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!


1 Answer 1


The definition you provide for relative compactness is the definition of compactness. We say a set is relatively compact if it's closure is compact. In terms of sequences, we have the following definition:

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.

The other parts of your answer are correct.

  • $\begingroup$ 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. $\endgroup$
    – analysis1
    May 7, 2021 at 13:53
  • $\begingroup$ Oh yes. It has a convergent subsequence in $X$, but not in $S$. I thought that's what you were saying. $\endgroup$ May 7, 2021 at 13:55
  • 1
    $\begingroup$ Alright, great, thank you very much for the help and your proof! $\endgroup$
    – analysis1
    May 7, 2021 at 13:57

You must log in to answer this question.

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