The set $A=\{ x \mid \|x\|=1 \}\subset \mathbb{R}^n$ is compact in $\mathbb{R}^n$ under Euclidean norm.
If I can show this is closed and bounded then we are done. It seems bounded but I am not sure how to show it is closed.
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$\begingroup$ Suppose that you have a convergent (in $\Bbb R^n$) sequence of points on the unit sphere. CAn you say anything on the norm of the limit? $\endgroup$– TZakrevskiyNov 29, 2015 at 15:45
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$\begingroup$ Are you asking about the compactness of every subset of A or about the compactness of A itself? $\endgroup$– DidNov 29, 2015 at 16:24
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3 Answers
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Just note that every norm is continuous (because 1-lipschitz from the second triangle inequality $|\,||x||-||y||\,|\leq||x-y||$) and then $A=||\cdot||^{-1}(\{1\})$ is a closed set as a reciprocal image of the closed set $\{1\}$ of $\mathbb{R}$ by the continuous map $||\cdot||.$
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Edit: The following answers a former version of the question, referring to a subset of $S^{n-1}$.
The statement in the title, appearing again in the body of the question, is wrong. A subset $$A\subset S^{n-1}\subset {\mathbb R}^n$$ is compact iff it is closed. Consider, e.g., the set $$A:=\bigl\{x\in{\mathbb R}^n\>\bigm|\>\|x\|=1, \ x_n>0\bigr\}\ .$$ The point $(1,0, \ldots,0)$ is an accumulation point of $A$, but does not belong to $A$.
answered Nov 29, 2015 at 16:17
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$\begingroup$ @CB note that the question was edited after this answer and before your comment $\endgroup$ Nov 30, 2015 at 3:17
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I assume you are not familiar with the topological concept of continuity (that's the answer everyone thinks about). Now, obviously, $A$ is bounded so it suffices to prove that it is in fact closed, that is, its complement is open.
Let $p\in\mathbb R^n\setminus A.$ We shall prove that $p$ is an interior point of $\mathbb R^n\setminus A.$
As $p\notin A$ then $\|p\|\neq1.$ Define $\varepsilon:=|\|p\|-1|.$ Then the open ball $B_{\varepsilon}(p)$ is contained in $\mathbb R^n\setminus A$ (here I used the triangle inequality) Therefore $p$ is an interior point of the complement of $A$ which means that $\mathbb R^n\setminus A$ is open and hence $A$ is closed. Thus $A$ is compact.
