Family of Sets Proof Theorem: Counterexample & Error

[Dansuer]

Homework Statement

Incorrect Theorem: Suppose F and G are familes of sets. If $\bigcup$F and $\bigcup$G are disjoint, then so are F and G

a) What's wrong with the following proof of the theorem?

Proof. Suppose $\bigcup$F and $\bigcup$G are disjoint. Suppose F and G are not disjoint. Then we can choose some set A such that A$\in$F and A$\in$G. Since A$\in$F, A$\subseteq$F, so every element of A is in $\bigcup$F. Similarly, since A$\in$G, every element of Ais in $\bigcup$G. But then every element of A is in both $\bigcup$G and $\bigcup$F, and this is impossible since $\bigcup$F and $\bigcup$G are disjoint. Contraddiction.

b)Find a counterexample to the theorem.

Homework Equations

The Attempt at a Solution

I've found a counterexample. If F and G are the empty set. F and G are not disjoint but $\bigcup$F and $\bigcup$G are.
i can't find why the proof is wrong though.

Thank you.

 

[micromass]

Hi Dansuer!

I'm afraid that your counterexample is not correct. If F and G are both empty, then F and G are disjoint (in that F doesn't contain something that G contains and vice versa).

You're on the right track however, you'll need to do something with the empty set. What if both F and G contain the empty set?

 

[Dansuer]

Ooops, yes that's what i meant. I wrote it wrong.
And i think that the problem in the proof is in the part that says that every element of A is in $\bigcup$F and $\bigcup$G. What if A has no element?

but i can't point on what it's wrong.

 

[micromass]

Ooops, yes that's what i meant. I wrote it wrong.

Ah yes, I actually thought so

And i think that the problem in the proof is in the part that says that every element of A is in $\bigcup$F and $\bigcup$G. What if A has no element?

You're close, but it's not yet it. The problem is indeed that A can have no elements. But even if A has no elements that it is still true that "every element of A is in $\bigcup$F and $\bigcup$G." So that is not our problem. Agreed?

Takee a look at the very last sentence of the proof:

But then every element of A is in both $\bigcup$G and $\bigcup$F, and this is impossible since $\bigcup$F and $\bigcup$G are disjoint. Contraddiction.

Can you elaborate why this is impossible? (keep in mind that A can be empty).

 

[Dansuer]

$\bigcup$F and $\bigcup$G are disjoint means that for every x, x is either not an element of $\bigcup$F or $\bigcup$G or both. But since every x$\in$is an element of both, it's impossible. Now i start to see it. Since we are talking about the elements of A, A can be empty and there is no contradiction.

Now, if we were not talking about the elements of A, but about every element x in general. Would it be a contradiction ?

 

[micromass]

$\bigcup$F and $\bigcup$G are disjoint means that for every x, x is either not an element of $\bigcup$F or $\bigcup$G or both. But since every x$\in$is an element of both, it's impossible. Now i start to see it. Since we are talking about the elements of A, A can be empty and there is no contradiction.

Indeed, the contradiction is that there is an element a in both $\bigcup F$ as $\bigcup G$. And this a is chosen in A. But if A were empty, then this is not possible. Thus there is no contradiction!

Now, if we were not talking about the elements of A, but about every element x in general. Would it be a contradiction ?

Well, you don't need an element in A, you just need an element in both $\bigcup F$ as $\bigcup G$ to reach a contradiction. But how would you choose such an element if you didn't choose it in A?

 

[Dansuer]

thank you very much for your help
i get it now