# Showing that a set X with a trivial topology is not metrizable

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).

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)

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

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