# Normed space and finite dimensional space [closed]

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 $dim E<\infty$

## closed as off-topic by Daniel W. Farlow, C. Falcon, Namaste, Shailesh, JonMark PerryMay 14 '17 at 0:11

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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.
• 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