Proving A×B ≠φ for A≠φ and B≠φ

In summary, the conversation discusses proving the statement "If A \neq \phi and B \neq \phi, then A\times B \neq \phi" with the given hint. There are two questions raised about whether it is enough to just use the hint or if the hint needs to be proven. The proof provided follows the given hint and uses the definition of the empty set to show that A\times B is not empty. There is a discussion about whether the proof needs to cite the axiom of non-empty sets and the definition of the empty set. Ultimately, it is determined that the proof is acceptable.
  • #1
bobby2k
127
2
I am supposed to prove:

If A [itex]\neq[/itex] [itex]\phi[/itex] and B [itex]\neq[/itex] [itex]\phi[/itex] then
A[itex]\times[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]

The HINT in the back of the book gives:
A [itex]\neq[/itex] [itex]\phi[/itex] [itex]\wedge[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]
[itex]\Rightarrow[/itex] [itex]\exists[/itex]a[itex]\subseteq[/itex]A [itex]\wedge[/itex]b[itex]\subseteq[/itex] B so that (a,b) [itex]\subseteq[/itex] A[itex]\times[/itex] B

I have 2 questions

1.Is it enough for the proof to write the hint, and then say "hence A [itex]\times[/itex]B" is not empty.

2. Or do I have to prove the hint?
If I have to prove the hint, is this a correct way to prove it?
[itex]A \neq \phi \wedge B \neq \phi [/itex] -Premise-(1)
[itex]A \neq \phi [/itex] - 1 and simplification-(2)
[itex]B \neq \phi [/itex] - 1 and simplification-(3)
[itex] \exists a \in A[/itex] - 2 and since A is not empty it has atleast one element-(4)
[itex] \exists b \in B[/itex] - 3 and since B is not empty it has atleast one element-(5)
[itex] (\exists a \in A) \wedge (\exists b \in B)[/itex]- 4 and 5- (6)
[itex](a,b) \in A \times B[/itex] - 6, and definition of [itex]A \times B [/itex] -(7)
[itex] A \times B \neq \phi [/itex] - 7 and definition of [itex]\phi[/itex] (8)

Is step (4) and (5) ok? I haven't read an axiom that specifies that every non-empty set has ateleast one element. Do I maybe have to use the face that there is only one empty set. And since there is only one empty set, every other set must have atleast one element, or else they would be the empty set?

Should step (7) have more explanation?

Also should existencial generalistion or specification have been cited anywhere in the proof?
 
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  • #2
What is the definition of the empty set that you are supposed to know and use?
 
  • #3
voko said:
What is the definition of the empty set that you are supposed to know and use?

Do you mean in step 8?
There I used that the "definition" of the empty set is that it contains no elements. And since we in step (7) showed that A [itex]\times[/itex]B contains an element, it cannot be the empty set.
 
  • #4
I think your proof is OK.

Even though I think just the hint alone is proof enough.
 

Related to Proving A×B ≠φ for A≠φ and B≠φ

What does "Proving A×B ≠φ for A≠φ and B≠φ" mean?

"Proving A×B ≠φ for A≠φ and B≠φ" is a statement in mathematical logic that means showing that the product of two sets, A and B, is not equal to the empty set when both A and B are not empty sets. In other words, it is proving that the intersection of two non-empty sets is also not empty.

Why is it important to prove A×B ≠φ for A≠φ and B≠φ?

Proving A×B ≠φ for A≠φ and B≠φ is important because it is a fundamental concept in set theory and logic. It helps to establish the relationship between two sets and is used in many mathematical proofs and theories. Additionally, understanding this concept can also aid in solving problems in various fields such as computer science and statistics.

How do you prove A×B ≠φ for A≠φ and B≠φ?

To prove A×B ≠φ for A≠φ and B≠φ, you can use a proof by contradiction method. This involves assuming that A×B=φ and then showing that this leads to a contradiction, which proves the original statement to be true. Another approach is to use set operations and logical reasoning to show that the intersection of two non-empty sets cannot be empty.

What are some real-life applications of proving A×B ≠φ for A≠φ and B≠φ?

Proving A×B ≠φ for A≠φ and B≠φ can have practical applications in fields such as statistics, where the concept of independence between two events is based on the intersection of two sets being non-empty. In computer science, understanding this concept is important for database queries and data analysis. It is also used in game theory and decision-making processes.

Can A×B ever be equal to φ for A≠φ and B≠φ?

No, A×B can never be equal to φ for A≠φ and B≠φ. This is because the definition of the intersection of two sets requires that both sets have at least one common element. Therefore, if A and B are both non-empty sets, their intersection will also always be non-empty and cannot be equal to the empty set.

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