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).


4 Answers 4


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)

  • $\begingroup$ I had thought that this was kind of the right idea after researching a bit, but I was not sure. Thank you very much (: $\endgroup$ Commented Apr 19, 2011 at 23:16
  • 1
    $\begingroup$ It's a bit weird to say "you're almost there" if he hasn't really done anything yet though. $\endgroup$
    – Myself
    Commented Apr 19, 2011 at 23:20
  • $\begingroup$ @Myself: I mistakently read "I think I'm supposed to use the Hausdorff property" when he actually said the opposite. $\endgroup$ Commented Apr 19, 2011 at 23:22
  • $\begingroup$ I thought that as well, since what I wrote was the problem itself. But the answer has been really helpful. $\endgroup$ Commented Apr 19, 2011 at 23:22
  • $\begingroup$ 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)? $\endgroup$
    – gary
    Commented Apr 20, 2011 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.

  • $\begingroup$ Fortunately, the proof I came up with take care of that. Thanks for pointing it out thought. $\endgroup$ Commented Apr 20, 2011 at 3:15
  • $\begingroup$ That'd hold for $\underline X = (\{\emptyset\}, \emptyset)$ too as it is also a (discrete & indiscrete) Topology. $\endgroup$
    – Aelx
    Commented Jul 31, 2022 at 8:49

The discrete topology is metrizable, $d(x,y)=1$ for all $x\neq y$. The indiscrete (if $|X|>1$) is not metrizable since, e.g. points can't be separated. In a metric space, two distinct points $x$ and $y$ have disjoint neighborhoods, e.g. $$ U_x=\{z \in X : d(x,z)<\delta/2\}, \quad U_y = \{z\in X : d(y,z)<\delta/2\} $$ where $\delta = d(x,y)$.


For indiscrete topologic space:

Assume that $(X,\tau)$ is metrizable and let G be the familiy of all open sets generated by the metric d. Then we say that $G= \tau$

Let $x,y\in X$ and $d(x,y)=2r$ $(r>0)$ for $x\neq y$. Then

$B(x,r)\cap B(y,r) = \emptyset$. On the other hand $B(x,r)$ and $B(y,r)$ are non-empty sets that belong to G and because of $G=\tau=\{ \emptyset,X\}$, we say that $B(x,r)=B(y,r)=X$ which leads us a contradiction. So our assumption is wrong, $(X,\tau)$ is not metrizable.


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