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

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?

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.

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.

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    FYI: this can be done directly using definition of convexity: $(x,y,t) \in (S \cap T)^2 \cap [0, 1] \implies \left((x,y,t) \in S^2 \cap [0, 1]\; \land (x,y,t) \in T^2 \cap [0, 1]\right) \implies \left(tx + (1-t)y \in S\; \land tx + (1-t)y \in T\right) \implies tx + (1-t)y \in S\cap T.$ – 2017-02-03

2 Answers 2

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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 $S\cap T$.

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    You're absolutely correct. For some reason, my brain was malfunctioning and misinterpreting what it means to have the intersection of a set. For my calculations to be true, we require $A$ AND $B$ to be elements of both $S$ and $T$. Thank you for the assistance. :) – 2017-02-02
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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 $