# Compactness of normed linear spaces

For a normed linear space $(X,||\cdot||)$ I want to show

a)For a continuous map $f : X \to \mathbb{R}$ the set $$f^{-1}(r):=\{x\in X: f(x)=r\} \subset X$$ is closed.

b) $$S(0,r)=\{x\in X: ||x||=r\}\subset X$$ is closed and bounded.

c) Provide an example that $S(0,r)$ is in general not compact.

I believe these questions lead on from eachother but I have a couple of things which confuse me.

My attempt at a):

Suppose we let $y$ be a limit point of $f^{-1}(r)$. So there is a sequence $\{y_n\}$ such that $y_n \in \{ x \in X : f(x)=r\}$ for all $n$ and $\lim_{n \to \infty} y_n = y$. Since $f$ is continuous we then have $f(y)= \lim_{n \to \infty}f(y_n)=\lim_{n \to \infty} r= r.$ Hence $y \in f^{-1}(r)$, so $f^{-1}$ contains all of its limit points and is closed.

I understand this is a proof and generally 'show that' questions have a more direct approach so perhaps this is the wrong approach. Any feedback on my attempt is much appreciated.

Also in regards to b) and c), as far as I understood a compact space is one which is closed and bounded. But then b) and c) would contradict eachother so is this not the case all the time?

• You have not understood what a compact space is. Review the definition carefully. – Giuseppe Negro Mar 26 '18 at 21:57

a) Your approach is correct, but it's simpler to say that, since $\{r\}$ is closed and $f$ is continuous, $f^{-1}(r)$ is closed.
b) It is closed by a) and it is bounded because each of its elements has norm $r$.
c) If $X$ is, say, the space of all bounded sequences of real numbers and if the norm of a sequence $(a_n)_{n\in\mathbb N}$ is $\sup_{n\in\mathbb N}|a_n|$, then the set $S(0,r)$ is closed and bounded, but not compact.
For a counterexample, consider the $l^{2}$ with the unit closed ball $S=\{x\in l^{2}: \|x\|_{2}=1\}$. Consider the sequence $(e_{n})$ with the $n$-th coordinate $1$ and $0$ otherwise. One finds no any convergent subsequence, so $S$ is not sequentially compact, which in turns that not being compact.