Definition 6.1 A topological space is said to be second countable if the topology
admits a countable basis of open sets.
A space with this property is also said to satisfy the second countability axiom.
Example 6.2 The Euclidean lineR has a countable basis: the family of open intervals
with rational endpoints
{ ]c, d[ | c, d ∈ Q}
is countable and forms a basis for the Euclidean topology. Any open set inR, in fact, is the union of open intervals, and for any]a, b[ we have
]a, b[ ={ ]c, d[ | a ≤ c < d ≤ b, c, d ∈ Q}.
Example 6.3 Any subspace in a second-countable space is second countable: ifB is
a countable basis for X and Y ⊂ X, then {A ∩ Y | A ∈ B} is a countable basis for Y .
Example 6.4 The product of two second-countable spaces is still second countable.
It’s easy show that ifA and B are bases of X and Y respectively,
C = {A × B | A ∈ A, B ∈ B}
is a basis of X× Y . If A and B are countable, so is C.
It’s also quite simple to find examples of quotients of second-countable spaces that do not inherit the property, like the one of Exercise6.1.
We recall (Definition3.14) that a subset is dense if it intersects every non-empty open set.
Definition 6.5 A topological space is called separable if it contains a countable
dense subset.
© Springer International Publishing Switzerland 2015
M. Manetti, Topology, UNITEXT - La Matematica per il 3+2 91, DOI 10.1007/978-3-319-16958-3_6
106 6 Sequences Lemma 6.6 Every second-countable space is separable.
Proof LetB be a countable basis of X; for any open set U ∈ B choose a point pU ∈ U. The set of such points E = {pU | U ∈ B} is countable, and dense because
it meets every basis element.
Lemma 6.7 Any separable metric space has a countable basis.
Proof Let(X, d) be a separable metric space and choose a countable dense subset E ⊂ X. It is enough to show that the countable family of open balls
B = {B(e, 2−n) | e ∈ E, n ∈ N}
is a basis. Take U open and x ∈ U, then choose an integer n ∈ N so that
B(x, 21−n) ⊂ U. As E is dense, there exists an e ∈ E ∩ B(x, 2−n); symmet-
rically, x belongs to the ball B(e, 2−n), so from the triangle inequality we have
B(e, 2−n) ⊂ B(x, 21−n) ⊂ U.
Example 6.8 Denote by 2(R) the set of real sequences {an}, n ≥ 1, such that
nan2< +∞, and set a = ∞ n=1 a2 n
for any a = {an} ∈ 2(R). Since for every a, b ∈ R we have (a + b)2+ (a − b)2=
2(a2+ b2), immediately 2(R) is a vector subspace. Given sequences {an}, {bn} ∈ 2(R), the triangle inequality
N n=1 (an− bn)2≤ N n=1 a2 n+ N n=1 b2 n
holds, for any N > 0. Passing to the limit for N → ∞ gives a − b ≤ a + b, hence2(R) is metrised by the distance d(a, b) = a − b.
Now let’s consider the countable set E ⊂ 2(R) of eventually null, rational sequences. Given any sequence a = {an} ∈ 2(R) and any real number ε > 0, we
may find an integer N > 0 and N rational numbers b1, . . . , bNsuch that
n>N
an2< ε, (an− bn)2< ε
Consider the sequence b= (b1, b2, . . . , bN, 0, 0, . . .): it satisfies a − b2= n≤N (an− bn)2+ n>N an2< 2ε .
This proves that E is dense, the metric space2(R) is separable, and by Lemma6.7
2(R) has a countable basis.
Proposition 6.9 In a second-countable space every open cover admits a countable
subcover.
Proof Let X have a countable basis B and consider an open cover A. For every x∈ X choose an open set Ux ∈ A and an element Bx ∈ B such that x ∈ Bx ⊂ Ux.
The open familyB= {Bx | x ∈ X} is a subcover of B, and so it is countable. We can
find a countable subset E ⊂ X such that B= {Bx | x ∈ E}, whence {Ux | x ∈ E}
is a countable subcover ofA.
Definition 6.10 A topological space satisfies the first countability axiom, or is first countable, if every point admits a countable local basis of neighbourhoods.
If a space is second countable then it is first countable: the open sets in a countable basis that contain a given point forms a local basis at that point.
Lemma 6.11 Metric spaces are first countable.
Proof Let x be a point in a metric space. The open balls{B(x, 2−n) | n ∈ N} are a
local basis of neighbourhoods around x. The next example shows that not all first-countable spaces are metrisable.
Example 6.12 The lower-limit lineRl(cf. Example3.9) is separable and first count-
able, but doesn’t have countable bases. In particular it can’t be metrisable. The subset of rational numbers is dense inRl, makingRlseparable. For any real a, the countable
open family{[a, a + 2−n[ | n ∈ N} is a local basis around a. Take an open basis B ofRl; for any x ∈ Rl the open set[x, +∞[ contains x, and so there is U(x) ∈ B
such that x ∈ U(x) ⊂ [x, +∞[. In particular, if x < y then x ∈ U(y) and hence
U(x) = U(y). Therefore the map Rl → B, x → U(x), is 1-1 and B is not countable.
Theorem 6.13 Let X be a second-countable, locally compact Hausdorff space.
There exists an exhaustion of X by compact sets, i.e. a sequence
K1⊂ K2⊂ · · ·
of compact sets covering X and such that Kn ⊂ Kn◦+1for any n.
Proof LetB be a countable basis, Bc⊂ B the subfamily of open sets with compact
closure. Since X is locally compact and Hausdorff,Bccovers X . CallA the collection of finite unions of elements ofBc. By constructionA is at most countable, every
108 6 Sequences
A ∈ A has compact closure and for any compact set K ⊂ X there’s an element A ∈ A such that K ⊂ A. Fix an onto map g : N → Bc, then define recursively a
sequence of compact sets K1⊂ K2⊂ · · · by
K1= g(1), Kn= An∪ g(n),
where Anis one of the elements inA that contain Kn−1. This makes the family{Kn}
an exhaustion by compact sets.
Exercises
6.1 (♥) Given real numbers x, y ∈ R, we define x ∼ y iff x = y or x, y ∈ Z. Prove
thatR/∼ doesn’t satisfy the first axiom of countability.
6.2 LetB be a basis in a second-countable space. Prove that there exists a countable
subfamilyC ⊂ B giving a basis for the topology.
6.3 Consider a continuous and onto map f: X → Y . Show that if X is separable
then also Y is separable. If, in addition, f is open and X second countable, then Y has a countable basis.
6.4 Prove that the product of two separable spaces is separable. Give an example of
a separable space containing a non-separable subspace. (Hint: Exercise3.62.)
6.5 Let f: X → Y be a continuous, closed and onto map such that f−1(y) is
compact for every y ∈ Y . Prove that Y has a countable basis if X has a countable basis.
6.6 (K) Take disjoint copies Ra of the real line, parametrised by a real number a ∈ ] − π/2, π/2[, and define continuous maps
fa: (Ra− {0})× ] − 1, 1[ → R2, fa(x, t) = cos(a) x − t sin(a), sin(a) x + t cos(a) .
Now consider the disjoint union ofR2with the stripsRa× ] − 1, 1[, and denote by
X = R
2∪
aRa× ] − 1, 1[
∼
the quotient under the smallest equivalence relation∼ such that
Prove that:
1. X is connected, separable and Hausdorff; each point has an open neighbourhood homeomorphic toR2;
2. X contains a discrete subset whose cardinality is more than countable. In partic- ular, X isn’t second countable;
3. X can’t be written as countable union of compact sets.