Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am wondering how to prove that the intersection of an infinite number of convex sets is convex.

I can prove that the intersection of two convex sets is convex, and I believe that I can simply do an induction on this result, but I've heard that it would be wrong to do this since I am working with infinity.

I guess another way to think of this question is whether or not I have to take something special into consideration since the word "infinite" is involved.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Let $S_{\alpha}$, $\alpha \in \Gamma$ be an infinite collection of convex sets.

Let $S = \displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$ be the intersection of these sets.

Consider $x,y \in S$.

$x,y \in S \Rightarrow x,y \in S_{\alpha}$, $\forall \alpha \in \Gamma$.

Since $S_{\alpha}$ is convex $\forall \alpha \in \Gamma$, any convex combination of $x,y$ i.e. $\lambda x + (1 - \lambda) y \in S_{\alpha}$, $\forall \alpha \in \Gamma$, $\forall \lambda \in [0,1]$

Hence, every convex combination is in $S_{\alpha}$, $\forall \alpha \in \Gamma$ and hence it is also in $\displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$

Hence, $S = \displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$ is convex.

share|improve this answer
    
Thanks for this. I knew there was something that I did not account for. What exactly does the $\Gamma$ represent? –  Elements Mar 11 '11 at 4:34
    
$\Gamma$ is an Indexing set. If you are dealing with countable intersections, then you can take $\Gamma = \mathbb{N}$ –  user17762 Mar 11 '11 at 4:37

Suppose $x$, $y$ is in $F=\bigcap^{\infty}_{i=1}F_{i}$. We argue that the segment $xy$ is in $F$ because it is in all $F_{i}$. Hence by definition $F$ is convex. There is no problem with $\infty$ in here.

share|improve this answer
    
@user7887: The intersection needn't be a countable intersection. –  user17762 Mar 11 '11 at 4:28
    
oh, I see the subtlety. –  Kerry Mar 11 '11 at 4:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.