A simple warm-up routine in set theory

     Since time usually flows only in one direction and this is my first post, I will tackle some aspects from the foundations of mathematics: set theory. Actually this post is trying to solve(hopefully is solving) an interesting problem that I found in The Art and Craft of Problem Solving by Paul Zeitz.

Putnam 1995:

Let S be a set of real numbers that is closed under multiplication(that is, if a and b are in S, then so is ab). Let T and U be disjoint subsets of S whose union is S. Given that the product of any three (not necessarily distinct) elements of T is in T and that the product of any three elements of U is in U, show that at least one of the two subsets T, U is closed under multiplication.

     You may want to consider trying to solve this yourself before reading any further. Also I have to mention that this is not it's standard proof, it is just a proof that I thought of, there may be other, similar or totally different, proofs. If you've found a different solution or you found a leach in mine please let me know.

      Solution:

 ∀ a, b, c in T we have abc ∈ T

∀ a', b', c' in U we have a'b'c' ∈ U

T ∪ U = S and T ∩ U = ∅

I'm going to use proof by contradiction to show that T or U is closed under multiplication.

So lets assume T and U are not closed under multiplication. This means that there exists two elements a, b in T such that ab is not in T. But if ab is not in T, then it must be in U because S is closed under multiplication.  The same goes for two elements a' and b' in U : their product is not in U because we supposed U is not closed under multiplication, so it must be in T. If we multiply all of them we get aba'b'  which must be an element of T since a, b, a'b' are three elements of T and their product must be in T. By the same reasoning aba'b' is an element of U since ab, a', b' are three elements of U.

So abba'b' is in  T ∩ U, but T ∩ U is empty and we reached a contradiction which means that the assumption that both T and U are not closed under multiplication is false.

     I have chosen this problem because despite it's simplicity it displays some important ideas:

-proof by contradiction

-symmetry

-reducing the problem by ignoring less important information(like the fact that the elements of S are real numbers)