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Prove that for each set $X$, the topological space $\left(X,2^X\right)$ is metrizable, where $2^X$ is the power set.

What I'm not sure is what are the conditions for a topological space to be metrizable (this is, exists a metric space $(X,d)$ that gives rise to the topological space $(X,\tau)$). As far as I've understood, it suffices to show that:

(i) If $A\in\tau$, then $A$ is an open set of $(X,d)$.

(ii) If $A$ is an open set of $(X,d)$, then $A\in\tau$.

Are these necessary and sufficient conditions? In this case, my attempt was:

Let $d$ be a function defined as $d(x,x)=0$ and $d(x,y)=1$ if $x\neq y$, then:

(i) Let $A\in2^X$. If $x\in A$, let $\delta:=\frac12$. Then $B(x,\delta)\subset A$, and thus $A$ is open.

(ii) Suppose $A\subset X$ open (it's easy to see that all sets $A\subset X$ are open). Then by definition of power set, $A\in 2^X$.

Using this metric, it seems that all topological spaces $(X,\tau)$ verify the condition (i), don't they? So a way to prove that a topological space is not metrizable is finding an open set $A\in X$ such that $A\not\in\tau$?

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    $\begingroup$ A linguistic nitpicks: It suffices to show that there exists a metric $d$ such that (i) and (ii) hold. You can't write conditions (i),(ii) without first introducing what $d$ is. $\endgroup$
    – anon
    Jul 8, 2015 at 17:46
  • $\begingroup$ Thanks! The book I'm reading only describes in few lines what is "metrizable", without a formal definition, and so I wasn't sure about it. $\endgroup$
    – user246336
    Jul 8, 2015 at 18:06

2 Answers 2

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First, you are right in identifying a metric that shows that this space is metrizable. But your proposed method of showing that a space isn't metrizable wouldn't work because the space could be metrizable, but with a different metric than the one you describe here (consider $\mathbb R$ with the usual topology). The usual way to show that a space is not metrizable is to pick any one of a number of theorems which hold if a space is metrizable and show that that theorem does not hold in the space you are considering.

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  • $\begingroup$ I see... I'm just starting to read about topology and so I don't know yet these theorems you say, I'll have to wait for a bit. Thanks for your help! $\endgroup$
    – user246336
    Jul 8, 2015 at 18:05
  • $\begingroup$ One of the theorems is used in the answer provided by @blazs is that a metrizable topological space is Hausdorff (though certainly, there are non-metrizable spaces which are Hausdorff, so this won't work in every case). $\endgroup$
    – Plutoro
    Jul 8, 2015 at 18:12
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Your proof that $(X, 2^X)$ is always metrizable is correct.

Here's an example of a topological space that is not metrizable.

Consider the set of real numbers equipped with the following topology $\mathcal{T}$: for each $a\in\mathbb{R}$ let $(a, \infty)$ belong to $\mathcal{T}$. Note that we also need to add $\emptyset$ and $\mathbb{R}$ to $\mathcal{T}$.

This is clearly a topological space. It is not metrizable because in a metric space it is always possible to find a pair of disjoint open balls; however, in $\mathcal{T}$ every two nonempty sets have a nonempty intersection.

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    $\begingroup$ It's always helpful to know examples when one starts studying something completely new. Thanks! $\endgroup$
    – user246336
    Jul 8, 2015 at 18:12

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