# Are countable intersections of convex sets convex?

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$. (actually it works for arbitrary intersections. – Davide Giraudo Jul 14 '12 at 11:40 They don't even have to be countable. I'm tempted to say even a proper class, as opposed to a set, of convex sets would have a convex intersection. But I don't know where you'd find any such class that's not a set. – Michael Hardy Jul 14 '12 at 17:04 ## 2 Answers Suppose$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 - 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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