
#1
Jan3012, 07:20 PM

PF Gold
P: 183

From the way its written, case 1 shows that A[itex]\subseteq[/itex]BUA while case 2 shows that B[itex]\subseteq[/itex]BUA; therefore, the conclusion would be AUB[itex]\subseteq[/itex]BUA. Maybe I'm missing something here..? 



#2
Jan3012, 09:46 PM

Mentor
P: 16,543

You are right. But if we substitute A and B, then we also get a proof for the other inclusion. That is: a proof for the other inclusion follows from proving the first inclusion.




#3
Jan3012, 10:03 PM

PF Gold
P: 183

A→B and B→A therefore A=B was a totally different proof. And some how, he accomplishes the same thing without using this because of something to do with his description of an "arbitrary x". 



#4
Jan3012, 10:06 PM

Mentor
P: 16,543

Simple proof questionFirst we prove (as in the OP) that [itex]E\cup F\subseteq F\cup E[/itex] for ALL sets E and F. This is what the OP does, right?? Now, we want to prove that [itex]A\cup B=B\cup A[/itex] for all sets A and B. Well [itex]\subseteq[/itex] follows if we take E=A and F=B. [itex]\supseteq[/itex] follows if we take E=B and F=A. So equality holds. 



#5
Jan3012, 10:55 PM

PF Gold
P: 183

Would you agree that in case 1: he essentially showed that A⊆BUA? Would you also agree that in case 2: he essentially showed that B⊆BUA? He believed that they didn't. Why would he think that? At any rate, I agree with you here; however, he seemed to be making a different argument (during the discussion). 



#6
Jan3012, 11:00 PM

Mentor
P: 16,543

Formally, you indeed need to provide justification for both inclusions. However, I wasn't present at the discussion, so I can't really say what your professor was trying to say. All I can say is that I think you have a good understanding of this situation and that what you say is correct. 


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