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Let $X=C([0,1])$ be the Banach space of continuous real valued functions on $[0,1]$ (with the $\sup$-norm).

I am wondering if $X$ can be written as a countable union of compact sets $K_1 \subset K_2\subset K_3 \dots$?

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up vote 5 down vote accepted


  1. If $X$ is an infinite-dimensional Banach space then the closed unit ball is not compact.

  2. Apply the Baire Category Theorem to $X = \bigcup_{n=1}^\infty K_n$ to find an open set with compact closure.

  3. Combine 1. and 2. to derive a contradiction.

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Thanks for the quick answer. Could you explain more the second point? Which version of the Baire Category theorem are you using and how does it imply $\bigcup K_n$ is locally compact? – Cantor Feb 2 '13 at 18:04
The following version of Baire: If a complete metric space is written as a countable union of closed sets, then at least one of them has non-empty interior. Suppose the interior of $K_n$ is nonempty, so it contains a small open ball $B_r(x_0)$ whose closure is compact since it is contained in $K_n$. Translate $x_0$ to zero, scale with $1/r$ to deduce that the closed unit ball is compact. – Martin Feb 2 '13 at 18:07

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