# Showing relatively compact and finite dimension

I have a question where I have to show:

1. Prove that a normed linear space is finite-dimensional if and only if every bounded subset is relatively compact.
2. Show that the spaces $$C[0,1]$$ and $$L^{2}(0,2 \pi)$$ are infinite dimensional.

Now for the first part I'm not entirely sure if I understand the concept of relative compactness well. To show a set is relatively compact is it enough to take a sequence in it and show that it has a convergent subsequence? Would that imply that the closure of the set is compact (which is what I understand is the definition of relative compactness or precompactness)? If this is true here is my proof.

1. Take any bounded subset B of a finite dimensional space X. Let $$x^{(n)}$$ be a sequence in B. Then due to equivalence of norms on finite dimensional spaces we have an $$M$$ and $$K$$ such that:

$$K \geq \sup_n ||x^{(n)}|| \geq \sup_n M \max_{k=1,2..N} \{|\alpha_k^{(n)}|\} > |\alpha_k^{(m)}|\quad \forall k$$

Where $$x^{(n)} = \sum_{i\leq N} e_i \alpha^{(n)}_i$$ where $$\{e_i\}_{i\leq N}$$ is a normalized finite basis. Then $$|\alpha_k^{(m)}|$$ has a convergent subsequence by Bolzano Weierstrass and we can inductively find a subsequence of $$x^{(n)}$$ with convergent coordinates. Now since convergence in coordinates implies convergence for finite dim sequences we have found a convergent subsequence and I think we are done(?).

For the other direction we can see that the closed unit ball is compact so the space must be finite dimensional.

2. For C[0,1] I take a sequence in the unit ball $$f_n(t) = t^n$$, suppose it has a convergent subsequence in sup norm then the sequence must also converge pointwise, but then the sequence must converge to $$1_{\{1\}}$$ which is not in C[0,1] hence can't be in closure of the unit ball either.

For Lp we take again a sequence in the closed unit ball given by $$f_n(x) = 1_{[0, 1/n]} x n$$ again if we have a convergent subsequence then that convergent subsequence has a subsequence that converges pointwise (to the same limit), but that limit clearly must be 0, but $$||f_n(x)|| = \sqrt{1/2}$$ hence contradiction again, so closure of unit ball is not compact. Thank you in advance for the help!

Definition: $$A$$ is $$\textit{relatively compact}$$ if for each sequence in $$\overline{A}$$, there is a convergent subsequence in $$\overline{A}$$.
I think there is a much simpler proof for 1. Suppose $$V$$ is finite dimensional, then there exists $$T: V\rightarrow \mathbb{R}^n$$ an isometric isomorphism for some $$n$$. Let $$A\subset V$$ be bounded and note that $$T^{-1}\cdot T(\overline{A})=\overline{A}$$. Note that $$T(\overline{A})$$ is closed and bounded in $$\mathbb{R}^n$$, so it's compact. Therefore, $$T^{-1}\cdot T(\overline{A})$$ is compact.
• Hi! In my proof I was referecing the following theorem: Let $(X, d)$ be a metric space and $S \subset X$. The following are equivalent: i. $S$ is relatively compact. ii. For all sequences $\left(s_{n}\right)_{n \in \mathbb{N}}$ in $S$, there is a subsequence $\left(s_{n_{k}}\right)_{k \in \mathbb{N}}$ which converges in $X$. bearing that in mind would my proof be correct? I took a sequence in the bounded subset and showed it has a convergent subsequence. May 7, 2021 at 13:53
• Oh yes. It has a convergent subsequence in $X$, but not in $S$. I thought that's what you were saying. May 7, 2021 at 13:55