# Proof of sets

• Feb 1st 2007, 06:44 PM
luckyc1423
Proof of sets
I got help with one proof already, I am just lost when it comes to proofs....here is another on...

Problem:

Prove that the empty set is unique. That is, suppose that A and B are empty sets and prove that A=B
• Feb 1st 2007, 06:58 PM
ThePerfectHacker
Quote:

Originally Posted by luckyc1423
I got help with one proof already, I am just lost when it comes to proofs....here is another on...

Problem:

Prove that the empty set is unique. That is, suppose that A and B are empty sets and prove that A=B

If we are working in the Zermelo-Frankael Set theory model, we will use the axiom of extensionality.
"Two sets are equal if and only if for any element of A if and only if is an element of B".
Since A has no elements the statement,
"If x in A then x in B" is true.
Similarly,
"If x in B then x in A" is true.

These statements are true because the hypothesis of the conditional is false. And a false impling a true of a false impling a true is always true. Thus these statements are both true.
• Feb 3rd 2007, 04:29 AM
Soroban
Hello, luckyc1423!

Quote:

Prove that the empty set is unique.
That is, suppose that \$\displaystyle A\$ and \$\displaystyle B\$ are empty sets. .Prove that \$\displaystyle A=B\$.

Here's a rather primitive approach . . .

Let \$\displaystyle A = \emptyset\$ and \$\displaystyle B = \emptyset.\$

When are two sets \$\displaystyle A\$ and \$\displaystyle B\$ unequal?

They are unequal if there is an element in \$\displaystyle A\$ which is not in \$\displaystyle B\$
. . . . . . . . . . .or if there is an element in \$\displaystyle B\$ which is not in \$\displaystyle A\$.

(1) Is there an element in \$\displaystyle A\$ which is not in \$\displaystyle B\$?
. . . Since \$\displaystyle A = \emptyset\$, the answer is No.

(2) Is there an element in \$\displaystyle B\$ whichis ot in \$\displaystyle A\$?
. . . Since \$\displaystyle B = \emptyset\$, the answer is No.

Therefore: .\$\displaystyle A = B\$