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Let $Y$ be a metric space and $X$ be a complete metric space. Define a linear map $T: Z\to Y$ where $Z\subset X$. Assume that $T$ is isometry and $T(Z)$ is relatively compact in $Y$. Show that $Z$ is relatively compact in $X$.

My proof is as follows but I am not sure if this works... It seems that I did not use the completeness of $Y$.

Since $T(Z)$ is relatively compact, then for any $n\in \mathbb{N}$, there is a finite cover of $T(Z)$ by open balls of radius $1/n$ with centre $\{y_1,y_2,\dots, y_N\}$. That is $$T(Z)\subset \cup_{i=1}^N B_{y_i}(\frac{1}{n})$$

Then for every $x\in Z$, there exists $y_x\in \{y_1,y_2,\dots, y_N\}$ so that $$ |\|x\|-\|y_x\||=|\|Sx\|-\|y_x\||\le \|Tx-y_x\|_Y<\frac{1}{n} $$

But how to find a finite cover of $A$ with centre in $X$ but not $Y$...

Hermi
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  • What do you mean by "linear map"? Also, if $T$ is a constant map then $T(Z)$ is always relatively compact, but obviously there's no reason to say the same for $Z$. Are there some extra conditions you are not stating? Or perhaps you want it the other way around? (That is, if $Z$ is relatively compact then show $T(Z)$ is?) – David Gao Dec 06 '23 at 04:42
  • I assume you mean a normed vector space as opposed to just a metric space? – spaceisdarkgreen Dec 06 '23 at 04:42
  • @spaceisdarkgreen Yes! – Hermi Dec 06 '23 at 18:08
  • @DavidGao We say $T$ is isometry if $T\in L(X, Y)$ be such that $|Tx|=|x|$.. – Hermi Dec 06 '23 at 18:13
  • So you meant $X$ is a normed linear space, $Y$ is a Banach space, $T$ is a linear isometry from $X$ to $Y$, and $Z$ is a subset of $X$? – David Gao Dec 06 '23 at 19:22
  • Assuming that’s what you meant, this is still false. Say $X = c_{00}$ equipped with the 2-norm, $Y = l^2$, and $T: X \rightarrow Y$ is the natural inclusion map. Let $Z’ = \prod_n [0, 1/2^n] \subset Y$ and $Z = Z’ \cap X$. Then the closure of $T(Z)$ in $Y$ is $Z’$, which is compact. But the closure of $Z$ in $X$ is $Z$ itself, which is not compact. – David Gao Dec 06 '23 at 19:30
  • @DavidGao Sorry... That should be $X$ complete space... My mistake. – Hermi Dec 08 '23 at 07:20
  • So just to clarify again, you meant that $X$ is a Banach space, $Y$ is a normed linear space, $T: X \rightarrow Y$ is an isometric linear map (you wrote $T$ as a map defined on $Z$, but $Z$ is not a linear space so it doesn’t make sense to then say $T$ is a linear map, so I’m instead assuming you meant $T$ is defined on the entire space $X$), $Z \subseteq X$ is a subset. You want to prove if $T(Z)$ is relatively compact in $Y$ then $Z$ is relatively compact in $X$. Is that correct? – David Gao Dec 08 '23 at 07:46
  • If that’s what you meant then you can just note that because $X$ is complete and $T$ is isometric, we have $T(X)$ is closed in $Y$. So the closure of $T(Z)$ in $Y$ is contained in $T(X)$. But then $T$ is a homeomorphism onto its range, so the closure of $Z$ in $X$ is homeomorphic via $T$ to the closure of $T(Z)$ in $Y$ and is thus compact as $T(Z)$ is relatively compact in $Y$. – David Gao Dec 08 '23 at 07:51
  • @DavidGao Can we also prove it by definition? I mean found one $\epsilon$-net to cover $Z$? Because $T(Z)$ is relatively compact in $Y$, we have one $\epsilon$-net $N$ for $T(Z)$. But how to ensure the inverse map of this $N$ well-defined? – Hermi Dec 08 '23 at 23:44
  • What is the definition of relatively compact you're using? That for any $\epsilon > 0$ there exists finitely many $\epsilon$-ball covering the space? That is not equivalent to what I've been using, that the closure of the set is compact, unless the ambient space is complete. If that is the definition you are using then just, in your terminology, for each $y_i$ choose $x_i \in Z$ such that $T(x_i) = y_i$. Then the $\epsilon$-balls around $x_i$ cover $Z$ and you're done. – David Gao Dec 09 '23 at 00:14
  • @DavidGao Yes. But the issue is that if we choose a $\epsilon$-ball cover $T(Z)$, there will be some balls with center outside $T(Z)$, right? Can we choose these balls with centers insider $T(Z)$? Then we can get $T^{-1}$ of these $\epsilon$-ball cover $Z$. That means $Z$ is relatively compact. – Hermi Dec 09 '23 at 00:18
  • I was under the impression that your definition of relatively compact already requires the centers of the $\epsilon$-balls to be in $T(Z)$. If that's not part of your definition, you can fix this by choosing finitely many $\epsilon/2$-balls covering the space. Then for each $\epsilon/2$-ball, if it does not intersect $T(Z)$, discard. If it does intersect $T(Z)$, then choose any $x$ in the intersection and the $\epsilon$-ball around $x$ contains the original $\epsilon/2$-ball... – David Gao Dec 09 '23 at 00:24
  • Do this for all the finitely many $\epsilon/2$-balls and you get a finite collection of $\epsilon$-balls with centers in $T(Z)$ covering $T(Z)$. This reduces the question to the case where the centers are in $T(Z)$ and you can apply what I mentioned before. – David Gao Dec 09 '23 at 00:24
  • @DavidGao Hi David, can you please take a look at this questoin: https://math.stackexchange.com/questions/4851938/revisited-question-lebesgue-measure-of-symmetric-difference-is-0? I do not understand the answer here... – Hermi Feb 05 '24 at 06:33

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