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This exercise is from a "challenge" list that my analysis teacher gave us to do:

Let $X$ be a set. There are two trivial topologies:

Indiscrete (not sure if this is the actual name, as I am translating from another language) Topology: the open sets are exactly $X$ and the empty set.

Discrete Topology: every subset of $X$ is open.

Deduce that the indiscrete topology is not metrizable and the discrete topology is metrizable.

My problem is that this is a "first/second" course in real analysis and no one in the class has seen topology and metric spaces beyond the basics, so I pretty much have no idea of how to do this. Suggestions on how to prove this (proofs itself are not needed) will be very much appreciated.

P.S.: I've looked myself at some topology books and I think I'm not supposed to use anything like Hausdorff spaces and stuff like that (not even sure if that is the actual way to go).

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3 Answers 3

up vote 9 down vote accepted

You're almost there.

For a space to have a metric, you must be able to distinguish any two points, that is: $d(x,y)=0$ if and only if $x=y$. But the indiscrete topology has way too few open sets for this to be possible, i.e. there cannot be any $\epsilon$-balls separating $x$ from $y$.

For the discrete topology, here's a hint: the discrete metric.

(alternatively: every metric space is Hausdorff)

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I had thought that this was kind of the right idea after researching a bit, but I was not sure. Thank you very much (: –  Leonardo Fontoura Apr 19 '11 at 23:16
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It's a bit weird to say "you're almost there" if he hasn't really done anything yet though. –  Myself Apr 19 '11 at 23:20
    
@Myself: I mistakently read "I think I'm supposed to use the Hausdorff property" when he actually said the opposite. –  Fredrik Meyer Apr 19 '11 at 23:22
    
I thought that as well, since what I wrote was the problem itself. But the answer has been really helpful. –  Leonardo Fontoura Apr 19 '11 at 23:22
    
Just another perspective: in metric spaces, all open balls B(p,r) centered at p, of fixed radius r>0 are open sets; I don't think too hard to show. Your topology should include all such sets. Can this be the case with just two open sets (with non-trivial cases of singleton spaces)? –  gary Apr 20 '11 at 0:38

That $X$ has more than one element is implicit in the above answers. If $X$ has exactly one element the discrete and indiscrete topologies coincide and are metrizable.

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Fortunately, the proof I came up with take care of that. Thanks for pointing it out thought. –  Leonardo Fontoura Apr 20 '11 at 3:15

the discrete topology is metrizable, $d(x,y)=1$ for all $x\neq y$. the indiscrete (if$|X|>1$) is not metrizable since points will not be closed.

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