metrizable space - MathOverflow

metrizable space

2012-01-05T11:35:43Z

Suppose $X=\cup K_n$ is a topological space, $K_n$ is a metrizable subspace in $X$ for every $n \in \omega$, then $X$ is a metrizable space?

In metrizable spaces, compactness is equivalent to $\sigma$-compactness?

One more: Is pseudocompactness hereditary with respect to $\sigma$-compact subspace?

Answer by Stephen S for metrizable space

2012-01-05T16:46:03Z

(This is just an expansion of my comment above.)

As Buschi Sergio points out, a topological space that is the union of countably many metrizable subspaces need not even be Hausdorff. My example of CW-complexes is intended to show that even when such a space is Hausdorff (and paracompact and submetrizable), it can still easily fail to be metrizable.

Every CW-complex is $\operatorname{F}_\sigma$-metrizable (that is, the union of countably many closed metrizable subsets). This is easy to see when there are only countably many cells, as the cell closures are compact and metrizable. But it's true in general, as I'll explain below. (I should perhaps clarify that by a "cell" of a CW-complex I mean what is sometimes called an "open cell". The cells partition the space.)

If a subset $A$ of a CW-complex is such that $A\cap e$ is compact for each cell $e$, then $A$ is closed and metrizable. (It's metrizable because it's the topological sum of the compact sets, which are metrizable.) Since each cell is $\sigma$-compact, we can express the CW-complex as a union of countably many closed metrizable subspaces of this form.

The simplest example of a CW-complex that is not metrizable consists of a single 0-cell and a countable infinity of 1-cells, forming a bouquet of circles. This is not metrizable as first-countability fails at the 0-cell.

Answer by Henno Brandsma for metrizable space

2012-01-05T19:25:37Z

No to all your questions. There are lots of countable non-metrizable spaces: an easy one is $\mathbb{N}$ in the cofinite topology, which can be written as a countable (disjoint) union of singletons (which are metrizable, and closed and compact). Using ultrafilter spaces (given an ultrafilter $\mathcal{F}$ on $\mathbb{N}$, define $X = \mathbb{N} \cup {\infty}$ where $\mathbb{N}$ is discrete and a neighbourhood of $\infty$ is of the form $A \cup {\infty}$, where $A \in \mathcal{F}$); such spaces are countable, hereditarily normal but not metrizable (not even first countable at $\infty$).

All these spaces above are $\sigma$-compact but not compact, and of course a discrete countable set like $\mathbb{N}$ is metrizable, $\sigma$-compact but not compact, as is $\mathbb{R}$, e.g.

Also $[0,1]$ is pseudocompact, but the $\sigma$-compact subset $\{ \frac{1}{n} \mid n \in \mathbb{N} \}$ is not (pseudo)-compact.