Let $(X,∥ · ∥)$ be a Banach space and consider the subset $C := {\{x ∈ X \ s.t. ∥x∥ = 1}\}$. Then:

  1. Is $ C $ a closed subset of $X$?
  2. Is $ C $ compact in $X$?
  3. Is $ C $ nowhere dense in $X$?

The first question seems trivial, norm is continuous w.r.t. strong convergence so if $x_n \to x$ then $∥x_n∥ \to ∥x∥$ and C is closed.

For what concerne the second, clearly $C$ is (closed and) bounded (because norm is finite) but this imply compactness only if $X$ is a FINITE dimensional space, and this is not specified. So i can rewrite the solution as " finite balls are compact iff $dim(X)< \infty$ ". There is a way to prove it without any assumption?

For the 3 question i have no idea, there is some theorem that can help?


1 Answer 1

  1. What you wrote is correct, but I think that the shortest way of proving that $C$ is closed consists in noting that $C=f^{-1}\bigl(\{1\}\bigr)$, with $f(x)=\|x\|$.
  2. Right.
  3. Yes, it is nowhere dense. Since $C$ is closed, asserting that it is nowhere dense is the same thing as asserting that $\mathring C=\emptyset$. If $x\in C$ and $r>0$, it is easy to see that $\|tx-x\|<r$ for some $t\in(0,1)$. But $\|tx\|=t<1$, and therefore $tx\notin C$. So, $B_r(x)\not\subset C$, and so $\mathring C=\emptyset$.
  • $\begingroup$ yes, your solution of the first question is more "elegant", but also the one that i provided seems correct right? The explanation of the 3 question is really perfect, thanks a lot! $\endgroup$
    – maru0032
    Sep 14, 2021 at 13:50
  • $\begingroup$ Yes, I forgot to write that your solution to the first problem was correct. It's done now. $\endgroup$ Sep 14, 2021 at 13:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .