Explaining an example of a topology not induced by a metric

Question

My TA asked us whether every topology can be induced by a metric. He answered "no" and gave an example:

For every set $X$ that satisfies $|X|>1$, its trivial topology (meaning $\{X,\emptyset \}$) isn't induced by any metric. His proof: let $x \neq y \in X$ and denote $d(x,y)=r$. So the set $B_\frac{r}{2}(x) = \{x\}$ is an open set which is not $X$ and not the empty set.

What I don't understand is why necessarily $B_\frac{r}{2}(x) = \{x\}$? If we look at $\Bbb{R}$ and the standard absolute value metric $B_\frac{r}{2}(x)$ is never a singleton. In other words, he only looked at metrics in which $B_\frac{r}{2}(x)$ is a singleton. What am I missing?

Answer 1

You can't really conclude that $B_{r/2}(x) = \lbrace x \rbrace$, but you can conclude that $B_{r/2}(x)$ contains $x$, but not $y$. This eliminates it both from being $\emptyset$ and from being $X$, and hence eliminates it from being open. But, if open balls are always open in the induced topology, so we have a contradiction.

Answer 2

You are right, the proof is inaccurate. It would be better to say that $B_{r/2}(x)$ is neither $\varnothing$ (since it contains $x$) nor $X$ (since it does not contain $y$), so this metric does not generate the trivial topology.