unit sphere with respect to the max. norm is compact

Question

Given $||x||_\infty=\max_i x_i$, where $x\in \mathbb{C}^n$, I want to show $\{x|\ ||x||_\infty=1\}$ is compact.

My try:

1) This set is closed, because it contains its boundary point $[1,...,1]$.

2) This set is bounded, because $|x|\leq \sqrt{n}$.

Is my proof acceptable?

Answer 1

Given $x^{N}=(x_{N,1},...,x_{N,n})$ with $\|x^{N}\|_{\infty}=1$ and $x^{N}\rightarrow x=(x_{1},...,x_{n})$.

For $\epsilon>0$, choose an $n_{0}$ such that $\|x^{n_{0}}-x\|<\epsilon$, then $\|x\|\geq\|x^{n_{0}}\|-\|x^{n_{0}}-x\|=1-\epsilon$, and $\|x\|\leq\|x-x^{n_{0}}\|+\|x^{n_{0}}\|<\epsilon+1$, so $\|x\|=1$, this shows that the set is closed.

For boundedness, each $x$ with $\|x\|=1$ must satisfy $|x_{i}|\leq 1$ for each $i=1,...,n$, so $\|x\|_{2}^{2}=\displaystyle\sum_{i=1}^{n}|x_{i}|\leq n$, so $\|x\|_{2}\leq\sqrt{n}$, as what you have written.

Answer 2

Basically the idea is correct.

BUT: You need to show that the set contains ALL its boundary points, not just one!

You could also use the fact that the norm is continuous and that the preimage of the closed the $\{1\}$ so has to be closed too.