Suppose that $X=\bigcup_{n=1}^\infty K_n$ 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 subspaces?

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    $\begingroup$ No; a classical example is $\mathbb{R}^\mathbb{N}$ as inductive limit of the finite dimensional spaces $\mathbb{R}^n$, $n\in \mathbb{N}$. Also, $C_c(\Omega)$ as inductive limit of the Banach spaces $C_K$ (=functions with support in $K$) along the family of compact subspaces $K$ of $\Omega$. $\endgroup$ – Pietro Majer Jan 5 '12 at 11:45
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    $\begingroup$ Let $X=I\coprod I$ where $I=[0,1]$ the real interval by natural topolgy, then any element of $X$ is $(0, x)$ (as $x$ is in the frist copy of $I$) or $(1, x)$ (....second copy $I$). On $X$ change the topology considering the neighbors of $(0,0)$ as the usual neighbors join with a neighbor of $(1,0)$, and this family become the neighborrds of $(1,0)$ too. (formally let $f: X\to [-1, 1]: (1, x)\mapsto x,\ (0, x)\mapsto -x$ and let the topology induced by $f$ on the set $X$, where $[-1,1]$ has the natural topology). THen $X$ is a union of two subspace (metrixable of course) but $X$ isnt $T_2$. $\endgroup$ – Buschi Sergio Jan 5 '12 at 12:15
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    $\begingroup$ Every CW-complex is the union of a sequence of closed metrizable subspaces, but there are many examples of CW-complexes that aren't metrizable (a wedge of countably infinitely many circles being perhaps the simplest). $\endgroup$ – Stephen S Jan 5 '12 at 13:11
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    $\begingroup$ Why do you all post just comments, although they are answers? $\endgroup$ – Martin Brandenburg Jan 5 '12 at 13:23
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    $\begingroup$ @Diagonal: A sequence in the TVS $R^N$ is Cauchy (resp. convergent) if and only if, it belongs to some $R^n$ and it is Cauchy (resp., convergent) there. This implies that $R^N$ is sequentially complete. On the other hand, $R^N$ is the countable union of closed proper subspaces $R^n$ (they are closed because they are complete). Hence, $R^N$ is not metrizable by the Baire category theorem. Note that the same argument works for $C_c(\Omega)$ and $\mathcal{D}(\Omega)$ etc. $\endgroup$ – Pietro Majer Jan 5 '12 at 13:24

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.


(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.


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