Prove that, If $S$ and $T$ are convex sets, $S \cap T$ is a convex set.

Question

Prove that, If $S$ and $T$ are convex sets, $S \cap T$ is a convex set.

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As I understand it, a convex set is a set such that, for any two points, $A$ and $B$, from the set, the line, $AB$, lies within the set.

The proposition is that, If $S$ and $T$ are convex sets, $S \cap T$ is a convex set; however, this does not seem correct to me. I do not see how the intersection of two sets, although independently convex, results in another single set that is necessarily convex?

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My Workings

A (hypothesis): $S$ and $T$ are convex sets.

B (conclusion): $S \cap T$ is a convex set.

B1: Given two points $A$ and $B$ from the set $S \cap T$, the line $AB$ lies within the set.

B2: For all points $A$ and $B$ from the set $S \cap T$, the line $AB$ lies within the set.

B2 rephrases B1 using the universal quantifier.

A1: Let $A \in S$ and $B \in T$ such that $A \in S \cap T$ and $B \in S \cap T$.

A2: ... ?

At this point, it is not clear to me that, for all points $A$ and $B$ from the set $S \cap T$, the line $AB$ lies within the set; therefore, I have struggled to continue formulating a proof.

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I would greatly appreciate it if someone could please take the time to explain what I am misunderstanding and how to correctly continue with my proof.

Answer 1

You are on the right track. Note that $A\in S\cap T$ and $B\in S\cap T$ implies that $A\in S$ and $B\in S$. Since $S$ is convex, line $AB$ lies within $S$. By similar argument line $AB$ lies within $T$. Hence line $AB$ lies within $S\cap T$.

Answer 2

Check that your statement A1 implies that both A lies in both S and T (and hence in the intersection). Likewise for B. This would imply that AB would lie in both S and T,right?

A more precise way to argue would be as follows:

Let A and B be such that A,B $\epsilon S \cap T$. This implies,in particular, both A and B lie in S.

S being a convex set,this implies AB lies in S.

Arguing similarly for T(instead of S),we can claim that AB lies in T Hence AB lies in $S \cap T $