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Let $(X,D)$ be a metric space. Suppose that $x$ and $y$ are two distinct points of $X$. Prove that there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$ and $U\cap V = \emptyset$.

I know that we must show that there exists a $p_1>0$, $p_2>0$ such that for any $x,y\in X$, $x\neq y$ we have that $x\in N(x,p_1)$ and $y\in N(y,p_2)$.

Also, we must have that $x\in U$ and only in $U$ and $y\in V$ and only in $V$.

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Well, $x$ and $y$ are points in a metric space. Given two points in a metric space, what can you do with them? – Michael Albanese Sep 25 '13 at 2:52
Recent duplicate. – Did Sep 25 '13 at 8:25
up vote 1 down vote accepted

Let $r=d(x,y)$ and \begin{align*} U=&\,\left\{z\in X\,\bigg|\,d(z,x)<\frac{r}{2}\right\},\\ V=&\,\left\{z\in X\,\bigg|\,d(z,y)<\frac{r}{2}\right\}. \end{align*} Note that $r>0$ since $x\neq y$.

Claim 1 $x\in U$ and $y\in V$.

Proof: Note that $d(x,x)=0<r/2$, so that $x\in U$. Similarly, $d(y,y)=0<r/2$, so that $y\in V$. $\blacksquare$

Claim 2 $U$ and $V$ are open.

Proof: Each of these sets is a ball around a point, so that it is open. $\blacksquare$

Claim 3 $U\cap V=\varnothing$.

Proof: To obtain a contradiction, suppose that $z\in U$ and $z\in V$. Then, by definition, $d(z,x)<r/2$ and $d(z,y)<r/2$. But then, the triangle inequality implies that $$r=d(x,y)\leq d(x,z)+d(z,y)<\frac{r}{2}+\frac{r}{2}=r,$$ which is a contradiction. $\blacksquare$

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Great explanation, I will make sure I can replicate this and understand it. – Julius Jackson Sep 25 '13 at 8:23

I tried to post this as a comment, but I don't have enough points:

Use the fact that for $ x\neq y, d(x,y)\neq 0$. And what are the open sets ( or a basis for them ) in a metric space ?

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It’s a decent hint, so it’s fine as an answer. If you wanted to offer a slightly larger hint, you might point out that the triangle inequality will definitely be needed. – Brian M. Scott Sep 25 '13 at 3:08
Thank you for mentioning the triangle inequality, I had forgotten about this property. – Julius Jackson Sep 25 '13 at 8:23

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