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We have a metric space $\boldsymbol{(X,d)}$ and $\boldsymbol Y$ is included in $\boldsymbol{X}$, a subset.

How can we prove that $\boldsymbol Y$ is totally bounded if and only if the closure of Y is totally bounded?

Any help would be greatly appreciated.

mimi
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  • Are you missing an assumption? (e.g., $X$ is totally bounded) – angryavian Oct 14 '18 at 21:22
  • Hi, and welcome to the Math.StackExchange. Why is this question interesting for you? What did you tried? These question are simply examples of providing context for your question: this will help other members help you. – Daniele Tampieri Oct 14 '18 at 21:23
  • I corrected the question ! Sorry for the confusion – mimi Oct 14 '18 at 21:45

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The implication "$Y$ totally bounded then $\overline{Y}$ is totally bounded", is proved here among other places. (it's one of the "related" links on the right part of the web page; always good to pay attention to those..)

A subset of a totally bounded set is still totally bounded and this takes care of the other direction.

Henno Brandsma
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