$\{z: |z|\leq R\}$ is compact ($z$ is a complex number, Proof verification)

Asked Aug 15 '20 at 02:40

Active Aug 15 '20 at 03:07

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Here's my proof: for each $z=x+iy=(x,y)$ ($x,y$ are real numbers), $|z|=|x^2+y^2|\leq R$ so, $\{z:|z|\leq R\}$ is equal to $\{(x,y):|x^2+y^2|\leq R\}$ then, by Heine-Borel Theorem in $\mathbb{R}^2$, $\{z:|z|\leq R\}$ is compact.

Is it correct? (i think it's not) if that proof is incorrect, then, how should i prove it?

sorry for my english

edited Aug 15 '20 at 03:07

Alejandro Tolcachier

1,938

asked Aug 15 '20 at 02:40

Choco Mint

I think you might need to be more explicit showing closure. – user361424 Aug 15 '20 at 03:15
It's not a proof, what you have shown,, it's just equivalent term. – A learner Aug 15 '20 at 03:25

$\{z: |z|\leq R\}$ is compact ($z$ is a complex number, Proof verification)

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