Show that $C$ is a convex subset of $\mathbb R^4$

Question

So I'm given the following information: Let $f:\mathbb R^4\rightarrow \mathbb R$ be given by

$$f(x_1,x_2,x_3,x_4)=(x_1-x_3)^2+(x_2-x_4)^2$$

and $C\subseteq \mathbb R^4$ by

$$C=\{(x_1,x_2,x_3,x_4)\in \mathbb R^4 \mid x_1^2+(x_2-2)^2\leq 1,\ x_3-x_4\geq 0\}$$

I'm asked to show that $C$ is a convex subset of $\mathbb R^4$.
I have the following definition:

A convex subset $C\subseteq \mathbb R^d$ is a subset that contains the line segment between any two of its points $x, y \in C$, i.e., $$(1-t)x+ty\in C$$ for every number $t$ with $0\leq t\leq 1$.

I'm just not sure how to put these things together, anything to help me get started is greatly appreciated.

Answer 1

HINT.-$V=(x,y,z,w)\in C\iff x^2+y^2-2y\le-3$ and $z\le w$.

Let $V_1$ and $V_2$ be in $C$; we need to show that $V_3=(1-t)V_1+tV_2=((1-t)x_1+tx_2,(1-t)y_2+ty_2,(1-t)z_1+tz_2,(1-t)w_1+tw_2)\in C$ and this is verify if and only if $$\begin{cases}((1-t)x_1+tx_2)^2+((1-t)y_2+ty_2)^2-2(t(1-t)y_2+ty_2)\le -3\\(1-t)z_1+tz_2\le(1-t)w_1+tw_2 \end{cases}$$It is the only way to proceed. What remains is a little bit of algebraic manipulation using the fact that $0\le t\le1$ and $$ x_1^2+y_1^2-2y\le-3,\space\space z_1\le w_1\\x_2^2+y_2^2-2y\le-3,\space\space z_2\le w_2$$

Answer 2

Let $A = \{(x_1,x_2) \in \mathbb R^2 \mid x_1^2+(x_2-2)^2\leq 1\}$ and $B = \{(x_3,x_4)\in \mathbb R^2 \mid x_3-x_4\geq 0\}$. Then $C = A \times B$. We show that $A, B$ are convex; this implies that their Cartesian product in convex.

$A$ is a closed disk with center $(0,2)$ and radius $1$; this is convex.

$B$ is the closed half-plane lying above the diagonal line $x_4 = x_3$; this is also convex.

You can prove the convexity of $A,B$ also formally. See Piquito's answer.