Weakly Lindelöf metrizable spaces are separable

Question

In Handbook of set-theoretic topology (Kunen & Vaughan, 1984), chapter 1 about Cardinal functions, Theorem 8.1 states that many of the cardinal functions are equal in the case of metrizable spaces. Among them, $w(X) = d(X) = wc(X)$, where:

• $w(X)$ is the weight of $X$,
• $d(X)$ is the density of $X$,
• $wc(X)$ is the weak covering number of $X$, that is, the smallest infinite cardinal $\kappa$ such that every open cover of $X$ has a subcollection of cardinality $\le\kappa$ whose union is dense in $X$.

The corresponding special names when these cardinal functions are countable are:

• $X$ is second countable
• $X$ is separable
• $X$ is weakly Lindelöf

In particular, a metrizable space is second countable iff it is separable iff it is weakly Lindelöf.

For the proof, the equality $w(X)=d(X)$ for metrizable spaces is well-known.

For the weak covering number, it is easy to see that the inequality $wc(X)\le w(X)$ holds in any topological space.

For the reverse, the Handbook's Theorem 8.1 leaves the verification to the reader. Can anyone provide a proof?

Answer 1

Perhaps we can show $d(X) \leq wc(X)$. For $n \in \omega$, let $\mathscr U_n = \left\{ B\left(x,2^{-n}\right) : x \in X \right\}$ and apply the weak covering number to produce $D_n = \{ x_{n,\alpha} : \alpha < wc(X) \}$ so that $W_n := \bigcup \left\{ B\left(x,2^{-n} \right) : x \in D_n \right\}$ is dense. Now, $D := \bigcup_{n\in\omega} D_n$ is of cardinality $\leq wc(X)$ since we are dealing with infinite cardinals.

We wish to show that $D$ is dense. So let $x \in X$ and consider any $\varepsilon > 0$. Let $n \in \omega$ be so that $2^{-n} < \varepsilon$. Since $W_{n+1} = \bigcup \left\{ B\left( y, 2^{-(n+1)} \right) : y \in D_{n+1} \right\}$ is dense, there is some $y \in W_{n+1}$ so that $d(y,x) < 2^{-(n+1)}$; since $y \in W_{n+1}$, there is some $\alpha < wc(X)$ so that $d(x_{n+1,\alpha} , y) < 2^{-(n+1)}$. By the triangle inequality, we see that $d(x_{n+1,\alpha} , x) < 2^{-n} < \varepsilon$. Since $x_{n+1,\alpha} \in D$ and since $x$ and $\varepsilon$ were arbitrary, we see that $D$ is dense.

Answer 2

Let $wc(X)\leq \kappa$ be infinite cardinal so that every open cover of $X$ has a subfamily of cardinality $\leq \kappa$ which union is dense in $X$.

Let $\mathcal{U}_n = \{B(x, 1/n) : x\in X\}$ and let $\mathcal{U}_n'\subseteq \mathcal{U}_n$ be such that $\bigcup \mathcal{U}_n'$ is dense and $|\mathcal{U}_n'|\leq \kappa$.

Pick $x_U\in U$ for each $U\in\mathcal{U}_n'$, and let $Y = \{x_U : U\in\mathcal{U}_n', n\in\mathbb{N}\}$. Clearly $$|Y|\leq \left|\bigcup_{n\in\mathbb{N}}\mathcal{U}_n'\right|\leq\sum_{n\in\mathbb{N}} \left|\mathcal{U}_n'\right| \leq \sum_{n\in\mathbb{N}}\kappa = \aleph_0\cdot \kappa = \kappa$$

If $U\subseteq X$ is open and non-empty, we can choose $B(x, 1/n)\subseteq U$. Then $\bigcup\mathcal{U}_{3n}'\cap B(x, \frac{1}{3n})\neq \emptyset$.

Then there is $z\in X$ such that $w\in B(z, \frac{1}{3n})\cap B(x, \frac{1}{3n})$ for some $w$ and $B(z, \frac{1}{3n})\in\mathcal{U}_{3n}'$. Let $y\in Y\cap B(z, \frac{1}{3n})$. Then $d(y, x) \leq d(y, z)+ d(z, w)+d(w, x) < \frac{1}{n}$, so $y\in B(x, 1/n)\subseteq U$.

The above argument proves that $d(X)\leq wc(X)$. Then $\mathcal{B} = \{B(x, 1/n) : x\in Y, n\in\mathbb{N}\}$ is a basis for $X$ and $|\mathcal{B}|\leq \kappa$, so that $w(X)\leq wc(X)$.