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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.

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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.

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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.

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@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

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