I need help with the following exercise of an introductory course in functional analysis.
Let $\mathbb{K}=\mathbb{R}\text{ or }\mathbb{C}$. Consider $l^1_N=\left\{x\in\mathbb{K}^N\colon||x||_1=\sum_{n=1}^N|x_n|< \infty\right\}$. We need to show that the closed unit ball ($B_1=\left\{x\in l_N^1\colon ||x||_1\leq 1\right\}$) is compact.
(I know that if a space is finite dimensional then it unit closed ball is compact. But here we are asked for a direct proof in this space without using this general result).
As a suggestion, it's enough to prove tha $B_1$ is sequentially compact (i.e. every sequence in $B_1$ has a convergent subsequence). I don't know how to construct such a subsequence exactly.
Thanks in advance.
Best Answer
Since we are working with $\mathbb R$ or $\mathbb C$ and those spaces are complete, it is sufficient to show that $B_1$ is closed and bounded.
For every $x\in B_1$, $\lVert x\rVert\le 1$ so $B_1$ is bounded.
To show that $B_1$ is closed we will show that its complement is open.
Let $p=(p_1,\dots ,p_N)$ be a point in $l_{N}^{1}$ that is not in $B_1$. Then $\lVert p\rVert\gt 1$ so $\lVert p\rVert=1+d$ for some $d\gt 0$.
Let $B_{\text{d/2}}(p)$ be the open ball of radius $\frac{d}{2}$ around $p$ and let $q\in B_{\text{d/2}}(p)$.
Then $$\lVert p\rVert=\lVert p-q+q\rVert\le\lVert p-q\rVert+\lVert q\rVert\implies\lVert q\rVert\ge\lVert p\rVert-\lVert p-q\rVert=1+d-\lVert p-q\rVert$$
Therefore, $$\lVert q\rVert\ge 1 + d - \lVert p-q\rVert$$
We chose $q$ so that $\lVert p-q\rVert\lt\frac{d}{2}$ so $$-\lVert p-q\rVert\gt -\frac{d}{2}\implies 1+d-\lVert p-q\rVert\gt 1+d-\frac{d}{2}=1+\frac{d}{2}\gt 1$$
Therefore, $\lVert q\rVert\gt 1$ so $q$ is in the complement of $B_1$ and since that is true for every $q$ in the open ball of radius $\frac{d}{2}$, the open ball around $p$ is contained in the complement of $B_1$. Since that is true for all $p\not\in B_1$, the complement of $B_1$ is open and $B_1$ is closed.