and there are
