3
$\begingroup$

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

  • 0
    Hint: Pick $\epsilon$ cleverly and apply the triangle inequality. If you don't know how to pick $\epsilon$, draw a picture in the Euclidean plane, and see if that helps you pick a good $\epsilon$. – 2017-02-01

4 Answers 4

6

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.

  • 0
    I would argue that there isn't really a benefit to this method, since showing there is no limit point to a single point uses the same reasoning as directly showing that its complement is open. – 2017-02-02
  • 0
    But to show that a point has no limit points, you have to show that every other point has a neighborhood (in a metric space, a ball) about it that misses the original point. – 2017-02-02
  • 0
    @Carmeister - heh. Beat me by 3 seconds. – 2017-02-02
3

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 $B_r(y) = \left \{ z\in M | D(z,y)\lt r\right \}$

2

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.

2

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)