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How do we define total boundedness for a subset $Y$ of metric space $(X,d)$?

If $(X,d)$ is a metric space, then $X$ is totally bounded if $\forall \epsilon>0 \;\; \exists n(\epsilon) \in N$ such that $\exists x_1,x_2\ldots x_n$ in X such that

$X= \cup_{i=1 }^{n} B_{\epsilon}(x_i)$

Now let $Y\subseteq (X,d)$, then how is total boundedness for $Y$ defined?

the centre of the balls must lie in only $Y$ or can they also lie in $X$ ? Also, the balls should be open in $Y$ or can we have that the balls be open in $X$?

chesslad
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1 Answers1

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The usual definition (see, for example, here: https://en.wikipedia.org/wiki/Totally_bounded_space ) is that $Y$ is totally bounded in $X$ if there exists a family of subsets of $X$ of arbitrary size that is a finite cover for $Y$. So the answers to your questions are:

  1. No; the centres of the ball can lie outside of $Y$ but must still lie in $X$
  2. The balls must be open in $X$.
postmortes
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  • please see this https://math.stackexchange.com/questions/3405463/every-subset-y-of-a-totally-bounded-metric-space-x-d-is-also-totally-bound/3405472#3405472 – chesslad Oct 23 '19 at 09:58