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

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?

• No, it doesn't mean that at all. Saying $L^\infty$ is nowhere dense says that the closure $\overline{L^\infty}$ has empty interior, which is much stronger than saying $L^\infty$ has empty interior. – David C. Ullrich Jun 26 at 15:26
• Proper subspaces are like lines and planes through the origin in $\mathbb{R}^3$. They don't contain any open balls of $\mathbb{R}^3$. – mechanodroid Jun 26 at 15:30
• @mechanodroid And hence proper subspaces have empty interior. The OP already knows that - what does this have to do with the question of whether $L^\infty$ is nowhere dense? – David C. Ullrich Jun 26 at 15:33

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 $$L^1$$.

• thanks for the answer. So inside $L^{\infty}$ I can't take any ball? – banach-alaoglu-zielony Jun 26 at 15:42
• What you can do (and it may be useful to do) is take $B\cap L^\infty$ for any ball $B$ in $L^1$. But this will never be all of $B$. – Aweygan Jun 26 at 15:44

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.

• Subspaces are only closed by definition if you define them to be closed, which is certainly not standard. – Aweygan Jun 26 at 15:39
• For the theorem stated about subspace not having interior, the term subspace has to to be defined as closed subspace. I think it is pretty common to assume closedness (for ex. In the well known textbook by Fomin and Kolmogorov) – GReyes Jun 26 at 15:46
• @GReyes It seems however a lot of times we would want to discuss dense subspaces of say $L^p$, as you said above, but this would be meaningless terminology if we assumed subspaces were closed by default. – rubikscube09 Jun 26 at 15:51
• @GReyes: No, it doesn't have to be. I don't even see how it would help. (The converse is true, though: you can certainly show that having nonempty interior implies being closed (in fact, clopen) --- this is true for any subgroup of a topological group. Of course this is vacuous for vector subspaces.) – tomasz Jun 26 at 15:53
• @GReyes Any proper subspace (closed or otherwise) has empty interior. Nowhere in the proof is closedness needed. – Aweygan Jun 26 at 16:03