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I noticed that $\mathbb R^n$ is not compact under any norm. I read some sources saying that any norm on $\mathbb R^n$ is equivalent to the Euclidean norm. That is, any norm in $\mathbb R^n$ induces the same topology.

And I also read some sources saying that for any set $(X, \mathcal T_{\text{indiscrete}})$ is compact. But indiscrete topology is not metrizable.

I wonder if there is a metric that makes $\mathbb R^n$ compact?

Benjamin
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    Let $f$ be a bijection from $\mathbb R^{n}$ to $[0,1]$ and define $d(x,y)=|f(x)-f(y)|$. – Kavi Rama Murthy Jul 17 '20 at 07:33
  • @KaviRamaMurthy How this makes $\Bbb{R}^n compact$?if $f$ is homeomorphism then you can say $\Bbb{R}^n$ is compact. Why are you letting $f$ be a bijection from $\mathbb R^{n}$ to $[0,1]$ and define $d(x,y)=|f(x)-f(y)|$? – Unknown Jul 17 '20 at 07:42
  • @AmanPandey You are confused. The question has nothing to do with the usual topology of $\mathbb R^{n}$. No topology is given and there is no question of constructing a homeomorphism. The space I have constructed is isometric to $[0,1]$ with usual metric and hence it is compact. – Kavi Rama Murthy Jul 17 '20 at 07:45
  • @KaviRamaMurthy Ah! I got it. – Unknown Jul 17 '20 at 07:50
  • @KaviRamaMurthy You should make that an answer. – badjohn Jul 17 '20 at 08:05
  • @badjohn This question is a duplicate I posted this as an answer just a few days ago. That is why I am not posting an answer. – Kavi Rama Murthy Jul 17 '20 at 08:07
  • @KaviRamaMurthy I didn't notice that. – badjohn Jul 17 '20 at 08:54

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There's a bijection $f:\mathbb{R}^n\to [0,1]$ which we can use to induce a metric on $\mathbb{R}^n$ making it compact: $d(x,y)=|f(x)-f(y)|$. Under this definition $f$ becomes a homeomorphism. More generally, the same applies to any set $X$ of cardinality $|\mathbb{R}|$.

On the other hand every compact metric space is separable. And so a compact metric space $X$ is an image of $l(A)=\{(x_n)_{n=1}^\infty\ |\ x_i\in A\}$ for some countable set $A$. Therefore a compact metric space has cardinality of at most $|\mathbb{R}|$. Unlike topological case, where compact spaces can be of arbitrary size.

So is every set of cardinality at most $|\mathbb{R}|$ compact metrizable? Clearly finite (under any metric) and countable sets (bijection with $\{0\}\cup\{1/n\ |\ n\in\mathbb{N}\}$) are. Under the continuum hypothesis there is no other choice. But without the continuum hypothesis? I do not know the answer.

freakish
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