How to apply metrization theorems for proving non-metrizability of a topological space?

Question

I was studying the following result: a $T_o$-space is metrizable if and only if there is a compatible normal sequence of open covers. Now my question is how to apply this result for proving a space to be non-metrizable? Can anyone explain with the help of example?

Answer 1

You have to show either that the space is not $T_0$ or that it does not have a compatible normal sequence of open covers. How you do that will in general depend on the space. Here’s one example that’s non-trivial but still relatively straightforward.

Let $X$ be the Sorgenfrey line. Suppose that $\langle\mathscr{G}_n:n\in\Bbb N\rangle$ is a compatible sequence of normal open covers of $X$. Then in particular we know that $\{\operatorname{st}(x,\mathscr{G}_n):n\in\Bbb N\}$ is a local base at $x$ for each $x\in X$, where $\operatorname{st}(x,\mathscr{G}_n)=\bigcup\{G\in\mathscr{G}_n:x\in G\}$. It follows that for each $x\in X$ there is an $n(x)\in\Bbb N$ such that $\operatorname{st}\left(x,\mathscr{G}_{n(x)}\right)\subseteq[x,x+1)$. For each $k\in\Bbb N$ let $X_k=\{x\in X:n(x)=k\}$; $X$ is uncountable, so there is an $m\in\Bbb N$ such that $X_m$ is uncountable.

For each $x\in X_m$ there are a $G_x\in\mathscr{G}_m$ and a $k(x)\in\Bbb N$ such that

$$\left[x,x+2^{-k(x)}\right)\subseteq G_x\;.$$

For each $\ell\in\Bbb N$ let $X_m(\ell)=\{x\in X_m:k(x)=\ell\}$; there must be an $\ell\in\Bbb N$ such that $X_m(\ell)$ is uncountable.

Since $X_m(\ell)$ is uncountable, it has an uncountable subset $Y$ such that $$(y-\epsilon,y)\cap Y\ne\varnothing\tag{1}$$ whenever $y\in Y$ and $\epsilon>0$. Fix $y\in Y$; by $(1)$ there is an $x\in Y$ such that $y-2^{-\ell}<x<y$. But then

$$y\in\left[x,x+2^{-\ell}\right)\subseteq G_y\in\mathscr{G}_m\;,$$

so $x\in G_y\subseteq\operatorname{st}(y,\mathscr{G}_m)\subseteq[y,y+1)$, which is absurd, since $x<y$. This contradiction shows that no such sequence of open covers can exist and hence that $X$ is not metrizable.

Note that I did not need to use the fact that the covers form a normal sequence: they can’t even be compatible (in the sense that $\{\operatorname{st}(x,\mathscr{G}_n):n\in\Bbb N\}$ is a local base at $x$ for each $x\in X$).

Of course in this case there are easier ways to show that $X$ is not metrizable; probably the simplest is to show that $X\times X$ is not normal.