# Is $C[0,1]$ equipped with $\lVert \cdot \rVert_1$ a countable union of nowhere dense sets?

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?

• 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.) Jul 28, 2015 at 21:07

$$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$
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.
• 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.
• @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 $X$. Jul 28, 2015 at 21:01