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Let $X:=C[0,1]$ is an metric space and $d(x,y)=\sup_{t\in[0,1]} |x(t)-y(t)|$ for $x,y\in X$. Then is $\bar B(0,1)$ closed and compact in $X$? I can prove that it is closed set by proving the complement is open. I think that it shouldn't be a compact set but not sure how to prove it. I know the definition of a compact set, that's any infinite open cover of a set has a finite subcover of that set.

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From Riesz's Theorem we have known that

"A normed space $X$ is a finite dimensional if and only if its closed unit ball is compact".

We can prove that $C[0,1]$ is a normed space with the norm given by $$ \|x\|=d(x,0), \quad \forall x\in C[0,1]. $$ Since $C[0,1]$ is an infinite dimensional space, its closed unit ball is not compact.

In $C[0,1]$ we can choose an infinite linear independent vectors as: $\{1, x, x^2, \ldots, x^n, \ldots\}$. Hence $C[0,1]$ is an infinite dimensional space.

For more detail, please see here

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I am just a new learner of basic topology, i am not sure what is that theorem, is there any way make use of some basic topology definition and theorem to prove it? – Mathematics Oct 22 '12 at 10:56
@Mathematics: You can see the following references for a direct proof.… – blindman Oct 22 '12 at 11:26
@Mathematics: Dear Sir. Are you agree with my answer? – blindman Oct 22 '12 at 12:44
i am wondering how he construct that counter example from the property of the compactness and the metric is not the same as my question. – Mathematics Oct 22 '12 at 12:58
@Mathematics: You should modify the example to obtain your own solution. – blindman Oct 22 '12 at 13:42

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