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$C_0([a, b], \mathbb{R})$ is the space of real-valued continuous function on $[a, b]$. The hint says think Baire. So I assume that $C_0([a, b], \mathbb R)$ is $\sigma$-compact. Then it is the countable union of compact, and hence closed, subsets $M_i$. Since $C_0([a, b], \mathbb R)$ is a complete metric space, by Baire's theorem at least one of the $M_i$'s has a non-empty interior. But I got stuck here. I think I can derive a contradiction from this, but I'm not sure how. Any hints on how to solve this (using contradiction or not) would be much appreciated.

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See also: math.stackexchange.com/q/292930 –  Martin Feb 5 '13 at 20:10

2 Answers 2

up vote 2 down vote accepted

Hint: Show that the closed unit ball is not compact. Now conclude that if $U$ is a compact set then it has an empty interior.

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I definitely see why the closed unit ball is not compact. But how can you conclude that if $U$ is compact then it has an empty interior? –  Aden Dong Feb 5 '13 at 16:35
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@AdenDong: If $U$ had a non-empty interior then there was some $x\in U$ such that there was an open ball $B\subseteq U$ such that $x\in B$. The closure of $B$ is a closed subset of $U$ and therefore compact. Show that this closure is homeomorphic to the closed unit ball (by scaling and translating) and derive contradiction. –  Asaf Karagila Feb 5 '13 at 16:37
    
I see. Thank you very much! –  Aden Dong Feb 5 '13 at 16:38

Hint : show that a subspace of a vector space that has not empty interior is equal to the whole space.

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