A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0hard to write because saying it says the same thing. but i think the way to do it is to show that each is contained in the other pick some element on the left, say \(x\) and show that it is in the right an vice versa if it is in the left, then either \(x\in B\) or \(x\in \cap A_{\alpha\in I}\) meaning \(\exists\beta\in I\) such that \(x\in A_\beta\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0then if \(x\in B\) it must be in \(A_{\alpha}\cup B\) for any \(\alpha\) and if \(a\notin B\) then by the proceeding it must be in some \(A_{\beta}\) oh crap i got my modifier wrong!!

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0if \(x\in \cap A_{\alpha \in I}\) it is in ALL \(A_{\alpha}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0so \(\forall \alpha \in I\) we know \(x\in A_{\alpha}\) and therefore \(x\in \cap (A_{\alpha \in I}\cup B)\)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.