Family of Sets Proof Theorem: Counterexample & Error

In summary: Ooops, yes that's what i meant. I wrote it wrong.In summary, the theorem is incorrect because it states that if F and G are the empty set, then F and G are not disjoint. However, this is not the case if F and G contain the empty set.
  • #1
Dansuer
81
1

Homework Statement



Incorrect Theorem: Suppose F and G are familes of sets. If [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are disjoint, then so are F and G

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

Proof. Suppose [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are disjoint. Suppose F and G are not disjoint. Then we can choose some set A such that A[itex]\in[/itex]F and A[itex]\in[/itex]G. Since A[itex]\in[/itex]F, A[itex]\subseteq[/itex]F, so every element of A is in [itex]\bigcup[/itex]F. Similarly, since A[itex]\in[/itex]G, every element of Ais in [itex]\bigcup[/itex]G. But then every element of A is in both [itex]\bigcup[/itex]G and [itex]\bigcup[/itex]F, and this is impossible since [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]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 [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are.
i can't find why the proof is wrong though.

Thank you.
 
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  • #2
Hi Dansuer! :smile:

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?
 
  • #3
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 [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G. What if A has no element?

but i can't point on what it's wrong.
 
  • #4
Dansuer said:
Ooops, yes that's what i meant. I wrote it wrong.

Ah yes, I actually thought so :smile:

And i think that the problem in the proof is in the part that says that every element of A is in [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]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 [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G." So that is not our problem. Agreed?

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


Dansuer said:
But then every element of A is in both [itex]\bigcup[/itex]G and [itex]\bigcup[/itex]F, and this is impossible since [itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are disjoint. Contraddiction.

Can you elaborate why this is impossible? (keep in mind that A can be empty).
 
  • #5
[itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are disjoint means that for every x, x is either not an element of [itex]\bigcup[/itex]F or [itex]\bigcup[/itex]G or both. But since every x[itex]\in[/itex]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 ?
 
  • #6
Dansuer said:
[itex]\bigcup[/itex]F and [itex]\bigcup[/itex]G are disjoint means that for every x, x is either not an element of [itex]\bigcup[/itex]F or [itex]\bigcup[/itex]G or both. But since every x[itex]\in[/itex]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 [itex]\bigcup F[/itex] as [itex]\bigcup G[/itex]. 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 [itex]\bigcup F[/itex] as [itex]\bigcup G[/itex] to reach a contradiction. But how would you choose such an element if you didn't choose it in A?
 
  • #7
thank you very much for your help:biggrin:
i get it now
 

Related to Family of Sets Proof Theorem: Counterexample & Error

What is the Family of Sets Proof Theorem?

The Family of Sets Proof Theorem, also known as the Union Proof Theorem, is a mathematical concept used to prove the union of two or more sets. It states that the union of a family of sets is equal to the set of all elements that belong to any of the sets in the family.

What is a counterexample in relation to this theorem?

A counterexample is a specific example that disproves a statement or theorem. In the context of the Family of Sets Proof Theorem, a counterexample would be a specific set or collection of sets that does not follow the theorem, thus disproving its validity.

Why is it important to understand counterexamples in relation to this theorem?

Understanding counterexamples allows for a deeper understanding of the theorem and its limitations. By recognizing counterexamples, we can identify and correct any errors in our proof or understanding of the theorem.

What are some common errors made in the proof of this theorem?

One common error is assuming that the union of two sets is always equal to the union of the individual elements in those sets. This is not always true, as the union of two sets may result in duplicate elements being removed. Another error is assuming that the order of the sets in the union does not matter, when in fact it can affect the resulting set.

How can one avoid making errors in the proof of this theorem?

To avoid errors, it is important to carefully examine the definition of the theorem and understand the conditions under which it holds true. It is also helpful to work through multiple examples and to double check your proof for any assumptions that may not be true in all cases.

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