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Lets look at the following question. Why would the proof be incorrect?

Let U be any set. Prove that there exists an $A \in \mathcal{P}(U)$ such that for every $B \in \mathcal{P}(U)$, $A \cap B = B$.

Proof. Our goal is $\exists A \in \mathcal{P}(U) \forall B \in \mathcal{P}(U)(A \cap B = B)$. We choose $A = B$. Clearly $B \cap B = B$.

I'm quite certain that the correct answer is Let A = U. What I'm struggling with is explaining why we can't Let A = B. Or is it fine to choose A = B?

fesodes
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2 Answers2

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The problem is that it says for every B. If you say A = B, then A is actually changing with every chosen B, so you aren't referring to a specific A.

johnnyb
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  • I've thought about that as well but then lets take a look at another proof.
    Prove that there exist a real number x such that for every real number y, $y^2x = y-x$.

    Our goal would be $\exists x \in \mathbb{R} \forall y \in \mathbb{R}(y^2x = y-x)$. The answer would be to choose $x = \frac{y}{y^2+1}$

    In this case x is also changing with every chosen y. Is it because x is more specific in this case?

     – fesodes Jan 14 '20 at 03:09
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    I think the example you give is faulty. Can you show where this is done? I can see how someone might say "prove that for every real number y there exists an x such that $y^2x = y - x$." In that case the result would be correct. But as you have written it, There is no real number $x$ that satisfies $y^2x = y - x$ for every real number $y$. – johnnyb Jan 14 '20 at 03:36
  • You are correct. I mixed up 2 questions from section 3.6 of the book "How to prove it". It should be as you described "prove that for every real number y there exists an unique x such that...". Thanks so much! – fesodes Jan 14 '20 at 03:49
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As you say, you want to show $$\exists A \in \mathcal{P}(U) \, \forall B \in \mathcal{P}(U)(A \cap B = B).$$

It might help to say this in words: Prove there exists an $A \in \mathcal{P}(U)$ such that for all $B \in \mathcal{P}(U)$ we have $A \cap B = B$.

The clause "for all $B \in \mathcal{P}(U)$ we have $A \cap B = B$" is a property that you want $A$ to have. Your goal is to choose an $A$ so that this clause is true. Thus your proof must begin

Let $A = $ __. Then, for all $B \in \mathcal{P}(U)$, we have $A \cap B = \ldots$ (fill in blanks) $ \ldots = B$.

The variable $B$, then, should not be introduced until you have already defined what $A$ is, so that the clause is true for all $B$.

Jair Taylor
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  • Thanks for making it so clear. I was confused at first since I thought there was a counter argument but then as johnnyb pointed out, that counter argument was faulty. I simply misread a question that had a solution for it in the Appendix section. – fesodes Jan 14 '20 at 04:07