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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?

ruakh
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user2157416
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  • 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 $\epsilon$. – Lee Mosher Feb 01 '17 at 21:39

4 Answers4

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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.

Med
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  • 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. – Carmeister Feb 02 '17 at 00:12
  • 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. – Paul Sinclair Feb 02 '17 at 00:12
  • @Carmeister - heh. Beat me by 3 seconds. – Paul Sinclair Feb 02 '17 at 00:12
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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 $B_r(y) = \left \{ z\in M | D(z,y)\lt r\right \}$

Giulio
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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.

Wolfram
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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 $B_r(y)\subset M\setminus\{x\}$.

szw1710
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