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Suppose I am asked to show that some topology is not metrizable. What I have to prove exactly for that ?

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    $\begingroup$ Use any theorem of the form "if a topology is metrizable then it satisfies property BLAH". For example, if a topology is metrizable then it is Hausdorff, even more it is normal, it satisfies the first-countability property, and so on. Look up some theorems in a topology book. Disprove any one of these properties and you have disproved metrizability. $\endgroup$ – Lee Mosher Aug 5 '14 at 13:08
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Recall that if $(X,d)$ is a metric space, then the metric induces a topology. But different metrics can induce the same topology, for example $(X,d')$ where $d'(x,y)=2d(x,y)$, is a different metric, but induces the same topology.

We say that a topological space is metrizable, if there is a metric which induces the topology.

So to show that a space is not metrizable, you have to show that there is no metric which can induce this topology. This is often done by refuting certain consequences of metrizability. For example we can prove that metric induced topology is always Hausdorff, and first-countable.

If a space is not first-countable, it's not a metric space. If it is not Hausdorff it's not a metric space.

If you have more information on the space, then you can use other conditions as well, e.g. connected metric space with two points is uncountable. If the space is countable and connected then it is not metrizable.

And so on and so forth.

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If a topology is metrizable, then the "diagonal sequence trick" is available. This means that if you have a sequence $$ x_{(n)} \to x,\qquad n \to \infty $$ and each term of the sequence is the limit of another sequence, belonging to a "good" set $G$: $$ G\ni x^{k}_{(n)}\to x_{(n)}, \qquad k\to \infty $$ then you can construct a "diagonal" sequence $x^{k(n)}_{(n)}\in G$ which converges to the original limit point $x$. In particular, $x$ is in the sequential closure of the good set $G$.

This fact is usually phrased in topological terms as "topological closure and sequential closure are the same". So the first thing to try when proving that a topology is not metrizable is showing that the diagonal sequence trick does not work.

P.S.: I have in mind von Neumann's proof that the weak topology on $L^2(\mathbb{S}^1)$-space is not metrizable. It goes as follows. Let $$G=\left\{ e^{int}+n e^{imt}\ | n,m\in \mathbb{N}_{\ge 1}\right\}$$ be the "good" set in $L^2(\mathbb{S}^1)$. The question is whether we can or we cannot reach $0$ by taking one weak limit in the good set. The answer is negative because, roughly speaking, to reach $0$ we need to let $n$ go to infinity, otherwise the term $e^{int}$ would never vanish in the weak limit. But letting $n$ go to infinity the term $ne^{imt}$ diverges in norm, hence it is not bounded in weak sense as well. So $0$ is not in the sequential closure of $G$.

However, if we let $n$ be fixed and we take a limit in $m$, then the term $ne^{imt}$ vanishes in the weak limit. With the above terminology we have $$ e^{int}+ne^{imt}=x^m_{(n)}\rightharpoonup x_{(n)}=e^{int}, \qquad m\to \infty.$$ If we now let $n\to \infty$ we get $$ x_{(n)}=e^{int}\rightharpoonup 0,$$ but we have already seen that $0$ is not in the sequential closure of $G$. So the diagonal sequence trick is not available in $L^2(\mathbb{S}^1)$-space with weak topology. Hence that topology is not metrizable.

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Certainly, if the topology is not Hausdorff, it is not metrizable.

From Wikipedia:

A compact Hausdorff space is metrizable if and only if it is second-countable.

You might find more on that Wikipedia page.

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  • $\begingroup$ $X$ endowed with $(X,\varnothing)$ as topological structure is not metrizable. $\endgroup$ – Zbigniew 8 hours ago

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