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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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1 Answer 1

up vote 13 down vote accepted

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.

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Extend this a little further and you've proved metric spaces are Hausdorff. –  dls Nov 30 '11 at 19:55
    
Thanks, that's awesome! How can you be sure that such an open set exists in all metric spaces? What if the open ball around x must include other elements, and the sum of those open balls must go back to X? –  dhz Nov 30 '11 at 20:01
    
@dhz Open balls come with every metric space -- they're how you define the (base for the) topology! –  Dylan Moreland Nov 30 '11 at 20:03
    
@DylanMoreland What I meant was, how can you be sure that such an open set exists in all metric spaces of more than one element? What if the open ball around x must include other elements, an the sum of those open balls add back to X? Not sure if that makes sense, but basically I'm having trouble going from ball B to open subset U. –  dhz Nov 30 '11 at 20:08
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@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

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