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let $B(p_0,r)$ be a open ball entered at $p_o$ with a radius of $r$. This ball could be written as union of infinitely many closed balls. Therefore union of closed balls can generate an open ball.

How can I improve my proof here? Am I on the right track? Tips/Tricks please?

Thanks.

Andrés E. Caicedo
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    Provided you know that $B(p_0,r)$ is not closed, I would say it is ok. Another way is: any subset is union of points, but not all infinite subsets are closed. –  Oct 15 '16 at 13:55
  • This is not a proof of your title. –  Oct 15 '16 at 13:55
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    You should give the sets in the union you're describing explicitly. Alternatively, if you can show that each point in your ball is contained in a closed subset of the ball, then the ball is the union of all of those closed subsets, one for each point. – Vik78 Oct 15 '16 at 13:56
  • points approach is great, thanks –  Oct 15 '16 at 14:07
  • In the space $\mathbb R$ the set ${1}\cup{\frac12}\cup\frac13}\cup\cdots$ isn't closed, is it? – bof Feb 04 '17 at 07:30

1 Answers1

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To prove this , you need only a counter example.

Consider the metric space , real line $\;\mathbb{R}\;$ with the usual metric. Consider the open ball $\;B(0,1)\;$ We can easily prove that $\;\;B(0,1)\;=\;\displaystyle \bigcup_{n=1}^{n=\infty}\;B[\frac{1 } {n} ,1-\frac{1}{n}]\;.\;$

This example shows that union of closed balls in a metric space need not be a closed ball. It also proves that the union of closed sets need not be a closed set.

Tips: Consider $\;\mathbb{R}\;$with usual metric. If $\;a<b\;,\;$we have $\;\;(a,b)\;=\;\displaystyle \bigcup_{n=1}^{n=\infty}\;[\;a+\frac{1}{n}\;,\;b-\frac{1}{n}\;]\;\;$ and $\;\;[a,b]\;=\;\displaystyle \bigcap_{n=1}^{n=\infty}\;(\;a-\frac{1}{n}\;,\;b+\frac{1}{n}\;)\;\;$ Using the second identity we can show that in a metric space, intersection of open balls need not be an open ball and the intersection of open sets need not be an open set.

The same idea can be extended to the case of n-dimensional Euclidean space $\;\mathbb{R}^{n}\;\;$with the usual metric also.

If $\;r>0\;,\;a\;\in\;\mathbb{R}^{n}\;\;$then we have $\;\;B\;(a,r)\;=\;\displaystyle \bigcup_{n=1}^{n=\infty}\;B\;[\;a\;,\;r-\frac{1}{n}\;]\;\;$ and $\;\;B\;[a,r]\;=\;\displaystyle \bigcap_{n=1}^{n=\infty}\;B\;(\;a\;,\;r+\frac{1}{n}\;)\;\;$

user114403
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