# Simple proof question

by SixNein
Tags: proof, simple
PF Gold
P: 194
 Prove AUB=BUA Let xεAUB xεA or xεB (Definition of union) case 1: xεA xεBUA (Def of union) since x is arbitrary, must be true for all x. (inclusion) therefore, AUB=BUA Case 2: xεB xεBUA (Def of union) since x is arbitrary, must be true for all x. (inclusion) therefore, AUB=BUA
Now, I was told that the above proof was valid by a professor. But I don't see how it could be valid as it is written. The only proof I can arguably see here is a proof that AUB$\subseteq$BUA.

From the way its written, case 1 shows that A$\subseteq$BUA while case 2 shows that B$\subseteq$BUA; therefore, the conclusion would be AUB$\subseteq$BUA.

Maybe I'm missing something here..?
 Mentor P: 18,346 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.
PF Gold
P: 194
 Quote by micromass 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.
See I tired to point this out in class. The professor argued that my argument of
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".

Mentor
P: 18,346
Simple proof question

 Quote by SixNein See I tired to point this out in class. The professor argued that my argument of 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".

First we prove (as in the OP) that $E\cup F\subseteq F\cup E$ for ALL sets E and F. This is what the OP does, right??

Now, we want to prove that $A\cup B=B\cup A$ for all sets A and B.
Well
$\subseteq$ follows if we take E=A and F=B.
$\supseteq$ follows if we take E=B and F=A.
So equality holds.
PF Gold
P: 194
 Quote by micromass OK, what about this: First we prove (as in the OP) that $E\cup F\subseteq F\cup E$ for ALL sets E and F. This is what the OP does, right?? Now, we want to prove that $A\cup B=B\cup A$ for all sets A and B. Well $\subseteq$ follows if we take E=A and F=B. $\supseteq$ follows if we take E=B and F=A. So equality holds.

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).
Mentor
P: 18,346
 Quote by SixNein Let me ask you this: 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).
I agree with you here.

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.
PF Gold
P: 194
 Quote by micromass I agree with you here. 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.
I just needed some extra eyes on it. I could have been wrong.

The class is being taught out of the computer science department. I honestly don't think this would have been an issue in the mathematics department.