Is $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$ compact? Is the following set compact? How can I show it? $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$
Clearly it is closed as it contains its boundary, but I can not show that it is bounded..

Substituting, we find that $x^2+y^2-x-y = (x-\frac{1}{2})^2 - \frac{1}{4} + (y-\frac{1}{2})^2 - \frac{1}{4} =  \frac{3}{4}$, ie $(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 = \frac{1}{4}$, so the projection of $C$ on the z-plane is a circle, which is bounded as $|x|\le K$, $|y|\le K$ for some constant $K$.
Now, going back to the first equation we have that $|z| = |1-x-y| \le |1| + |x| + |y| \le 1+2K$, so $z$ is bounded as well.
We conclude that $C$ is bounded, and since you already know how to show that it is closed, we conclude that it is compact.
