It’s part of the human intuition to answer ‘two’ when asked ‘How many pieces make up the space X = R − {0}?’ This happens because we are able to distinguish two parts in X , namely X−= {x < 0} and X+= {x > 0}, that in our mind match the word ‘piece’ of the question.
Topological structures allow to define in a mathematically precise way the notions of connected space and connected component, that correspond to the naïve con- cepts of ‘one single piece’ and ‘piece of cake’ if the cake is already sliced.
Definition 4.1 A topological space X is called connected if the only subsets that
are both open and closed are∅ and X. A non-connected topological space is called
disconnected.
Lemma 4.2 On a topological space X the following properties are equivalent:
1. X is disconnected;
2. X is the disjoint union of two open, proper subsets; 3. X is the disjoint union of two closed, proper subsets.
Proof (1) ⇒ (2), (3). Let A ⊂ X be open, closed and non-empty. If A = X the
complement B= X − A is open, closed and non-empty and X is the disjoint union of A and B.
(2) ⇒ (1). Suppose A1∪ A2 = X with A1, A2open, non-empty and disjoint.
Then A1= X − A2is closed, too.
(3)⇒ (1). If C1∪ C2 = X, where C1, C2are closed, non-empty and disjoint,
C1= X − C2is open.
Example 4.3 The space X = {x ∈ R | x = 0} is disconnected. The non-empty
subsets
X−= X ∩ ] − ∞, 0[, X+= X ∩ ]0, +∞[
are open in the subspace topology, disjoint and their union is X .
One says a topological subspace is connected if it is so for the induced topology.
Lemma 4.4 Let X be a topological space and A ⊂ X an open and closed subset.
For any connected subspace Y ⊂ X either Y ⊂ A or Y ∩ A = ∅.
Proof The intersection Y∩ A is open and closed in Y . As Y is connected, necessarily Y ∩ A = Y (hence Y ⊂ A) or Y ∩ A = ∅. Example 4.5 Let X be a topological space, K ⊂ X a closed subset and U ⊂ X an
open subset containing K . If X and U− K are connected, also X − K is connected. In fact, suppose by contradiction X − K = A ∪ B, with A, B open, disjoint and non-empty. Since U − K is a connected subspace of X − K , by Lemma4.4either
U− K ⊂ A or U − K ⊂ B; to fix ideas suppose U − K ⊂ A and U − K ∩ B = ∅.
Then A∪ K = A ∪ U, U ∩ B = ∅ and hence X = (A ∪ K ) ∪ B = (A ∪ U) ∪ B is the union of open, disjoint, non-empty sets, against the assumption.
4.1 Connectedness 65 Theorem 4.6 The interval[0, 1] is connected in the Euclidean topology.
Proof Take C and D non-empty closed subspaces of[0, 1] such that C ∪ D = [0, 1];
we claim C ∩ D = ∅. Suppose, to fix ideas, 0 ∈ C, and write d ∈ [0, 1] for the infimum of D. It will be enough to prove d ∈ C ∩ D. As D is closed, d must belong in D, so if d = 0 we are done. If d > 0 we set E = C ∩ [0, d]. As E is closed and contains[0, d[ it follows d ∈ E, so d ∈ C.
Theorem 4.7 Let f: X → Y be a continuous map. Then X connected implies f (X)
connected.
Proof Let Z ⊂ f (X) be a non-empty, open and closed subset in f (X). By definition
of subspace topology there exist an open A ⊂ Y and a closed C ⊂ Y such that
Z = f (X) ∩ A = f (X) ∩ C. But f is continuous, so f−1(Z) = f−1(A) is open,
and f−1(Z) = f−1(C) is closed. As X is connected, f−1(Z) = X and therefore
Z = f (X).
Definition 4.8 A topological space X is path(wise)-connected if, given any two
points x, y ∈ X, there is a continuous mapping α: [0, 1] → X such that α(0) = x and α(1) = y. Such an α is called a path from x to y.
The continuous image of a path connected space is clearly path connected. The intuitive notion of connectedness of a graph seen in Chap.1 corresponds to path connectedness.
Lemma 4.9 Any path connected space is connected.
Proof Let X be a path connected space and A, B ⊂ X open, non-empty and such
that A∪ B = X: we want to prove A ∩ B = ∅. Choose x ∈ A, y ∈ B and a path
α: [0, 1] → X such that α(0) = x and α(1) = y. The open sets α−1(A), α−1(B)
are non-empty, their union is[0, 1] and so there exists t ∈ α−1(A) ∩ α−1(B). The point α(t) belongs to A ∩ B, which is thus not empty. We shall prove later on, as consequence of Proposition10.5, that open connected subsets inRnare path connected. If n≥ 2, there exist closed, connected subsets in Rn that are not path connected (Exercise4.20). The most wicked—at least in our view— example of connected but path disconnected subset is described in Exercise8.6.
Lemma 4.10 Let A and B be path connected subspaces in a topological space. If
A∩ B = ∅, then A ∪ B is path connected.
Proof It suffices to show that if x ∈ A and y ∈ B (or conversely) there is a path α: [0, 1] → A ∪ B from α(0) = x to α(1) = y. Pick z ∈ A ∩ B and choose paths β : [0, 1] → A and γ : [0, 1] → B such that β(0) = x, β(1) = γ(0) = z, γ(1) = y.
Now define α as follows:
α(t) =
β(2t) if 0≤ t ≤ 1/2.
Example 4.11 The spacesRnand the spheres Snare path connected for every n> 0. Indeed, a possible path joining any two points x, y ∈ Rnis the standard parametri- sation of the segment between x, y: α(t) = tx + (1 − t)y. The sphere Sn, n > 1, is the union of two open sets homeomorphic toRn under stereographic projection,
and the claim follows now from Lemma4.10.
We recall that a subset A⊂ Rnis convex if t x+ (1 − t)y ∈ A for every x, y ∈ A and any t ∈ [0, 1]. Intervals, for instance, are precisely the convex subsets of R. Another example is the open ball
B(p, r) = {x ∈ Rn| p − x < r},
convex for any p ∈ Rnand any r > 0. In fact if x, y ∈ B(p, r) and t ∈ [0, 1], the triangle inequality gives
p − (tx + (1 − t)y) = t(p − x) + (1 − t)(p − y)
≤ t(p − x) + (1 − t)(p − y) = t p − x + (1 − t) p − y
< tr + (1 − t)r = r
so t x+ (1 − t)y ∈ B(p, r).
Corollary 4.12 Any convex subset ofRnis path connected and hence connected. Proof Let A be convex inRn; for any x, y ∈ A the map α: [0, 1] → A, α(t) =
t x+ (1 − t)y is continuous.
Corollary 4.13 For a subset I ⊂ R the following are equivalent conditions:
1. I is an interval, i.e. a convex subset; 2. I is path connected;
3. I is connected.
Proof Only(3) ⇒ (1) is non-trivial. If I ⊂ R is not an interval, there exist a < b < c
such that a, c ∈ I but b ∈ I . The open disjoint sets I ∩ ] − ∞, b[ and I ∩ ]b, +∞[ aren’t empty and disconnect I .
We are now in the position of showing the first applications of connectedness.
Example 4.14 Let f: [0, 1] → [0, +∞[ be continuous and such that f (1) = 0.
There exists an x ∈ [0, 1] such that f (x) = x. In fact the range of [0, 1] → R,
x→ f (x) − x, is a connected set containing f (1) − 1 = −1 and f (0) − 0 ≥ 0. Example 4.15 The interval]0, 1[ is not homeomorphic to [0, 1[. Suppose the con-
trary, and that f: [0, 1[ → ]0, 1[ was a homeomorphism. Then f would induce, by restriction, a homeomorphism f: ]0, 1[ → ]0, 1[ − { f (0)}. The contradiction emerges from the observation that ]0, 1[ is connected whereas ]0, 1[ − { f (0)} is disconnected.
4.1 Connectedness 67
To prevent a common mistake let us point out that connectedness doesn’t forbid the existence of open, non-empty subspaces homeomorphic to closed subspaces. Here’s an example: in the Euclidean topology the intervals [0, 1[ and [1, 2[ are homeomorphic; yet the former is open and the latter closed as subspaces of the connected space[0, 2[.
Lemma 4.16 Take n > 0 and a continuous map f : Sn → R. There exists x ∈ Sn
such that f(x) = f (−x). In particular f cannot be one-to-one. Proof Consider the continuous map
g : Sn→ R, g(x) = f (x) − f (−x).
As n> 0, the sphere Snis connected, the image g(Sn) is connected in R and hence convex. Choose an arbitrary point y∈ Sn. The interval g(Sn) contains g(y), g(−y) and therefore also the convex combination
1 2g(y) + 1 2g(−y) = 1 2( f (y) − f (−y)) + 1 2( f (−y) − f (y)) = 0. Hence 0∈ g(Sn) and an x ∈ Snmust exist such that g(x) = 0.
Example 4.17 No subset inR can be homeomorphic to an open set in Rn, for any
n > 1. In fact every open subset in Rncontains a subset homeomorphic to the sphere Sn−1, so Lemma4.16applies.
Lemma 4.18 Let Y be a connected topological space and f: X → Y a continuous,
onto map such that f−1(y) is connected for every y ∈ Y . If f is open, or closed, then X is connected.
Proof Suppose f is open and let A1, A2⊂ X be open, non-empty and such that X =
A1∪ A2. As Y = f (A1)∪ f (A2) and Y is connected, there exists y ∈ f (A1)∩ f (A2),
so f−1(y) ∩ Ai = ∅, i = 1, 2. But f−1(y) is connected, so f−1(y) ∩ A1∩ A2= ∅,
and a fortiori A1∩ A2= ∅.
If f is closed the same proof works with the proviso of taking A1, A2
closed.
Theorem 4.19 The product of two connected spaces is connected.
Proof We have to show that if X and Y are connected, so is X × Y . For that it’s
enough to note that the projection X × Y → Y is continuous, onto, open and its fibres are homeomorphic to X . The claim descends from Lemma4.18. Using induction we immediately obtain that finite products of connected spaces are connected.
Exercises
4.1 (♥) Let p, q ∈ R2denote the points(1, 0) and (−1, 0). Which of the following subspaces inR2are connected?
A= {x ∈ R2| x − p < 1 or x − q < 1}; B= {x ∈ R2| x − p < 1 or x − q ≤ 1}; C = {x ∈ R2| x − p ≤ 1 or x − q ≤ 1}.
4.2 Which are the connected subspaces in a discrete topological space?
4.3 Let A, B be subspaces such that A ∪ B and A ∩ B are connected. Prove that if
A, B are both closed or both open, A and B are connected.
4.4 For any triple p, q, v ∈ Rn, the path
αp,q,v: [0, 1] → Rn, αp,q,v(t) = (1 − t)p + tq + t(1 − t)v,
is a parabolic arc (possibly degenerate) with endpoints p and q.
Prove that if p = q, for any x ∈ Rn− {p, q} there exists at most one vector v orthogonal to p− q and such that x belongs to the range of αp,q,v.
4.5 (♥) Prove that if n ≥ 2 the complement of a countable subset in Rn is path connected.
4.6 Take n ≥ 2 and f : Sn → R a continuous map. Call A the set of points
t ∈ f (Sn) such that the fibre f−1(t) has finite cardinality.
Show that A contains at most two points. Give examples of maps f where A has cardinality 0, 1 and 2.
4.7 Prove that the product of two path connected spaces is path connected. 4.8 A subset A inRn is star-shaped with respect to the point a∈ A if for every b∈ A the segment joining a with b is contained in A. Prove that star-shaped sets are
path connected.
4.9 Let X ⊂ R2be a bounded and convex subset. ShowR2− X is path connected.
4.10 (K, ♥) Let A be an open convex set in R2and p a point not in A. Prove that there exists a straight line through p that does not meet A.
4.11 (K) Let X be a Hausdorff space with the property that X − A is connected for
any finite subset A⊂ X. Prove that the space of configurations
Confn(X) = {(x1, . . . , xn) ∈ Xn | xi = xj for every i= j}