Prove AUB=BUA: Simple Proof Question

  • Thread starter SixNein
  • Start date
  • Tags
    Proof
In summary: 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.
  • #1
SixNein
Gold Member
122
20
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[itex]\subseteq[/itex]BUA.

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..?
 
Physics news on Phys.org
  • #2
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
micromass said:
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".
 
  • #4
SixNein said:
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 [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
micromass said:
OK, what about this:

First 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.

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).
 
  • #6
SixNein said:
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.
 
  • #7
micromass said:
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.
 

1. How do you prove AUB=BUA?

In order to prove AUB=BUA, we must show that both sets contain the same elements. This can be done by using the definition of set equality, which states that two sets are equal if and only if they have the same elements.

2. What is the simple proof for AUB=BUA?

The simple proof for AUB=BUA involves showing that every element in AUB is also in BUA, and vice versa. This can be done by using a Venn diagram or by using logical statements to compare the two sets.

3. Can you explain why AUB=BUA is a true statement?

Yes, AUB=BUA is a true statement because the two sets have the same elements. This is because the union of two sets is defined as the set of all elements that are in either set, and the intersection of two sets is defined as the set of all elements that are in both sets. Therefore, AUB=BUA is true because the two sets have the same elements.

4. What are some examples of sets that follow the rule AUB=BUA?

Some examples of sets that follow the rule AUB=BUA are the set of even numbers and the set of multiples of 3. Both of these sets have the same elements, so AUB=BUA is true for these sets.

5. Is AUB=BUA a commutative property?

Yes, AUB=BUA is a commutative property. This means that the order in which we list the elements in the set does not matter. The result will always be the same, as long as the two sets have the same elements.

Similar threads

  • Set Theory, Logic, Probability, Statistics
2
Replies
54
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
746
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
671
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top