4
$\begingroup$

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?

1 Answers 1

14

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.

  • 8
    Extend this a little further and you've proved metric spaces are Hausdorff.2011-11-30
  • 0
    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?2011-11-30
  • 0
    @dhz Open balls come with every metric space -- they're how you define the (base for the) topology!2011-11-30
  • 0
    @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.2011-11-30
  • 2
    @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.2011-11-30
  • 0
    @dhz An open ball is an open set.2011-11-30
  • 0
    @DylanMoreland Got it! And not too hard to prove too! Thanks for the help! I'm just starting to self study topology so am rough on the basics.2011-11-30