The "fancy math thing" is LaTex. On this forum surround the LaTex code with [ math ] and [ /math ] (without the spaces). To learn the code itself, look at

http://www.mathhelpforum.com/math-help/latex-help/
You want the word "intersection" of sets, not "unity" (and NOT "union" which was my first guess). Other than that, your English is excellent.

That "union of any

**three** is non-empty" is interesting. There are any number of examples of bad induction where the error is that you cannot go from n= 1 to n= 2 although it is easy to go from n to n+ 1 for n> 1.

And, of course, the whole bit about "convex subsets" is irrelevant. What is important is that if the intersection of any

**three** sets in a collection of sets is non empty then the intersection of all of them is non-empty.

Let

be a collection of n sets such that the intersection of any three of them is non-empty. Then n must be at least 3 for that to make sense.

Base case: n= 3. Since

only contains 3 sets if "the intersection of any 3 is non-empty" then certainly the intersection of all

**3** is non-empty.

Now suppose this is true for n= k. We will show it is true for n= k+1.

Suppose

is a collection of k+1 sets having the property that the intersection of any 3 of the sets is non-empty.

Let U be any one of the sets and Let

be

with U removed. Then

still has the property that the intersection of any three of its sets is non-empty and since we are assuming that the theorem is true for n= k, it is true that the intersection of all sets in

is non-empty. Let X be that intersection.

Let V be any set other than U and let

be

with V removed. By the same argument, the intersection of all the remaining sets is non-empty. Let Y be that intersection.

Now V is a member of

and U is a member of

. And, since k is at least three, there exist a third set, W, which is in both

and

(this is the crucial point and why we need that "intersection of any

**three** sets is non-empty) . All of the points in X are in both V and W and all of the points in Y are in both U and W. That means that W contains points of the intersection of X and Y is in W and so that interersection is non-empty. From that, it follows that the intersection of all sets in

is non-empty.