# Help Proving This Set-Based Theorem

• Mar 31st 2013, 07:42 PM
RobertXIV
Help Proving This Set-Based Theorem
Greetings,
I'm not sure how in-depth this community goes into discrete math (also, as far as I know discrete math varies depending on the school and location) but I'm having trouble proving this theorem.

S x T subset S x U /\ S =! 0set implies T subset U

where x is any arbitrary set operator, and /\ is and.

Thanks.. Hope I supplied enough information.
• Mar 31st 2013, 10:31 PM
emakarov
Re: Help Proving This Set-Based Theorem
Quote:

Originally Posted by RobertXIV
where x is any arbitrary set operator, and /\ is and.

Are you sure x is not Cartesian product?
• Apr 1st 2013, 10:29 AM
RobertXIV
Re: Help Proving This Set-Based Theorem
Indeed, emakarov is correct. Sorry about that one :P.
• Apr 1st 2013, 10:35 AM
emakarov
Re: Help Proving This Set-Based Theorem
Then do you need further help with this problem?
• Apr 1st 2013, 10:36 AM
HallsofIvy
Re: Help Proving This Set-Based Theorem
Any member of S X T is of the form (s, t) where x $\displaystyle \in$ S and t$\displaystyle \in$ T. If that is a subset of S X ( U $\displaystyle \cap$ S) then t$\displaystyle \in S$.
• Apr 1st 2013, 11:47 AM
RobertXIV
Re: Help Proving This Set-Based Theorem
I take it you mean s is an element of S, not x, correct?
If so, this helps.
Thank you!
• Apr 1st 2013, 12:14 PM
Plato
Re: Help Proving This Set-Based Theorem
Quote:

Originally Posted by RobertXIV
I take it you mean s is an element of S, not x, correct?
If so, this helps.

I think that the problem is: If $\displaystyle \left( {S \times T} \right) \subseteq \left( {S \times U} \right) \wedge S \ne \emptyset$ then $\displaystyle T \subseteq U$

If $\displaystyle T=\emptyset$ then $\displaystyle T \subseteq U$ else suppose that $\displaystyle t\in T$.

Given that $\displaystyle (\exists s\in S)$ so $\displaystyle (s,t)\in\left( {S \times T} \right)\subseteq\left( {S \times U} \right)$.

Thus because this meams that $\displaystyle t\in U$. this shows that $\displaystyle T \subseteq U$.