Is the set $\{(x, y, a)\in\mathbb{R}^3\mid x^2+y^2\le 4 \}$, where $a\in\mathbb{R}$ is a fixed parameter, a convex set?
I know that the set $\{(x, y)\in\mathbb{R}^2\mid x^2+y^2\le 4 \}$ is convex (in $\mathbb{R}^2$), but I don't know if it holds for $\mathbb{R}^3$, when the $z$ coordinate is fixed, and if it's true, why.
Of course it remains convex if we add irrelevant coordinates:
Suppose a convex set of $\Bbb R^2$ is defined by $C=\{(x,y)\mid \phi(x,y)\}$, where we have some condition $\phi$ (whose exact form is irrelevant).
Then defining $C' = \{(x,y,z)\mid \phi(x,y)\}$ is also convex:
let $\vec{x}=(x,y,z),\vec{u}=(u,v,w) \in C'$ and $t \in [0,1]$. So $(x,y) \in C$ and $(u,v) \in C$ by definition so $(tx +(1-t)u, ty+(1-t)v) \in C$ as $C$ is convex.
So it follows that $(tx +(1-t)u, ty+(1-t)v, p) \in C'$ whatever $p$ is, so in particular, $t\vec{u} + (1-t)\vec{v} = (tx +(1-t)u, ty+(1-t)v, tz + (1-t)w) \in C'$ as well, and $C'$ is convex.
Or alternatively, appeal to a theorem that a product of convex sets is convex (we multiply $C$ by $\Bbb R$ here).
If the third coordinate is fixed, do the same, $\{a\}$ is also convex, and the first proof can be easily adapted too.
For each value of $a \in \mathbb R$ you get a disc in the plane $\pi_a = \{(x,y,a): x,y \in \mathbb R\}$, and each disc is convex.
