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Set theoretical proof

Let AA and BB be sets such that A∩B=A∪BA \cap B = A \cup B. Prove that A=BA=B. [6 marks]

I'm not really sure how to go about answering questions of this type, so excuse how potentially stupid my approach may look. :colondollar:

Proof: Suppose that x∈Ax \in A and y∈By \in B such that x≠yx \neq y. Then x∉(A∩B)x \notin (A \cap B) and y∉(A∩B)y \notin (A \cap B). We have that x∈(A∪B)x \in (A \cup B) and y∈(A∪B)y \in (A \cup B). But since A∩B=A∪BA \cap B = A \cup B, we have that x∈(A∩B)x \in (A \cap B) and y∈(A∩B)y \in (A \cap B). This then forces the condition that x=yx=y. So x∈A≡y∈A≡x∈B≡y∈B.x \in A \equiv y \in A \equiv x \in B \equiv y \in B. So that A=BA=B.
(edited 9 years ago)
Original post by Zacken

Proof: Suppose that x∈Ax \in A and y∈By \in B such that x≠yx \neq y. Then x∉(A∩B)x \notin (A \cap B) and y∉(A∩B)y \notin (A \cap B).


I don't see how that (final quoted sentence) follows from your definition of x and y.
Reply 2
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Zacken
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Original post by ghostwalker
I don't see how that (final quoted sentence) follows from your definition of x and y.


I was thinking of something along the lines of 1∈A1 \in A and 2∈B2\in B but 1∉(A∩B)1 \notin (A \cap B) and 2∉(A∩B)2 \notin (A \cap B)
Original post by Zacken
Let AA and BB be sets such that A∩B=A∪BA \cap B = A \cup B. Prove that A=BA=B. [6 marks]

I'm not really sure how to go about answering questions of this type, so excuse how potentially stupid my approach may look. :colondollar:

Proof: Suppose that x∈Ax \in A and y∈By \in B such that x≠yx \neq y. Then x∉(A∩B)x \notin (A \cap B) and y∉(A∩B)y \notin (A \cap B).


This doesn't seem to make sense. What stops x∈A∩Bx \in A \cap B? This has to be true for some elements of AA.

You need to show that:

1. x∈A⇒x∈Bx \in A \Rightarrow x \in B
2. x∈B⇒x∈Ax \in B \Rightarrow x \in A

using the fact that A∩B=A∪BA \cap B = A \cup B
Original post by Zacken
I was thinking of something along the lines of 1∈A1 \in A and 2∈B2\in B but 1∉(A∩B)1 \notin (A \cap B) and 2∉(A∩B)2 \notin (A \cap B)


Those are specific elements which wouldn't necessarily be members of general sets. There are no specific elements you can use for a general set.

atsruser's suggestion would be the way I'd do it.
Reply 5
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Zacken
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Original post by atsruser
This doesn't seem to make sense. What stops x∈A∩Bx \in A \cap B? This has to be true for some elements of AA.

You need to show that:

1. x∈A⇒x∈Bx \in A \Rightarrow x \in B
2. x∈B⇒x∈Ax \in B \Rightarrow x \in A

using the fact that A∩B=A∪BA \cap B = A \cup B


I see, yeah. That was a bit of a stupid assumption on my part.

Would this be better?

Let x∈Ax \in A, then it follow that x∈(A∪B)x \in (A \cup B), this implies that x∈(A∩B)x \in (A \cap B), so x∈Ax\in A and x∈Bx \in B.

Let x∈Bx \in B, then it follow that x∈(A∪B)x \in (A \cup B), this implies that x∈(A∩B)x \in (A \cap B), so x∈Ax\in A and x∈Bx \in B.

Which shows that x∈A⇒x∈Bx\in A \Rightarrow x\in B and x∈B⇒x∈Ax\in B \Rightarrow x \in A.

Hence A=BA=B.
Reply 6
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Zacken
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Original post by ghostwalker
Those are specific elements which wouldn't necessarily be members of general sets. There are no specific elements you can use for a general set.

atsruser's suggestion would be the way I'd do it.


Yeah, that was a bit of a brain fart on my part. :tongue: Thanks!
Is my revised proof better? :smile:
Original post by Zacken
I see, yeah. That was a bit of a stupid assumption on my part.

Would this be better?


Yes, that's correct.

BTW you don't need the brackets around the A∩BA \cap B and so on. They're distracting.
Reply 8
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Zacken
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Original post by atsruser
Yes, that's correct.

BTW you don't need the brackets around the A∩BA \cap B and so on. They're distracting.


Yay! Great, thanks to you and Ghost. +Rep. :smile:

Right, brackets. Will keep in mind next time!

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