I've seen here Homeomorphism between a normed space and its open unit ball that there is a homeomorphism between the unit open ball and the whole normed linear space. But I was wondering, is there a bijection between such space and a closed ball (not necessarily a homeomorphism)? The function in the link seems to me not to be restricted just to open balls in the sense of being a bijection, altough restricted to open balls in the sense of being continuous. So my question is: is there a bijection between a closed ball and a normed space? Could it be a homeomorphism? Thank you very much!
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I've seen now one problem. So, if we consider the function $F : B \to X$, $F(x)=\text{tan}(\frac{\pi}{4} ||x||) x$, do you think it will work? – Guilherme Ottoni Jan 25 '22 at 16:27
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A bijection will exist because they have the same cardinality. There can't be a homeomorphism because of the boundary. – John Douma Jan 25 '22 at 17:01
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@JohnDouma For infinite dimensional spaces this can happen! – Claire Jan 25 '22 at 18:40
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First of all such a bijection allways exists: If $(V,\|\cdot\|)$ is the normed space, let $S_n=\{x\in V:\|x\|=\frac 1n\}$ for $n\in\mathbb N_{\geq1}$. Then the map which sends $S_{n+1}$ to $S_n$ via $x\mapsto\frac{n+1}nx$ and is the indentity otherwise is a bijection from the open to the closed unit ball of $V$.
Now if $V$ is finite dimensional, the closed unit ball is compact while $V$ is not, so there can't be a homeomorphism.
But for infinite dimensional $V$ this can change: The hilbert space $l_2$ is homeomorphic to it's closed unit ball, see for example the answer here.
Claire
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Wow! This is indeed very interesting! Thank you so much for the example! – Guilherme Ottoni Jan 27 '22 at 17:12