Proof by contradiction: the intersection of any collection of non-empty convex sets is convex Looking for feedback on how the below can possibly be simplified, or if it's even correct!
Let $S_{i}$ be a convex set for $i = 1, 2, ..., n$.
Suppose $S = \cap_{i=1}^{n} S_{i}$ and $\exists x_{1}, x_{2} \in S$ such that $\alpha x_{1} + (1-\alpha) x_{2} \notin S$, with $\alpha \in [0, 1]$
Then $\forall S_{i}$, $(x_{1}, x_{2} \in S_{i}) \land (\alpha x_{1} + (1-\alpha) x_{2} \notin S_{i}$).
This is a contradiction as by definition of a convex set:
$\exists x_{1}, x_{2} \in S_{i} \Rightarrow \alpha x_{1} + (1-\alpha) x_{2} \in S_{i}$, with $\alpha \in [0, 1]$.
Edit:
Responding to Tomasz' comment, here's an additional attempt at a proof:
Let $\{S_{d}\}_{d \in D}$ be some arbitrary collection of non-empty convex sets with index $d$ and $S = \bigcap_{d \in D}S_{d}$.
$\exists x_{1}, x_{2} \in S \Rightarrow x_{1}, x_{2} \in S_{d}, \forall d \in D$
Then by definition of a convex set, all points of the form $\alpha x_{1} + (1 - \alpha) x_{2}$ with $\alpha \in [0,1]$ are in $S_{d}$ $\forall d \in D$ and $\alpha x_{1} + (1 - \alpha) x_{2} \in S$.
 A: The proof is basically correct, except it is not really a proof by contradiction.
It would be more clear if you restated it as a direct proof (that it essentially is).
Further, there is an issue in that in the statement you write "any collection", while in your proof you act as if the collection was finite. This does not really affect the reasoning in any meaningful way, but you should be careful what you write.
Edit regarding the updated proof: it is essentially correct, but there are serious issues with the notation:

*

*The quantifier $\exists x_1,x_2$ does not make sense. You are checking that the following holds for each pair $x_1,x_2$. not assuming that a pair exists.

*Writing the quantifier $\forall d\in D$ after the variable it's supposed to bind is bad form.

*Overall, using too many symbols in mathematical prose is bad form and makes things hard to read (it's different when you're writing on blackboard and saying out loud what you mean, but when you want to communicate math via text alone, this is just not a good idea).

*It does not make sense to talk of "sets with index $d$". $d$ is just a symbol for an element of $D$. What would make sense would be to say that it is a family indexed by $D$ (capital). In fact, you can dispense with the index set altogether, and just write that you have a family $\mathcal S$ of convex sets and $S=\bigcap \mathcal S$, but that is largely a matter of taste.

A better way to write this is as follows.
Let $\mathcal S$ be some arbitrary collection of non-empty convex sets and $S = \bigcap \mathcal S$.
We will show that $S$ is convex. It suffices to show that for each $x_1,x_2\in S$ and $\alpha\in [0,1]$, the combination $\alpha x_1+(1-\alpha)x_2\in S$.
Fix an arbitrary $S'\in \mathcal S$. Then $S\subseteq S'$, so $x_1,x_2\in S'$, and since $S'$ is convex, $\alpha x_1+(1-\alpha)x_2\in S'$. But since $S'\in \mathcal S$ was arbitrary, we have that $$\alpha x_1+(1-\alpha)x_2\in \bigcap \mathcal S=S,$$ which completes the proof. $\square$
(By the way, there is no reason to assume that the collection consists of nonempty sets. Empty set is also convex. Otherwise, if you don't consider the empty set to be convex, then the proposition is false: the intersection of non-empty convex sets can be empty.)
A: A proof by contradiction should go as follows.
Assume that the intersection is not convex then there exist
$x,y \in\,\bigcap\,K_{i}$ such that for some $\lambda$ in
$[0,1]\,\,\, \lambda\,x+(1-\lambda)y \notin\,\bigcap\,K_{i}$
But $x,y \in K_{i}$ for all $i$ and hence $\lambda\,x+(1-\lambda)y\,\in \,K_{i}$
for all $i$ and hence $\lambda\,x+(1-\lambda)y\,\in \,\bigcap\,K_{i}$, contradiction!!
