[Math] Compactness of the unit sphere in finite-dimensional normed vector space

compactnessfunctional-analysismetric-spacesnormed-spacesspheres

In order to prove that norms defined on any finite-dimensional real (or complex) vector space $E$ are equivalents, I need to proof the compactness of the unit sphere $S_{\infty}=\{x\in E\,\vert\,||x||_{\infty}=1\}$ where the particular norm depends on a basis $\mathcal{B}$. Let $\dim E=n$.
In my notes, we define an isometry $\phi:E,||\cdot||_{\infty}\to\mathbb{C}^{n},||\cdot||_{\infty}:x\mapsto(x_{1},\dots,x_{n})$ and we say that, as it is an isometry, $\phi(S_{\infty})$ is compact implies $S_{\infty}$ is compact. I understand that $\phi(S_{\infty})$ is compact because bounded and closed. I don't understand the implication $\phi(A)$ compact $\implies$ $A$ compact.

Thanks in advance for your answers

Best Answer

It's just about showing that every sequence (bounded in the $||.||_\infty$ norm) $\{v_n\}\subset S_\infty$ of vectors has a convergent subsequence. For $w_n:=\phi(v_n)$, the corresponding sequence $\{ w_n \} $ has, by the virtue of compactness of $\phi(S_\infty)$, a convergent subsequence $\{w_{n_k}\}$ converging to some $w \in \phi(S_\infty)$. Now there is an obvious canditate for a subsequence of $\{v_n\}$ converging to $v := \phi^{-1}(w)$ and to show convergence one just needs to use the fact that $\phi$ is a linear isometry. This means that for any pair $x,y \in E $ of vectors, we have $||x-y|| = ||\phi (x-y)||=||\phi (x) -\phi (y)||$.

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