4.1The Countability Axioms

Suppose $(X, \Tau)$ is a topological space. We say some $x \in X$ admits a countable basis at $x$ if there exists some countable collection $\mathcal{B}$ of open neighborhoods of $x$ such that, for every open neighborhood $U$ of $x$, $U$ contains some element of $\mathcal{B}$. If all points in $X$ satisfy this property, we say that $X$ satisfies the first-countability axiom, or is first-countable.

Suppose $X$ is a metrizable topological space. For any $x \in X$, consider the collection of open neighborhoods $\mathcal{B} = \{B_n\}_{n \in \bZ_+}$ about $x$ such that $B_n = B(x, 1/n)$. Then, for any basic open neighborhood $U = B(x, \epsilon)$ about $x$, we have that there is some $n \in \bZ_+$ such that $1/n < \epsilon$, and so $B_n(x, 1/n) \subseteq U$, making $\mathcal{B}$ a countable basis. Thus $X$ is first-countable.

Let $X$ be a topological space.

1. Suppose $A \subseteq X$. If $(x_n) \subseteq A$ is a convergent sequence such that $x_n \to x$, then $x \in \bar{A}$. The converse is true if $X$ is first-countable.

2. If $f$ is a continuous function, then for any $x_n \to x$, we have that $f(x_n) \to f(x)$. The converse is true if $X$ is first-countable.

If a topological space $X$ admits a countable basis for its topology, then we say $X$ satisfies the second-countability axiom, or is second countable.

Note that $\bR$ admits a countable basis, where each element takes th eform $(a, b)$ for rational endpoints $a, b \in \bQ$. Similarly, $\bR^n$ admits a countable basis where each element takes the form $\prod_{k = 1}^n (a_k, b_k)$, where $a_k, b_k \in \bQ$ for every $k$. Even $\bR^\omega$ in its product topology admits a countable basis, with each set taking the form $\prod_{k = 1}^n (a_k, b_k) \times \prod_{k \in \bZ_+ \setminus \{1, \dots, n\}} \bR$ for $a_k, b_k \in \bQ$ for each $k$.

The uniform topology on $\bR^\omega$ is metrizable by the uniform metric $\bar{\rho}(x, y) = \sup_{n \in \bZ_+} \{ \bar{d}(x_n, y_n) \}$, where $\bar{d}(x_n, y_n) = \min(|x_n - y_n|, 1)$; hence, it is first-countable.

To show $\bR^\omega$ is not second-countable, we appeal to the lemma that if a space $X$ has a countable basis $\mathcal{B}$, then any discrete subspace $A \subseteq X$ is countable. Indeed, since each $\{a\}$ is open in $A$, there is some open set $U \subseteq X$ such that $U \cap A = \{a\}$. By the definition of a basis, there exists some $B_a \in \mathcal{B}$ with $a \in B_a \subseteq U$, yielding $B_a \cap A = \{a\}$. For any distinct $a, b \in A$, the basis elements $B_a$ and $B_b$ are not equal since $a \in B_a$ but $a \notin B_b$ (as $B_b \cap A = \{b\}$). The map $a \mapsto B_a$ is therefore an injection of $A$ into $\mathcal{B}$, proving $A$ must be countable.

Finally, consider the subspace $A \subseteq \bR^\omega$ comprising all sequences of $0$s and $1$s. This set $A$ is uncountable. However, for any distinct $a, b \in A$, they must differ in at least one coordinate, so $\bar{\rho}(a, b) = 1$. The subspace $A$ therefore inherits the discrete metric, giving it the discrete topology. By contrapositive, $\bR^\omega$ is not second-countable.

A subspace of first/second-countable spaces is first/second-countable. The countable product of first/second-countable spaces is first/second-countable.

A set $A \subseteq X$ is said to be dense if $\bar{A} = X$.

A space $X$ is said to be Lindelöf if every open cover of $X$ admits a countable subcover.

A space $X$ is said to be separable if there exists a dense subset of $X$ that is countable.

Suppose that $X$ is second-countable.

1. $X$ is Lindelöf.

2. $X$ is separable.

That is, second-countability implies being Lindelöf and separable.

Consider the set $\bR_\ell$ endowed with the lower limit topology, also known as the Sorgenfrey line. We will show that it is first-countable, separable, and Lindelöf, but not second-countable.

Of course, any $x \in \bR_\ell$ admits a countable basis $\{[x, x + 1/n)\}_{n \in \bZ_+}$, so $\bR_\ell$ is first-countable. It is also easy to see that $\bQ$ is dense in $\bR_\ell$, and so $\bR_\ell$ is separable.

We show $\bR_\ell$ is not second-countable. Suppose $\mathcal{B}$ is a basis for the lower limit topology. Then, for any $x \in X$, let $B_x$ be some basis element such that $x \in B_x \subseteq [x, x + 1)$. For any $x \ne y \in \bR_\ell$, note that $B_x \ne B_y$, since $[x = \min(B_x)] \ne [y = \min(B_y)]$. Thus the map from $\bR_\ell$ to $\mathcal{B}$, defined by $x \mapsto B_x$, is injective, and so $|\bR_\ell| < |\mathcal{B}|$, making the basis uncountable.

The proof of $\bR_\ell$ being Lindelöf is annoying.

Lindelöf-ness is not a property that is preserved under products. For example, even though $\bR_\ell$ is Lindelöf, we can quickly show $\bR_\ell \times \bR_\ell$ (also known as $\bR_\ell^2$, the Sorgenfrey plane) is not. Choose the basis that has sets of the form $[a, b) \times [c, d)$, and consider the set

which is essentially the graph of $f(x) = -x$ in $\bR_\ell^2$. It is quick to show $L$ is closed, so we have that $\bR_\ell^2 \setminus L$ is open. Then consider the open cover formed by $\bR_\ell^2 \setminus L$ adjoined with open sets of the form $[a, b) \times [-a, c)$. Every basis element (except the first) intersects $L$, which is uncountable, and so there is no countable subcover. Thus $\bR_\ell^2$ is not Lindelöf.

The ordered square $I_o^2$ is compact (and thus Lindelöf), but the subspace $I \times (0, 1)$ is not Lindelöf.

1. A space $X$ is first-countable at a point $x \in X$ (i.e., satisfies the first-countability axiom at $x$) if there exists a countable collection of open neighborhoods $\{B_n\}$ of $x$ such that, for any open neighborhood $U$ of $x$, there is some $n \in \bZ_+$ for which $B_n \subseteq U$.

2. From any countable basis $\{B_n\}$ at a point $x \in X$, one can derive a countable basis that is “nested” (i.e., $U_1 \supseteq U_2 \supseteq \cdots$). Indeed, let $U_n = B_1 \cap \cdots \cap B_n$. Then note $x$ is in each $B_k$, and so $x$ is in each $U_k$. Moreover, the finite intersection of open sets is open, so each $U_k$ is an open neighborhood of $x$. Obviously $\{U_n\}$ forms a nested chain of open neighborhoods of $x$.

3. First countability is a property that lets convergent sequences probe limit points and the continuity of functions. For a space $X$ and some $A \subseteq X$, if we have a sequence $(x_n) \subseteq X$ such that $x_n \to x$, then $x \in \bar{A}$ by definition of convergence in a topological space. The converse is not necessarily true in a general topological space, but if $X$ is first-countable, then it is immediate (simply take a countable basis at any $x \in X$, make it nested, then choose a sequence as one would in a metric space).

4. Similarly, if $f: X \to Y$ is continuous, then for any $(x_n) \subseteq X$ such that $x_n \to x$, we have that $f(x_n) \to f(x)$ (simply by the property that if $V$ is an open neighborhood of $f(x) \in f(X)$ for continuous $f$, there is some open neighborhood $U$ of $x$ such that $f(U) \subseteq V$). The converse is only true in a first-countable space, however. Using the previous theorem about sequences, one can prove that $f(\bar{A}) \subseteq \bar{f(A)}$.

5. A space $X$ is second-countable if it admits a countable basis.

6. A subset $D \subseteq X$ is dense if $\bar{D} = X$.

7. A space $X$ is Lindelöf if every open covering of $X$ admits a countable subcovering.

8. A space $X$ is separable if there exists a dense subset of $X$ that is countable.

9. Second-countability of a space $X$ implies the space being Lindelöf and separable. If $X$ is second-countable, it admits a countable basis $\{B_n\}_{n \in J}$. For any open covering $\mathcal{A}$ of $X$, note that for any $x \in X$, there is some $A \in \mathcal{A}$ such that $x \in A$. Since $A$ is open, there is a basis element $B_n$ such that $x \in B_n \subseteq A$. For all $x \in B_n$, choose this representative $A$ again when querying the cover for a member that contains an element. Associating $A$ with $n$ by $A_n$, note that $\{A_n\}$ is an open subcover of $X$, making $X$ Lindelöf. Furthermore, for each $n \in J$, choose some $x_n \in B_n$, and let $D$ be the countable set of all $x_n$. For any $x \in X$, any basic open neighborhood intersects $D$ (since it contains elements from every basic, open set), and so $\bar{D} = X$, making $X$ separable as well.

For a topological space $X$, we say a subset $A \subseteq X$ is $G_\delta$ if $A$ can be written as the countable intersection of open sets. Show that singletons are $G_\delta$ space in first-countable, $T_1$ topological spaces.

Show every compact metrizable space $X$ has a countable basis.

Show that if a separable space is metrizable, then it is second-countable.

Show that if a Lindelöf space $X$ is metrizable by a metric $d$, then it is second-countable.

We have shown that, in metrizable spaces, being Lindelöf, separable, and second-countable are all equivalent.

Show that if $X$ is separable, then every collection of disjoint, open sets in $X$ is countable.