Proving A×B ≠φ for A≠φ and B≠φ

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Homework Help Overview

The discussion revolves around proving that the Cartesian product of two non-empty sets, A and B, is also non-empty. The original poster is exploring the implications of the hint provided in their textbook and questioning the necessity of proving the hint itself.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster questions whether it is sufficient to state the hint as proof or if they must provide a detailed proof of the hint. They also seek clarification on specific steps in their reasoning and the definitions involved.

Discussion Status

Some participants have provided feedback on the original poster's proof structure, indicating that they believe the hint alone may suffice as proof. There is an ongoing exploration of the definitions and logical steps involved in the proof process.

Contextual Notes

Participants are discussing the definitions of the empty set and the implications of non-empty sets having elements. There is a mention of axioms related to set theory that may be relevant to the proof.

bobby2k
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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?
 
Last edited:
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What is the definition of the empty set that you are supposed to know and use?
 
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.
 
I think your proof is OK.

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

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