I have troubles with proving that the closure of a bounded set is bounded. Can somebody help me out with this problem? Thanks!
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What does it mean for the set to be bounded? – The very fluffy Panda May 17 '14 at 10:31
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A is bounded if there exists a M>=0 (real) such that d(x,y)<=M for all x,y in A. – Roos Jansen May 17 '14 at 10:38
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Well, now with this in mind. Can you a find a set that encloses A? – The very fluffy Panda May 17 '14 at 10:44
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Yes, there exists a x0 and a r>0 such that B(x0,r) ensloses A. – Roos Jansen May 17 '14 at 11:31
2 Answers
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Let $M$ be a bounded set. We have to prove that $\sup_{x,y\in \overline{M}}d(x,y)$ is finite.
Take $x,y\in \overline{M}$; by definition of the closure, there exists $x'$, $y'\in M$ such that $d(x,x')\lt 1$ and $d(y,y')\lt 1$. We thus have, by the triangle inequality, $$d(x,y)\leqslant d(x,x')+d(x',y')+d(y,y')\leqslant 2+\sup_{u,v\in M}d(u,v).$$ We are done, since the RHS does not depend of $x$ or $y$.
Davide Giraudo
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Let $(M,d)$ be a metric space and let $S$ be a bounded subset of $M$. Since $S$ is bounded, we can choose an open ball $B(x,r)$ to enclose $S$. Taking the closures, we see $$\overline{S}\subseteq\overline{B(x,r)}.$$ Surely, $\overline{B(x,r)}$ is the closed ball with center $x$ and radius $r$, and by enlarging it a little bit, we can see $\overline{S}$ contained in some other open ball, which shows its boundedness.
Boar
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