# Is it possible that in a metric space $(X, d)$ with more than one point, the only open sets are $X$ and $\emptyset$?

Is it possible that in a metric space $(X, d)$ with more than one point, the only open sets are $X$ and $\emptyset$?

I don't think this is possible in $\mathbb{R}$, but are there any possible metric spaces where that would be true?

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One of the axioms is that for $x, y \in X$ we have $d(x, y) = 0$ if and only if $x = y$. So if you have two distinct points, you should be able to find an open ball around one of them that does not contain the other.
@dhz Hm. I'm not sure what you mean. Of course, it may be, as in $\mathbf{R}$, that every open ball contains more than one element. But these sets of the form $B(x, r) = \{y \in X : d(x, y) < r\}$ are always open. –  Dylan Moreland Nov 30 '11 at 20:12