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Let's consider the space of all continuous function $C[0,1]$ on the intervall $[0,1]$. But instead of using the usual supremum norm we use the $L^1$-Norm $: \lVert f \rVert_1=\int_0^1 \lvert f(x) \rvert dx$ for $f \in C[0,1]$

It is well known that this space is not complete with respect to the $L^1$-Norm.

The Baire Category Theorem states that if a metric space is Banach then it cannot be a countable union of nowhere dense sets.

What can we say about the converse in this case? Is $(C[0,1],\lVert \cdot \rVert_1) $ a countable union of nowhere dense sets?

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    $\begingroup$ In a more sophisticated sense, the answer has to be "yes" because it certainly isn't Banach, and you didn't use the axiom of choice. If I've done my work correctly, in a model of ZF+DC in which every set of reals has the Baire property, we can show that every incomplete separable normed space is meager. (As a subset of its completion, it has the BP, so it is meager in its completion, and an elementary argument shows it must therefore be meager in itself.) $\endgroup$ Jul 28, 2015 at 21:07

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Yes,

$$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$

and $A_n$ is closed for each $n$ - if $\lVert g\rVert_\infty > n$, then there is a $\delta > 0$ and a non-degenerate interval $[a,b] \subset [0,1]$ such that $\lvert g(x)\rvert \geqslant n+\delta$ for all $x\in [a,b]$, and hence $\lVert g-f\rVert_1 \geqslant \delta\cdot (b-a)$ for all $f\in A_n$, and $A_n$ has empty interior - that follows for example by the open mapping theorem, if it had non-empty interior the two norms would be equivalent. But one can also argue elementarily that each $A_n$ has empty interior.

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  • $\begingroup$ This is a good argument, and shows something stronger than what I was going to show. I would still like to point out that "$C([0,1])$ with the $L^1$-norm topology is meager in itself" is a stronger statement than "$C([0,1])$ is meager in $L^1$ with its norm topology". I'm not sure which one the OP intended, but as you've shown, both are true. $\endgroup$
    – Ian
    Jul 28, 2015 at 18:36
  • $\begingroup$ @Ian: In some sense they are actually equivalent. You can show that if $X$ is a topological space and $A$ is a dense subset of $X$ then $A$ is meager in itself (in the subspace topology) iff it is meager in $X$. $\endgroup$ Jul 28, 2015 at 21:01

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