Which of these spaces is metrizable? The question:
Which of the following topological spaces are metrizable?


*

*Let $X$ be any non-empty set, and let the topology consist only of the empty set $\emptyset$ and the full space $X$.

*Let $X$ be any infinite set, and let the topology consist of the empty set together with all subsets of $X$ whose complements are finite.

*Let $X$ be the three-element set $\{a, b, c\}$, and let the topology consist of the following subsets of $X$: $\{\emptyset, \{a\}, \{a,b\}, \{a,c\}, X\}$.


(Hint: If a space is metrizable, then its open sets must have certain properties.)
My answer and query:
None of these seem to be metrizable. I am almost 100% sure that 1 cannot be metrizable. I think 2 isn't because I don't have any idea how to do it, and similarly for 3. However, I can't figure out the hint so I am not sure. Am I right or not?
(For reference, this is part of Problem 16-3 of Introduction to Topology and Modern Analysis by G.F. Simmons. The unquoted part of the question deals with obviously metrizable cases.)
 A: Given what you’re said in the comments, I think that you’re expected to realize that if $x$ and $y$ are distinct points in a metric space $\langle X,d\rangle$, and $0<\epsilon\le\frac12d(x,y)$, then $B(x,\epsilon)$ and $B(y,\epsilon)$ are disjoint open sets containing $x$ and $y$, respectively. (Later you’ll learn that spaces in which any two distinct points have disjoint open nbhds are called Hausdorff or $T_2$ spaces.)
This will allow you to rule out all three of the examples except for the special case of (1) in which $X$ has only one point; that special case is metrizable. You can write down the metric very easily: the only distance to be specified is $d(x,x)$, which must be ... ?
Actually, for (1) (when $X$ has more than one point) and (3) you really need only the fact that if $x\ne y$, and $0<\epsilon\le d(x,y)$, then $y\notin B(x,\epsilon)$, and for (2) you need only the fact that in any metric space with at least two points there are two disjoint non-empty open sets.
A: Something to think about:
All metric spaces are Hausdorff (a.k.a T2) and as the Hausdorff property is a topological invariant you can check if a space is not metrizable by checking if it's not Hausdorff. For example, part (3) of your question:
For $ a,b \in X $ we cannot find two open sets $ U,V \subset X $ such that $ a \in U$ and $ b \in V $ with $ U \cap V = \emptyset $, because the only choices are $ \lbrace a \rbrace$ and $\lbrace a, b \rbrace $ with $ \lbrace a \rbrace \cap \lbrace a b \rbrace = \lbrace a \rbrace $ . Thus $ X $ is not Hausdorff and hence cannot be metrizable.
Be careful though, the converse is not true! Not all Hausdorff spaces are metric spaces, so if you find a space is Hausdorff you still have work to do if you want to show it's a metric space.
