A proper subspace of a normed vector space has empty interior clarification

Question

So every proper subspace of a normed vector space has empty interior. I'm not asking for the proof, my problem is that this seems to me very strange. So if I have a normed vector space, in any proper subspace I can't take any ball inside the subspace?

For example suppose we work on a set with finite measure, $[a,b]$ for example. Let's take $L^{P}$ spaces over $[a,b]$. We know that now $L^{\infty}$ is included in $L^{1}$. So $L^{\infty}$ is a proper subspace of $L^{1}$. Now this means that $L^{\infty}$ is nowhere dense?

Answer 1

What this means is that $L^\infty$, as a subset of $L^1$, has empty interior (since $L^\infty$ is a proper subspace of $L^1$), as you have stated.

But $L^\infty$ is not nowhere dense in $L^1$. In fact, $L^\infty$ is dense in $L^1$. This is because any $L^1$ function can be approximated by simple functions, which are in $L^\infty$. Thus the closure of $L^\infty$ is $L^1$, which certianly has nonempty interior in $L^1$.

Answer 2

In the infinite-dimensional case, subspaces are closed by definition. $L^\infty$ is not closed in $L^1$ (its closure is all of $L^1$), so it is not an example of proper subspace.

$L^\infty$ is dense in $L^1$ (even compactly supported functions are). Of course ir is not nowhere dense, its closure is $L^1$!

And yes, as a subset, $L^\infty$ has empty interior.