Prove AUB=BUA: Simple Proof Question

[SixNein]

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\subseteqBUA.

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

Maybe I'm missing something here..?

 

[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.

 

[SixNein]

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".

 

[micromass]

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".

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.

 

[SixNein]

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.

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).

 

[micromass]

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.

 

[SixNein]

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.

ANyway, thanks for your time.