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Prove that the set $A=\{x\in \Bbb R^n:\|x\|_1=1\}$ is compact under the Euclidean Norm.

Attempt:Under the Euclidean Norm a set is compact iff it is closed and bounded.

Let $x=(x_1,x_2,\ldots ,x_n)$. Then $\|x\|_1=1\implies |x_1|+|x_2|+\ldots +|x_n|=1$

The map $(*):x\mapsto (|x_1|+|x_2|+\ldots +|x_n|)$ is continuous and hence $A=(*)^{-1}(1)$ is closed.

Now $(\sum_{i=1}^n|x_i|)\le (\sum_{i=1}^nx_i^2)(\sqrt n)\implies (\sum_{i=1}^nx_i^2)\ge \dfrac{1}{\sqrt n}$.

How to show that the set is bounded from here.

Will you please help.

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    $||(1/\sqrt{2}, \ 1/\sqrt{2})|| = 1$, but $1/\sqrt{2} + 1/\sqrt{2} = \sqrt{2} \neq 1$. – 2017-01-08

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It's easy to see that $|x_i| \leq 1$ for any $1 \leq i \leq n$ which implies that $\sum_i |x_i|^2 \leq n$ which means that $\|x\|_2 \leq \sqrt{n} < \infty$. Hence, the set is bounded and compact.

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If you want another proof ,you can proceed with sequences .

Prove that $A=\{x \in \mathbb{R}^n,\|*\|_1=1\}$ is compact by taking a sequence in $A$ and prove that it has a convergent subsequence.

You can use the fact that $\|*\|_1$ is equivalent with the euclidean norm in a finite dimensional space.