Proof verification of the fact that intersection of two convex regions is convex

Question

Here is the question:

Assume that the intersection of two convex regions of a plane is a nonempty set. Prove that it is also a convex region.

The definition of a convex region is given as:

A region $R$ in the plane is said to be convex if, for all points $A, B ∈$ $R$, we have $AB ⊂ R$. Otherwise, R is said to be a non-convex region.

Proof:
Let us assume that the intersection of two regions $R$ and $S$ is non-convex. Then, there must be points $A$, $B$ $∈$ $R\cap S$ such that (wlog) $AB$ is not a subset of $R$. Thus, we must have a point $X\in AB$ for which $X\notin R\cap S$. Now, there are three possible cases:
$(i)$ $X\in R-S$
$(ii)$ $X\in S-R$
$(iii)$ $X\in (R\cup S)'$
For case $(i)$, $X\notin S$ and thus $S$ is non-convex by the given definition, resulting in a contradiction. Same goes for (ii), (iii). Thus, the region of intersection is convex.

Answer 1

If $a,b\in R\cap S$, then $a\in R$ and $b\in R$, therefore $\underline{ab}\in R$.

If $a,b\in R\cap S$, then $a\in S$ and $b\in S$, therefore $\underline{ab}\in S$.

$\underline{ab}\in R$ and $\underline{ab}\in S$ imply $\underline{ab}\in R\cap S$.

Answer 2

With $AB$ you mean $\overline{AB}$ hence the line segment from $A$ to $B$.

A definition of "convex set" reads as follows:

A set $A\subseteq\mathbb{R}^n$ is convex when for every $x,y\in A$ we have that the line segment from $x$ to $y$ lies in $A$.

With other words: $A\subseteq\mathbb{R}^n$ is convex when for every two $x,y\in A$ we have that $tx+(1-t)y\in A$ for every $t\in [0,1]$.

With this definition a direct proof is very simple.

Let $A, B\subseteq\mathbb{R}^n$ be convex. Let $x,y\in A\cap B$.

We have to show that $tx+(1-t)y\in A\cap B$ for every $t\in [0,1]$.

This is immediate as $A$ and $B$ are convex. (I leave the easy verification of the details to you.)