Prove that set $\{(x,y,z)\in\mathbb{R}^3: x^2+y^2 \leq z, x+y+z=1\}$ is compact Prove that set $S=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2 \leq z, x+y+z=1\}$ is compact!
I'm not really sure how to go about this. I can prove the set is closed since it's an intersection of preimages (of closed sets) of continuous functions.
I see that if $z$ gets too big the inequality can't stand considering that $x+y+z=1$ but I just can't see a formal way to prove that. I'm pretty sure that if $z>2$ it's pretty much impossible for the inequality to hold.
I've noticed that $z\geq 0$ but other than that I just can't find a way to prove that the set is bounded. Thanks a lot!
Edit: I've fixed the inequality.
Also, the problem asks: for $f:R^3 \rightarrow R^2, f(x,y,z) = (x,y)$ what is $f(S)$ (also not really sure ohw to solve,considering I can't even prove it's compact.

Note that $$
(1-z)^2=(x+y)^2\leqslant 2(x^2+y^2)\leqslant 2z,$$we know $z$ is bounded. Therefore, it follows from the condition $x^2+y^2\leqslant z$ that $x^2+y^2+z^2$ is also bounded.
