0
$\begingroup$

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

The approach that I'm trying is to show that $d(x_1, X \setminus O_1) \geq d(x_1, O_2)$ and $d(x_2, X \setminus O_2) \geq d(x_2, O_1)$. But i'm not having any luck

$\endgroup$
3
  • 1
    $\begingroup$ Certainly choose open sets with radius $d(x_1,x_2)/2.$ $\endgroup$
    – IAmNoOne
    Oct 2 '16 at 2:36
  • $\begingroup$ How do you expect us to help you prove something about $O_1$ and $O_2$ when you haven't told us what $O_1$ and $O_2$ are? $\endgroup$
    – bof
    Oct 2 '16 at 2:54
  • 1
    $\begingroup$ "Two disjoint open sets $O_1$ and $O_2$ containing $x_1$ and $x_2$ respectively." Did you even read the question? $\endgroup$
    – 1233211
    Oct 2 '16 at 4:20
3
$\begingroup$

Since $x_1$ and $x_2$ are distinct, $d(x_1, x_2) > 0$ by definition of the metric. Set $y:= d(x_1, x_2)$. Then choosing the open sets to be

$B(x_i, \frac{y}{2}):= \{z \in X \ | \ d(z, x_i) < \frac{y}{2}\}$ for $i=1, 2$

we see that $B(x_1,\frac{y}{2}) \cap B(x_2, \frac{y}{2}) = \varnothing$ by the triangle inequality.

$\endgroup$
1
  • $\begingroup$ I'm sorry, but i'm still not 100% seeing it. Do you mean to use the triangle inequality to rewrite $d(x_1, x_2)$? How does this show that the intersection is empty? Thank you for your answer. $\endgroup$
    – 1233211
    Oct 2 '16 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.