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 April 12th, 2012, 01:46 PM #1 Newbie   Joined: Apr 2012 Posts: 1 Thanks: 0 Every totally bounded set in a metric space is bounded We discussed this theorem in class, which I zoned out since I was up too late writing a paper and I am totally lost. It seems logical that every totally bounded set in a metric space is bounded but I am have trouble understanding the theorem I randomly wrote down. Can anyone tell me in terms I can understand I would appreciate it.
 April 20th, 2012, 06:23 PM #2 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Every totally bounded set in a metric space is bounded Below are the following definitions: A set $S\subseteq X$ is $totally \, bounded$ if for each $\epsilon >0$there exist $x_1,x_2,\ldots , x_n\in X$ such that $S\subseteq \bigcup_{i=1}^nB_{\epsilon}(x_i)$. A set $S\subseteq X$ is $bounded$ if there exists $x\in X$ and $r>0$ such that $S\subseteq B_r(x)$. In the definition of total boundedness, we can make our radius as small as we like, but in the other definition, our radius isn't in under our control. Thus, total boundedness is a stronger condition, and it always implies boundedness. The converse however, it not true in general.
 April 21st, 2012, 01:56 PM #3 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Every totally bounded set in a metric space is bounded If you know that $S \subseteq \bigcup_{i=1}^{n}B(x_i,\epsilon)$ what means that $S$ is totally bounded I would suggest to choose: $S \subseteq B(x_1,\max\{d(x_i,x_j)|\forall x_i,x_j \in X\}+\epsilon)$, this means that $S$ is bounded.

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# totally bounded set implies bounded

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