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Prove that $S = \{(x,y,z,w)\in \mathbb{R} : \lvert\lvert (x,y)\rvert\rvert_1 + \lvert\lvert (z,w)\rvert\rvert_{\infty} \leq 1\}$ is a compact set!

Where :

$\lvert\lvert (x,y)\rvert\rvert_1 = \lvert x \lvert + \lvert y\rvert$

$\lvert\lvert (z,w)\rvert\rvert_{\infty} \} = \max\{{\lvert z \rvert, \lvert w \rvert}\}$

My attempt:

Since $\lvert x \lvert + \lvert y\rvert + \max\{{\lvert z \rvert, \lvert w \rvert}\} \leq 1$ it follows that each element individually has to be $\leq 1$ since each of the terms in the inequality is $\geq 0$. So $x,y,z,w\leq 1$.

That means S $\subset K(0,2)$ (open ball around $0$,radius of $2$ )

Now I'm not sure how to go about proving that $S$ is closed. I don't think I can prove it directly from a definition of a closed set since I'm not even remotely sure what this is supposed to look like. I could also try to prove it's closed by proving that the limit every convergent series in $S$ is an element of $S$, but I'm not exactly sure what to do.

If I define a convergent sequence $a_n = (x_n,y_n,z_n,w_n)$ with the limit $a=(x,y,z,w)$:

If $a_n$ converges so do the sequences $x_n,y_n,z_n,w_n$, and since $a_n \in S$, obviously $x_n,y_n,z_n,w_n \leq 1, \forall n \in \mathbb{N}$ so their limits are also $\leq 1$. But I'm not sure how to limit them further since I don't know what to do with those absolute values. (I would normally use the fact that the limit of their sum is the sum of their limits)

All hints appreciated!

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  • If you put $f (x,y,z,w)=\Vert (x,y)\Vert_1+\Vert (z,w) \Vert_\infty $, then $S $ is a preimage of (closed) set $[0,1] $ and $f $ is continuous, so $S $ must also be closed. –  Nov 12 '17 at 11:07

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The set $S$ is closed, because $S=f^{-1}\bigl((-\infty,1]\bigr)$, where$$f(x,y,z,w)=\bigl\|(x,y)\bigr\|_1+\bigl\|(z,w)\bigr\|_\infty.$$ Besides, $f$ is continuous and $(-\infty,1]$ is closed.

José Carlos Santos
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