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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.

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  • $\begingroup$ Such a space would have at most countably many closed sets. $\endgroup$ Commented May 21, 2021 at 15:52

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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 $x\in X$.

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  • $\begingroup$ Oh, of course you are right! Thank you, actually the solution was not so hard, but I somehow didn't grasp it ^^ $\endgroup$ Commented May 21, 2021 at 16:02

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