Understanding Helly's theorem

Question

The following is part of the inductive step in Wiki's proof for Helly's theorem. Why is it true?

In this new collection, every subcollection of d + 1 sets will have nonempty intersection

Answer 1

The whole inductive step in the proof is

Suppose $n > d + 2$ and that the statement is true for $n−1$. The argument above shows that any subcollection of $d + 2$ sets will have nonempty intersection. We may then consider the collection where we replace the two sets $X_{n−1}$ and $X_n$ with the single set $X_{n−1} \cap X_n$. In this new collection, every subcollection of $d + 1$ sets will have nonempty intersection. The inductive hypothesis therefore applies, and shows that this new collection has nonempty intersection. This implies the same for the original collection, and completes the proof.

First we have

The argument above shows that any subcollection of $d + 2$ sets will have nonempty intersection.

The argument referred to is the one given for the base case. It does not rely on the induction hypothesis. Now, we form a collection of $n-1$ sets consisting of $X_1$ through $X_{n-2}$ and $X_{n-1}\cap X_n$. If you choose $d+1$ elements from this collection and take their intersection, there are two possibilities:

• The $d+1$ elements chosen includes $X_{n-1}\cap X_n$. In this case there are $d$ other sets $X_j$ selected. So the intersection of all of them is the intersection of those $d$ sets, $X_{n-1}$, and $X_n$, or a collection of $d+2$ sets, which by indicated argument we know must have a non-empty intersection.
• Or the $d+1$ elements do not include $X_{n-1}\cap X_n$. Neither $X_{n-1}$ nor $X_n$ can be in the selected sets, so if we take the intersection of the $d+1$ sets and also intersect with $X_n$, this is a collection of $d+2$ sets, and again must have non-empty intersection. But $A \cap X_n \ne \varnothing$, then $A \ne \varnothing$. So once again, the intersection of the $d+1$ elements cannot be empty.

In either case, the intersection of $d+1$ elements is not empty.