# Let $(X,d)$ be a metric space with $x_1$,$x_2$ distinct points of $X$. There are two disjoint open sets containing these points

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

• Certainly choose open sets with radius $d(x_1,x_2)/2.$ Oct 2 '16 at 2:36
• 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?
– bof
Oct 2 '16 at 2:54
• "Two disjoint open sets $O_1$ and $O_2$ containing $x_1$ and $x_2$ respectively." Did you even read the question? Oct 2 '16 at 4:20

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.
• 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. Oct 2 '16 at 4:46