Proof of sets

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February 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
February 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.
February 3rd 2007, 04:29 AM
Soroban

Hello, luckyc1423!

Quote:

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

Here's a rather primitive approach . . .

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

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

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

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

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

Therefore: . $A = B$