Prove that the complement of a point in a metric space is open I need to prove that the complement of a point in a metric space is open.
My thoughts so far: 
Suppose $M$ is a metric space. Let $x\in M$ and let $U = M-\{x\}$ be the complement of $x$. My approach is to show that for every $y\in U$ there exists an open ball, $B_r(y)$, centered around $y$, such that $x\not\in B_r(y)$. 
The ball can be represented as $B_r(y)=\{z\in U : D(y,z)<\epsilon\}$ for some $\epsilon>0$. Assuming the distance between $x$ and $y$ is $D(x,y)=d$ I suppose we would have to show there exists a $z$ such that $D(y,z)<\epsilon<d$? Not really sure how to proceed. Am I on the right track?
 A: The easy way is to show that a point is closed. Then you know that the complement is open.
To show a point is closed, you must show that it contains all its limit points. But, there is no limit point to it and you automatically deduce that it is closed.
A: Let $y \in U$. Then $D(x,y) \neq0$ as $y \neq x$. So $\exists r \gt0 \in \Bbb R$ such that $D(x,y) = r$. Let $r' = \frac r 2$.
$x \notin B_{r'}(y)$ because $D(x,y) = r \gt r' \gt0$. So $B_{r'}(y) \subset U \Rightarrow U$ is open.
P.S.: We could use $r$ instead of $r'$, as $B_r(y) = \left \{ z\in M | D(z,y)\lt r\right \}$
A: You don't need to prove the existence of such $z$. The open ball with radius $D(x,y)$ and the center in $y$ does not contain $x$, that is, this ball is inside $M-\{x\}$. So $y$ is an internal point.
A: Let $y\ne x$, as you claim and $r=\frac{1}{2}D(x,y)$. If $z\in B_r(y)$, then by the triangle inequality $D(x,y)\le D(x,z)+D(y,z)<D(x,z)+r$. But $D(x,y)=2r$, which implies that $D(x,z)\ge r$, so $x\not\in B_r(y)$, which finishes the proof, because $B_r(y)\subset M\setminus\{x\}$.
