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Let $X$ be a Banach space $\{C_n\colon n\in\mathbb N\}$ a collection ofconvex sets in $X$. Is the set $$C=\bigcap_{n\in\mathbb N}C_n$$ convex?

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Taking intersections is very well behaved. Most properties are preserved. Just appeal to the definition. If $x,y\in C$ then $x,y\in C_n$ for each $n\in\mathbb N$ and use the convexity of the $C_n$s – Host-website-on-iPage Jul 14 '12 at 11:40
Yes: pick two points in the intersection $x$ and $y$. For each $n$ and $0\leq a\leq 1, $ax+(1-a)y\in C_n$. (actually it works for arbitrary intersections. – Davide Giraudo Jul 14 '12 at 11:40
They don't even have to be countable. I'm tempted to say even a proper class, as opposed to a set, of convex sets would have a convex intersection. But I don't know where you'd find any such class that's not a set. – Michael Hardy Jul 14 '12 at 17:04

Suppose $x,y\in C$ then $x,y\in C_n$ for all $n\in\mathbb{N}$. Now as $C_n$ is convex hence for all $0\leq\lambda\leq 1$ we have $\lambda x+(1-\lambda)y\in C_n$ for each $n$. So $\lambda x+(1-\lambda)y\in C$. So $C$ is convex

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In fact, any intersection of convex sets in this setting is convex. Suppose that $\mathcal{E}$ is a Banach space and that $\mathcal{K}$ is any family of convex sets. If $\bigcap\mathcal{K}$ is void, it is vacuously convex. Otherwise, take $x,y\in\bigcap\mathcal{K}$, $\lambda\in[0,1]$ and $K\in \mathcal{K}$. Since $K$ is convex, we have $$(1-\lambda)x + \lambda y\in K.$$ But this holds for any $K\in \mathcal{K}$, so $$(1-\lambda)x + \lambda y\in \bigcap \mathcal{K}.$$

There is no induction or sequential property here. In fact, the argument generalizes to any real vector space.

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