# Can C([0,1]) be written as a countable union of compact sets?

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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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.
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