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(I will delete this post after I get some suggestions on comments)

The following is a sketch of proof I thought of:

I may consider $\mathbb R^{m,n}$ being equivalent to Euclidean space $\mathbb R^{mn}$ and $\|\cdot\|$ as a continuous function $\mathbb R^{mn} \rightarrow \mathbb R$. For simplicity, let $f(\cdot) = \| \cdot \|$. Then, $f^{-1}([0,1])$ is closed in $\mathbb R^{mn}$ by continuity of $f$. To complete the proof, it suffices to show boundedness of $f^{-1}([0,1])$.

Let $\vec u_{ij} = (0, \cdots, u_{ij}, 0, \cdots, 0)$, where $u_{ij}$ is chosen in a way that $f(\vec u_{ij}) > 1$. Repeat the procedure for ${ij} \in \{1, \cdots, mn\}$, so I have appropriate choices of $u_{ij}$ for $ij \in \{1, \cdots, mn\}$. Then, construct a rectangle $R = \prod\limits_{ij}^{mn} R_{ij} \subset \mathbb R^{mn}$, where $R_{ij} = [-u_{ij}, u_{ij}]$. Then, I may conclude that $f^{-1}([0,1]) \subset R$ is compact by Heine-Borel.

I have two questions.

  1. Is there a simpler approach to show compactness of $f^{-1}([0,1])$?
  2. If I consider Banach space instead of $R^{mn}$, compactness may not hold anymore, right?
James C
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    @coboy that's not true. $e^{it}: [0,1) \rightarrow S^1$ is continuous. $S^1$ is compact, but $f^{-1}(S^1)$ is not compact. – James C Jan 11 '23 at 21:55
  • Oh my bad you are totally right it's true only for the image ! Sorry I will delete my comment. – coboy Jan 11 '23 at 21:58
  • It is not clear to me how you conclude that the image is in $R_{ij}$. Norms on finite dimensional spaces are equivalent, so by appropriate choice of norm $|A| = \max_{i,j} |A_{ij}|$ that there is some $B$ such that $|A_{ij}| \le B$ for all $i,j$ and all $A$. – copper.hat Jan 11 '23 at 22:01
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    No matter what norm ones uses in a finite dimensional space, they are all equivalent in the sense that they generate the same topology. Recall that closed and bounded sets in finite dimensional normed spaces are compact. – Mittens Jan 11 '23 at 22:02
  • I know they are equivalent. I'm just trying to penetrate his way of thinking. And he knows that closed and bounded sets in his space are compact: he is trying to exploit this fact. – Anne Bauval Jan 11 '23 at 22:08
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    James C, you are overcomplicated things: $f^{-1}([0,1])$ is bounded by definition: it is the closed unit ball for your norm. As for your last question: right, this is Riesz's theorem: the closed unit ball of a real normed vector space $V$ is compact iff $\dim V<\infty.$ – Anne Bauval Jan 11 '23 at 22:19
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    @AnneBauval Yes, I meant Heine-Borel. Thanks for the correction. – James C Jan 11 '23 at 22:31
  • @AnneBauval As you mentioned in your last comment, $f^{-1}([0,1])$ is bounded if I use metric induced by $| \cdot |$. However, in order to Heine-Borel, I need Euclidean setting; I needed to consider $f^{-1}([0,1])$ w.r.t to Euclidean topology since there is no guarantee that $f^{-1}([0,1])$ is also bounded in Euclidean space. – James C Jan 11 '23 at 22:35
  • You did not answer to my question about which $|~|_{m,n}$ you chose. (In fact, as commented by Oliver, all norms on your space are equivalent, which means that they not only generate the same topology: they have the same bounded sets. But you may ignore that general theorem and prove it only for your norm.) – Anne Bauval Jan 11 '23 at 22:37
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    @AnneBauval It is an arbitrary norm that satisfies any norm property. (Not necessarily $\sup \frac{|Av|_n}{|v|_m}$). And yes, I am reading equivalence of norms at this moment: https://math.stackexchange.com/questions/57686/understanding-of-the-theorem-that-all-norms-are-equivalent-in-finite-dimensional. Thank you for your comments. – James C Jan 11 '23 at 22:40
  • Your $f^{-1}([0,1]) \subset R$ does not hold for any norm. – Anne Bauval Jan 11 '23 at 22:52
  • @AnneBauval After rechecking my sketch of proof, indeed there is no guarantee that $f^{-1}([0,1]) \subset R$. However, I may modify slightly: define $M:= \sum\limits_{ij=1}^{mn} |u_{ij}|$. Then, instead of $R$ earlier, I define a square $S := \prod\limits_{ij=1}^{mn} [-M, M]$. I believe this works. – James C Jan 11 '23 at 23:01
  • No, it does not. E.g. (with $mn=2$) you may have $|(1,0)|=|(0,1)|=|(3,3)|=1.$ – Anne Bauval Jan 11 '23 at 23:08
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    @AnneBauval I understood the equivalence of norm in any finite-dimensional normed vector space, so the question became trivial. Thank you. – James C Jan 15 '23 at 01:23

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