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Every subset $Y$ of a totally bounded metric space $(X,d)$ is also totally bounded

$X$ is totally bounded $\rightarrow \forall \epsilon > 0 \; \exists n(\epsilon) \in N$ and $\exists x_1\ldots x_n \in X $ such that $$\cup_{i=1}^{n}B_\epsilon(x_i) = X \supseteq Y$$ so $Y$ is totally bounded.

Is this correct?

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There are two notions of total boundedness for a subset $Y$ of a metric space $X$. In one definition we just $Y$ is totally bounded and require points from $Y$ such that the $\epsilon$ balls around them cover $Y$. In the other notion we talk about $Y$ being a totally bounded subset of $Y$ (and say $Y$ is totally bounded in $X$) where is no such requirement.

To show that $Y$ with the restriction of the metric $d$ is totally bounded in its own right you have to obtain points from $Y$, but your points $x_i$ may not be in $Y$.

Use what you have done with $\epsilon$ changed to $\epsilon /2$. Without loss of generality assume that $Y$ has at least one point in common with each of the balls $B(x_i,\epsilon /2)$. If $y_i \in B(x_i, \epsilon /2)\cap Y$ show that the balls $B(y_i,\epsilon)$ cover $Y$.

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  • $\begingroup$ Please see this : math.stackexchange.com/questions/3405281/… please explain the ambiguity $\endgroup$ – Abhay Oct 23 '19 at 9:57
  • $\begingroup$ @Abhay I hope my edited version of answer clarifies everything. $\endgroup$ – Kavi Rama Murthy Oct 23 '19 at 10:07
  • $\begingroup$ so is it possible that a metric space be totally bounded in its own right but not in a bigger space? or is it possible that a subset of a metric space is totally bounded but not as its subspace? I am confused. Or is it that these two notions are equivalent ? $\endgroup$ – Abhay Oct 23 '19 at 10:12
  • $\begingroup$ They are equivalent. But the way your question is worded I believe you have prove that you get points from $Y$ rather than assume that two notions are equivalent. It is just a question of how the question is interpreted. @Abhay $\endgroup$ – Kavi Rama Murthy Oct 23 '19 at 10:16
  • $\begingroup$ thank you very much $\endgroup$ – Abhay Oct 23 '19 at 10:23

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