Proving Non-Empty Set Theory Equations: A Solution Using Induction Method

[CornMuffin]

Homework Statement

Let A = A1 x ... x An
B = B1 x ... x Bn
C = C1 x ... x Cn
Such that A,B,C are non empty, A=B\cup C and B\cap C = \emptyset

prove that there exists a k in {1,...,n} such that B_k\cap C_k = \emptyset
and for i\neq k, A_i = B_i = C_i

Homework Equations

The Attempt at a Solution

I figure the best way to prove this is by induction.

for n=1, there is nothing to show, so I will start with n=2

Assume B_1 \cap C_1 \neq \emptyset and B_2 \cap C_2 \neq \emptyset

then B \cap C = (B_1 \times B_2 )\cap (C_1 \times C_2) = (B_1 \cap C_1) \times (B_2 \cap C_2)

which implies that B \cap C \neq \emptyset since B_1 \cap C_1 \neq \emptyset and B_2 \cap C_2 \neq \emptyset

but this is a contradiction to our conditions, therefore, there exists a k in {1,...,n} such that B_k\cap C_k = \emptyset


it seems like this can be easily extended for any n



but i am having trouble proving the second part of the problem:
for i\neq k, A_i = B_i = C_i

I tried using proof by contradiction for n=2, but i couldn't get anywhere with that


Any hints on what I should try?

 

[micromass]

You need to prove two things:

B_i\subseteq A_i~\text{and}~A_i\subseteq B_i

The first is not so hard: pick x_i\in B_i, extend it to an element of B. And use that B\cup C=A.

The second is more difficult. Pick x_i\in A_i. Assume that it is not in B_i. Try to extend x_i to an element you know is not in B\cup C. This will be a contradiction. (hint: pick a x_k\in C_k).