Normed space and finite dimensional space

Question

i have this problem

let $E$ be a normed vector space and $A\subset E$ compact

How to prove that if $\overset{\circ}{A}\neq\emptyset$ then $dim E<\infty$

score 3 · Answer 1 · edited May 11 '17 at 11:48

3

This implies that the interior of $A$ contains a closed ball which is compact therefore the unit ball is compact, so the space is finite dimensional.

edited May 11 '17 at 11:48

JJR

answered May 11 '17 at 11:30

Tsemo Aristide

i don't understand how to prove, i think that i must start by: $\overset{\circ}{A}\neq\emptyset$ then there exists $x\in \overset{\circ}{A}$ it means there exist $r>0$ such that $x\in B(x,r)\subset A$ how to continue after that ? – Vrouvrou May 11 '17 at 17:35
Suppose that the adherence $\bar B(x,r)$ of $B(x,r)$ is compact, the map $h(u)=(u-x)/r$ is continue, this implies that $h(\bar B(x,r))=\bar B(0,1)$ is compact since the image of compact set by a continuous map is compact. – Tsemo Aristide May 11 '17 at 17:46
Why $\overline{B}(x,r)$ is compact ?is $\overline{B}(x,r)=B'(x,r)$ ???? – Vrouvrou May 11 '17 at 17:49
because it is a closed subset of the compact set $A$. – Tsemo Aristide May 11 '17 at 17:50
but if $B\subset A$ and $B\subset \overline{B}$ this not imply that $\overline{B}\subset A$ ? but how we conclude ????? – Vrouvrou May 11 '17 at 17:53
If $B\subset A$ and $A$ is closed $\bar B\subset \bar A=A$ and here $A$ is closed since a compact set is closed. – Tsemo Aristide May 11 '17 at 17:55
ahh ok, so you define $h: \overline{B}(0,r)\rightarrow B'(0,1)$ and then $B'(0,1)$ is compact so $E$ is a finite dimensional space ? – Vrouvrou May 11 '17 at 17:57
ok,......... this is it – Tsemo Aristide May 11 '17 at 18:01
then we have equivalence between finite dimentional space and the unite closed ball is compact ??? – Vrouvrou May 11 '17 at 18:02

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