Section 1.18 Countability axioms
We introduced the notions of
first countability and
second countability during our discussion of compactness in metric spaces (
1.16). In this section we will look more closely at these and related properties.
Example 1.18.2. Metric spaces are first countable.
Let \((X,d)\) be a metric space. For any \(x\in X\) the collection
\begin{equation*}
\mathcal{B}_x=\{B_{1/n}(x)\colon n\in \Z_+\}
\end{equation*}
is a countable local basis at \(x\text{.}\) Thus \(X\) is first countable.
Example 1.18.3. First countable cofinite spaces.
Let \(X\) be a topological space with the cofinite topology. Prove: \(X\) is first countable if and only if \(X\) is countable.
Solution.
Assume
\(X\) is countable. A nonempty open set of
\(X\) is of the form
\(U=X-\{x_1,x_2,\dots, x_n\}\) for some
\(n\text{.}\) It follows that the nonempty open sets of
\(X\) are in bijection with the set of all finite subsets of
\(X\text{.}\) By
Remark 1.18.1, this collection is countable. Thus
\(X\) has countably many open sets. Thus for any
\(x\in X\) the set of all open sets containing
\(x\) is countable.
Assume \(X\) is first countable. Take any \(x\in X\) and let \(\{V_n\}\) be a countable local basis at \(x\text{.}\) By definition we have \(V_n^c\) finite for each \(n\text{,}\) in which case \(\{x\}\cup \bigcup_{n=1}^\infty V_n^c\) is countable, being a countable union of countable sets. I claim \(X=\{x\}\cup \bigcup_{n=1}^\infty V_n^c\text{,}\) proving \(X\) is countable. Indeed, take any \(y\ne x\text{.}\) The set \(U=X-\{y\}\) is an open set containing \(x\text{.}\) Thus we have \(V_n\subseteq X-\{y\}\) for some \(n\text{,}\) in which case \(\{y\}\subseteq V_n^c\text{,}\) or equivalently, \(y\in V_n^c\text{,}\) as desired.
Definition 1.18.4. Sequentially closed/continuous.
Let \(X\) and \(Y\) be topological spaces.
Given a set \(A\subseteq X\text{,}\) its sequential closure is the set \(\operatorname{sc} A\) of all \(x\in X\) for which there is a a sequence \((x_n)_{n=1}^\infty\) of elements of \(A\) with \(x_n\rightarrow x\text{.}\) The set \(A\) is sequentially closed if \(\operatorname{sc} A=A\text{.}\)
A function \(f\colon X\rightarrow Y\) is sequentially continuous if it maps convergent sequences to convergent sequences: i.e., if \(x_n\rightarrow x\) in \(X\text{,}\) then \(f(x_n)\rightarrow f(y)\) in \(Y\text{.}\)
Theorem 1.18.5. First countable and sequential properties.
Let \(X\) be a first countable space.
For any \(x\in X\) there is a nested countable local basis \(B_x=\{V_n\colon n\in \Z_+\}\text{:}\) i.e., for all \(n\geq 1\) we have \(N_n\supseteq N_{n+1}\text{.}\)
A set \(A\subseteq X\) is closed if and only if it is sequentially closed.
A function \(f\colon X\rightarrow Y\) is continuous if and only if it is sequentially continuous.
Proof.
Given \(x\in X\) let \(\{U_n\colon n\in \Z_+\}\) be a countable local basis at \(x\text{.}\) For each \(n\geq 1\text{,}\) define \(V_n=U_1\cap U_2\cap\cdots \cap U_n\text{.}\) (Note that \(V_n\ne\emptyset\) since \(x\in V_n\text{.}\)) We have \(V_n\supseteq V_{n+1}\) for all \(n\geq 1\text{.}\) Furthermore, given any open \(U\ni x\text{,}\) we have \(V_n\subseteq U_n\subseteq U\text{.}\) Thus \(\{V_n\}\) forms a nested local basis.
First, it is easy to see that in any space we have \(A\subseteq \operatorname{sc} A\subseteq \overline{A}\text{.}\) Thus in any topological space, if \(A\) is closed, then we have \(A=\operatorname{sc} A=\overline{A}\text{,}\) and thus \(A\) is sequentially closed. Now assume \(A\) is sequentially closed, and take \(x\in \overline{A}\text{.}\) Using (1), we pick a a local basis \(\{V_n\colon n\in \Z_+\}\) at \(x\) that is nested: i.e., for all \(n\geq 1\text{,}\) \(V_n\supseteq V_{n+1}\text{.}\) Now, for each \(n\in \Z_+\) pick an element \(x_n\in V_n\cap A\text{.}\) We have \(x_n\rightarrow x\text{:}\) indeed, if \(U\ni x\) is any open set containing \(x\text{,}\) then \(V_N\subseteq U\) for some \(n\text{,}\) in which case \(x_k\in V_k\subseteq U\) for all \(k\geq N\text{.}\) Since \(A\) is sequentially compact, we have \(x\in A\text{.}\) Thus \(\overline{A}\subseteq A\text{,}\) and we conclude that \(A=\overline{A}\text{.}\)
The forward direction is true in any topological space, as we have shown before. Let’s prove the reverse direction. Assume \(f\colon X\rightarrow Y\) is sequentially continuous. Fix any \(x\in X\) and open set \(U\) containing \(f(x)=y\in Y\text{.}\) Assume by way of contradiction that there is no open set \(V\ni x\) such that \(f(V)\subseteq U\text{.}\) Let \(\{V_n\}_{n=1}^\infty\) be a nested local basis at \(x\text{.}\) Since \(f(V_n)\not\subseteq U\) for all \(n\text{,}\) we can find an \(x_n\in V_n\) such that \(f(x_n)\notin U\text{.}\) The sequence \((x_n)\) converges to \(x\text{:}\) indeed, given any open \(V\) containing \(x\) there is an \(N\) such that \(V_N\subseteq V\text{.}\) Since the local basis is nested, we have \(V_k\subseteq V_N\subseteq V\) for all \(k\geq N\text{,}\) and hence \(x_k\in V\) for all \(k\geq N\text{.}\) However, it is clear that \((f(x_n))\) does not converge to \(f(x)=y\text{,}\) since by construction no \(f(x_n)\) lies in the open set \(U\text{.}\) This contradicts the fact that \(f\) is sequentially compact. We conclude that \(f\) is continuous.
We now re-introduce second countability along with two related notions.
Definition 1.18.6. Second countable, Lindelöf, separable.
Let \(X\) be a topological space. A set \(A\subseteq X\) is dense if \(\overline{A}=X\text{.}\)
\(X\) is second countable if it has a countable basis.
\(X\) is Lindelöf if every open cover of \(X\) has a countable subcover.
\(X\) is separable if it has a countable dense subset.
Theorem 1.18.8. Second countable: strongest countability axiom.
Let \(X\) be a topological space.
If \(X\) is second countable, then \(X\) is first countable, Lindelöf, and separable.
If \(X\) is a metric space, then being second countable, Lindelöf, and separable are all equivalent properties.
Proof.
-
Assume \(X\) is second countable and let \(\mathcal{B}=\{B_n\}_{n=1}^\infty\) be a countable basis.
For each \(x\text{,}\) the set \(\mathcal{B}_x=\{B\in \mathcal{B}\colon x\in B\}\) is easily seen to be a countable local basis at \(x\text{.}\) This proves \(X\) is first countable.
Let \(X=\bigcup_{i\in I}U_i\) be an open cover of \(X\text{.}\) We will show that there is a countable subcover, and hence that \(X\) is Lindelöf. Let
\begin{equation*}
\mathcal{B}'=\{B\in\mathcal{B}\colon B\subseteq U_i \text{ for some } i\in I\}\text{.}
\end{equation*}
Since \(\mathcal{B}\) is countable, so is \(\mathcal{B}'\text{.}\) Write \(\mathcal{B}'=\{B_1', B_2',\dots\}\text{.}\) For each \(n\in \Z_+\text{,}\) pick an index \(i_n\in I\) such that \(B'_n\subseteq U_{i_n}\text{.}\) We claim that \(X=\bigcup_{n=1}^\infty U_{i_n}\text{,}\) and thus that a countable subcovering of \(\{U_i\}_{i\in I}\text{.}\) Indeed, for any \(x\in X\text{,}\) we have \(x\in B\subseteq U_i\) for some \(i\in I\) and \(B\in \mathcal{B}\text{.}\) It follows that \(B=B'_n\) for some \(n\text{,}\) in which case \(x\in B_n'\subseteq U_{i_n}\text{,}\) as desired.
Lastly we show that \(X\) is separable. For each \(n\in \Z_+\) let \(q_n\) be any element of \(B_n\text{.}\) I claim that the set \(Q=\{q_n\colon n\in \Z_+\}\) is dense, showing \(X\) is separable. Indeed, given any open set \(U\text{,}\) we have \(B_n\subseteq U\) for some \(n\text{,}\) in which case \(q_n\in U\text{.}\)
This is a (classic) homework exercise. Enjoy!
Example 1.18.12. \(\R^n\) is second countable.
Prove that \(\R^n\) is second countable with respect to the Euclidean topology.
Solution.
First observe that the set
\(\Q^n\) of all
\(n\)-tuples is countable. According to (2) of
Theorem 1.18.8 it would suffice to show that
\(\Q^n\) is dense in
\(\R^n\text{.}\) We take the hard route and show directly that
\(\R^n\) has a countable basis.
For each \(\boldx\in \Q^n\text{,}\) let \(\mathcal{B}_\boldx=\{B_{1/n}(\boldx)\colon n\in \Z_+\}\text{.}\) The set
\begin{equation*}
\mathcal{B}=\bigcup_{\boldx\in \Q^n}\mathcal{B}_\boldx\text{,}
\end{equation*}
being a countable union of countable sets, is countable. I claim it is a basis. It suffices to show that given any \(\boldy=(y_1,y_2,\dots, y_n)\in \R^n\) and \(\epsilon > 0\text{,}\) we can find a \(B_{1/m}(\boldx)\in \mathcal{B}\) such that \(y\in B_{1/m}(\boldx)\subseteq B_\epsilon(\boldy)\text{.}\) To this end pick \(m\in \Z_+\) such that \(1/m< \epsilon/2\) and pick \(\boldx\in \Q^n\) such that \(\abs{x_i-y_i}< 1/mn\text{,}\) so that \(d(\boldx, \boldy)< 1/m< \epsilon/2\text{.}\) It follows that \(\boldy\in B_{1/m}(\boldx)\subseteq B_{\epsilon}(\boldy)\text{,}\) as desired.
Example 1.18.13. \(\R_\ell\) is not metrizable.
Let \(X=\R_\ell\text{.}\) Show that \(X\) is first countable, Lindelöf, separable, but not second countable. Conclude that \(\R_\ell\) is not metrizable.
Solution.
For each \(x\in \R\text{,}\) the set \(\mathcal{B}_x=\{[x,q)\colon q\in \Q\}\) forms a countable local basis at \(x\text{.}\) Thus \(X\) is first countable.
It is easy to see further that \(\Q\subseteq \R_\ell\) is dense with respect to the left limit topology.
To see that \(X\) is not second countable, consider any basis \(\mathcal{B}\) of \(X\text{.}\) For each \(x\in \R\text{,}\) we can find a basis element \(B_x\in \mathcal{B}\) such that \(x\in B_x\subseteq [x,x+1)\text{.}\) Note that since \(\inf B_x=x\) for such a \(B_x\text{,}\) the map \(x\mapsto B_x\) defines an injection of \(\R\) into \(\mathcal{B}\text{:}\) \(B_x=B_y\implies \inf B_x=\inf B_y\implies x=y\text{.}\) This proves that \(\mathcal{B}\) is uncountable.
At this point, by (2) of
Theorem 1.18.8 we know that
\(\R_\ell\) is not metrizable, since it is separable but not second countable. See Munkres for a proof that
\(\R_\ell\) is also Lindelöf.
Theorem 1.18.14. First/second countable: subspaces, images, and products.
If \(X\) is first countable (resp. second countable) and \(Y\subseteq X\) is a subspace, then \(Y\) is first countable (resp. second countable).
If \(f\colon X\rightarrow Y\) is open, continuous and surjective, then if \(X\) is first countable (resp. second countable), \(Y\) is first countable (resp. second countable).
Let \(X=\prod_{i\in I}X_i\) be a product of topological spaces. The following statements are equivalent.
\(X\) is first countable (resp. second countable).
\(X_i\) is first countable (resp. second countable) for all \(i\in I\) and the set \(J\) of all \(j\in I\) such that \(X_j\) does not have the trivial topology is countable.
As a result, a countable product of first countable (resp. second countable) spaces is first countable.
Proof.
This is an easy consequence of the fact that for any basis (or local basis) \(\mathcal{B}\) of \(X\) the corresponding set \(\mathcal{B}_Y=\{B\cap Y\colon B\in \mathcal{B}\}\) is a basis (or local basis) in \(Y\text{.}\)
It is easy to see that such a function \(f\) maps bases (or local bases) in \(X\) to bases (or local bases) in \(Y\text{.}\)
Homework exercise. Enjoy!
Example 1.18.15. First countable: continuous image.
Give an example of a first countable space \(X\) and continuous function \(f\colon X\rightarrow Y\) such that \(f(X)\) is not first countable.
Solution.
Consider the identity function
\(\id\colon \R\rightarrow \R\text{,}\) where the codomain is given the cofinite topology and the domain is given the standard topology. (This is continuous since the standard topology is finer than the cofinite topology.) Since the standard topology is metrizable, the domain is first countable. On the other hand, we saw in
Example 1.18.3 that any uncountable set is not first countable with respect to the cofinite topology.
Example 1.18.16. \(\R^\R\) is not metrizable..
Let \(X=\R^\R=\prod_{r\in \R}\R\text{.}\) Show that \(X\) is not metrizable.
Solution.
According to (3) of
Theorem 1.18.14 an uncountable product of nontrivial spaces is not first countable. Thus
\(X\) is not first countable, in which case it is also not metrizable.
