# Prove ΦxΦ =Φ

1. Sep 3, 2009

### evagelos

Prove:

$$\emptyset\times\emptyset =\emptyset$$

2. Sep 4, 2009

### Preno

The empty set has no elements. Therefore, there is no ordered pair of elements of the empty set (using any of the common definitions of an ordered pair). There isn't really much to prove.

3. Sep 4, 2009

### HallsofIvy

Or, if you want to get "fancy", a proof by contradiction: Suppose that $\Phi \times \Phi$ is not empty. Then there exist a pair, (a, b) in $\Phi \times \Phi$ such that $a\in \Phi$ which contradicts the fact that $\Phi$ is empty.

4. Sep 4, 2009

5. Sep 4, 2009

### jgens

I'm not a mathematician, so perhaps my comment is naive, but the empty set is defined as the unique set which contains no elements. If $a$ were a member of the empty set, then the empty set would not be empty which is a contradiction. Maybe I'm misunderstanding something here?

6. Sep 4, 2009

### Dragonfall

You're not. Evagelos is wrong in assuming that there is no contradiction. It contradicts (assuming we're talking about ZF) the axiom of the empty set.

7. Sep 4, 2009

### Hurkyl

Staff Emeritus
HoI's statement is a theorem.

8. Sep 4, 2009

Halls' approach is, as others have stated, fine.

9. Sep 4, 2009

### techmologist

Yeah, I thought that was pretty much the definition of cartesian product: If A and B are sets, and C = AxB, then y is in C iff there is an a in A and a b in B such that y = (a, b).

If one of A or B is the empty set, then for every y, it is not in C.

10. Sep 4, 2009

### evagelos

Let Hall write a formal proof then to find out whether his proof is right or wrong.

Sorry Halls i know that you dislike formal proofs,but that is the only way,otherwise we will argue indefinitely

11. Sep 4, 2009

### evagelos

No ,i did not say that there is no proof by contradiction,but Hall's contradiction is not based somewhere

12. Sep 4, 2009

### Dragonfall

Like I said, his contradiction is based on the axiom of of the empty set, is it not?

What do you mean by "based on" then?

13. Sep 5, 2009

### CRGreathouse

The proof is so shockingly obvious that I'm surprised even you could ask for it. I can only assume this is a homework problem.

14. Sep 7, 2009

### CompuChip

I fail to see how Hall's proof is not a formal one.

Alternatively, you could give a direct proof: it is true that
$$\emptyset\times\emptyset \subseteq \emptyset$$
because for all $x \in \emptyset\times\emptyset$, $x \in \emptyset$ - similarly the other inclusion follows. (In fact, if $A \subseteq \emptyset$, then it is always the case that $A = \emptyset$).

15. Sep 7, 2009

### HallsofIvy

We have asked evagelos repeatedly what he means by a "formal" proof and he has never answered clearly. I suspect he means something like used in "Principia Mathematica".

16. Sep 7, 2009

### CompuChip

Wait, let me get my ruler and compass...

17. Sep 7, 2009

### Preno

I suppose if you wanna get fancy, you can always try a categorical proof (of isomorphism, that is).

Last edited: Sep 7, 2009
18. Sep 9, 2009

### evagelos

Here is a formal proof of $$\emptyset\times\emptyset = \emptyset$$,a direct one:

1)$$\forall A\forall B [ A\times B =\emptyset\Longleftrightarrow A=\emptyset\vee B=\emptyset]$$.....................................................................................a theorem in set theory

2) $$\emptyset\times\emptyset =\emptyset\Longleftrightarrow \emptyset = \emptyset\vee\emptyset =\emptyset$$..............................................................................from (1) and using Universal Elimination ,where we put $$A=\emptyset$$ and $$B=\emptyset$$

3)$$\emptyset=\emptyset\vee \emptyset =\emptyset\Longrightarrow \emptyset\times\emptyset =\emptyset$$...............................................................................from (2) and using biconditional Elimination

4)$$\forall x[ x=x]$$...........................................................................................................an axiom in equality

5)$$\emptyset =\emptyset$$.............................................................................................................from (4) and using Universal Elimination,where we put $$x=\emptyset$$

6)$$\emptyset =\emptyset\vee \emptyset =\emptyset$$......................................................................................from (5) and using idempotent law ( p ====> p or p)

7)$$\emptyset\times\emptyset =\emptyset$$....................................................................................from (3) and (6) and using M.Ponens.

Now if you wish, write a formal proof using contradiction

19. Sep 10, 2009