$S$ is convex and compact (proof) Assume the set:
$$
S = \left\{ (x,y,z)^T \in \mathbb{R}^3 : y = z, \, x^2 +2y^2 \leq 1 \right\}
$$
How do we prove it's convex and compact?
My attempt:


*

*Convexity:
Let $a \in [0,1]$ and assume $ax + (1-a)w$. For $S$ to be convex we must have:
$$
ax + (1-a)w \in S
$$
but whenever I try to substitute this point into the equations of the set, I can't yield a direct result.


*

*Compactness:
Obviously, $S$ is the intersection of an elliptic cylinder and a plane, so it's bounded and closed and thus compact.
Question: Is there a similar elaboration based on the geometry of $S$ for proving convexity?
 A: Tony's hint in the comments is apt: the intersection of convex sets is convex:

If $C=\cap_{i \in I} C_i$, all $C_i$ convex, then if $x,y \in C$, for each $t \in [0,1]$ and each $i$, $x,y \in C_i$ so $tx + (1-t)y \in C_i$ and as $i$ was arbitrary, $tx+(1-t)y \in C$.

And the plane $P=\{(x,y,z) \in \Bbb R^3\mid y=z\}$ is clearly convex, even a linear subspace.
And the elliptic cyclinder $E=\{(x,y,z) \in \Bbb R^3\mid x^2+2y^2 \le 1\}$ is also convex; an algebraic proof is not too hard. 
So $P \cap E$, your set of interest, is indeed convex. 
Closedness is the same argument: a plane is closed in $\Bbb R^3$ and so is $E$, as $f^{-1}[\{0\}]$ for $f(x,y,z)= y-z$ or $f^{-1}[(-\infty,1]]$ for $f(x,y,z)=x^2+2y^2$ resp., both of which are continuous.
$P$ and $E$ are both unbounded, but their intersection is bounded: if $(x,y,z) \in P \cap E$ then $x^2+y^2+z^2 = x^2 + 2y^2\le 1$ (as $y=z$ from being in $P$ and the final one from being in $E$.) So $\|p\|\le 1$ for all points $p \in P \cap E$. Then use Heine-Borel in $\Bbb R^3$ to conclude it is compact.
