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.
 A: Intuitive proof: (see figure below) It is the intersection of the interior of a paraboloid with a slant (not vertical) plane. It is thus the closed interior of an ellipse (E) (belonging to this plane), which is bounded. Closed and bounded: we have a compact set.
Out of this one can elaborate a rigorous proof using the explicit equation of the projection of ellipse (E) :
Setting : $x=r \cos(\theta), y=r \sin(\theta)$, one has $z \ge r^2$, thus plugging this expression into the plane equation, one has:
$$\tag{1}r \cos(\theta)+r \sin(\theta)+r^2-1 \geq 0$$
Let us show that (1) describes the interior of bounded curve with equation :
$$\tag{2}r^2+r(\cos(\theta)+\sin(\theta))-1 = 0$$
Quadratic equation (2) in $r$ has the following positive solution (we are looking for a positive $r$):
$$r=\dfrac{-(\cos(\theta)+\sin(\theta))+ \sqrt{5+\sin(2 \theta)}}{2}$$
This is the polar equation of the projection of (E) onto plane xOy ; it is clearly a bounded curve ($r < 3$). As the corresponding $z$ is also bounded, we have the result. 

A: 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.
A: Sorry, I have duplicated the previous answer... I erase it.
