(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let A = A1 x ... x An

B = B1 x ... x Bn

C = C1 x ... x Cn

Such that A,B,C are non empty, [itex]A=B\cup C[/itex] and [itex]B\cap C = \emptyset [/itex]

prove that there exists a k in {1,...,n} such that [itex]B_k\cap C_k = \emptyset [/itex]

and for [itex]i\neq k[/itex], [itex]A_i = B_i = C_i [/itex]

2. Relevant equations

3. 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 [itex] B_1 \cap C_1 \neq \emptyset [/itex] and [itex]B_2 \cap C_2 \neq \emptyset [/itex]

then [itex] 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) [/itex]

which implies that [itex]B \cap C \neq \emptyset [/itex] since [itex]B_1 \cap C_1 \neq \emptyset [/itex] and [itex]B_2 \cap C_2 \neq \emptyset [/itex]

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

it seems like this can be easily extended for any n

but i am having trouble proving the second part of the problem:

for [itex]i\neq k[/itex], [itex]A_i = B_i = C_i [/itex]

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

Any hints on what I should try?

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# Question on Set Theory

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