Show that a normed space is complete if and only if the closed unit ball is complete.
I know theorem about this: If a normed space X has the property that the unit closed ball is compact, then X is finite dimensional space. Is there any relation between the theorems?
Thank you for your helping .
[Math] normed space is complete if and only if the closed unit ball is complete
functional-analysisreal-analysis
Best Answer
A closed subset of a complete space is complete. So if $X$ is complete, its unit ball is too.
On the other hand, if the unit ball is complete and $\{x_k\}$ is a Cauchy sequence in $X$, there must exist a constant $M$ such that $\|x_k\| \le M$ for all $k$: Cauchy sequences are bounded. Let $y_k = \frac{x_k}{M}$ so that $\{y_k\}$ is Cauchy and $\|y_k\| \le 1$ for all $k$. Thus $y_k \to y$ in the unit ball, so that $x_k \to My$.