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The following question is from Fred H. Croom's book "Principles of Topology"

Let $(X,d)$ be a metric space and $x_1,x_2$ distinct points of $X$. Prove that there are disjoint open sets $O_1$ and $O_2$ containing $x_1$ and $x_2$, respectively.

How would I approach this problem?

My attempt thus far was to show that $x_1$ and $x_2$ respectively have infinitely many elements in their own neighborhoods: $O_1 $ and $O_2$.

Once you reach a small enough neighborhood for both distinct points, the intersection between the two open sets would be disjoint. To strengthen the claim, I wanted to show that the distance between the two sets would at one point be greater than $0$, proving they are disjoint.

Am I approaching this the right way? If not, how would I properly prove this?


I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

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  • $\begingroup$ The idea of getting small enough nbhds is right. You don’t need to make them so small that the distance between them is positive: it’s enough that they simply not overlap. See doglah’s answer. $\endgroup$ – Brian M. Scott Feb 11 '15 at 7:58
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I'll assume that you know that open balls in a metric space are open sets. Then since $x_1$ and $x_2$ are distinct $$ r = \frac{1}{2} d(x_1,x_2) > 0. $$

Then the balls $B_{r}(x_1)$ and $B_{r}(x_2)$ are disjoint open sets containing $x_1$ and $x_2$ respectively. If there were a point $p$ in both balls then $$ d(x_1,x_2) \leq d(x_1,p) + d(p,x_2) < r + r = d(x_1, x_2). $$ Clearly this is impossible, so the sets are disjoint.

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