# Compact and nowhere dense subset of a Banach space

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. 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 $$f(x)=\|x\|$$.
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 $$\|tx-x\| for some $$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$$.