I dont know how to solve this: "Show that $A$ is compact, where $A$ is the set of $x \in\mathbb R^n$ which $\|x\|=1,$ with $\|\cdot\|$ is the maximum metric in $\mathbb R^n.$"
My try: I know that in $\mathbb R^n$: a set is compact iff is bounded and closed. Well, I know that a set is closed if it is equals to its closure and i also know that the closure of A is equal to (boundary of $A$) union with $A.$ I tried to see what is the boundary of $A$ by the definition: $x$ is in boundary of $A$ iff for all open balls, $B,$ with center in $x$ and radius $> 0$ we have: ($B \cap A$) is not empty and ($B \cap A$'s complementar) is not empty...but i dont know how to get from here... :(
Can anyone hep me, please?
(Sorry for i dont know how to write with codes)
If you believe that $||\cdot||_{max}:\Bbb{R}^n \to \Bbb{R}$ is continuous, then $A=||\cdot||_{max}^{-1}(\{1\})$ is closed because $\{1\}$ is closed in $\Bbb{R}$. Now you only have to show that is is bounded with respect to the usual metric (which isn't difficult).