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).
 A: 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.
A: 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 $\delta = d(x,y)$.
A: 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)
A: 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.
