TITLE: Does any metric on the real numbers imply a topology with countable many open subsets?
QUESTION [5 upvotes]: a friend of mine came up with a question wich seems to be easy at first, but after some thinking I could not come up with a solution. The question is, weather you can define a metric $d$ on $\mathbb{R}$ such that the number of open sets via this metric is only countable.
My guess is that it is not possible. I thought that for any $\epsilon > 0$ the open Ball $B_{\epsilon}(x)$ should contain infinitely many other elements (at least for uncountable many real numbers $x$) so you can't just have the set with only one element as an open leading to uncountable many open subsets. But I am not quite sure if that leads to an answer, I just wanted to mention it.
I would be glad if someone is able to answer my question, thanks a lot already!
Hannes L.

REPLY [4 votes]: In a metric space, the complement of any singleton set is always open.  So if $X$ is any set equipped with a metric, there are at least $|X|$ open sets, since $X\setminus\{x\}$ is open for each $x\in X$.