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

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- Feb 1st 2007, 06:44 PMluckyc1423Proof 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 PMThePerfectHacker
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 AMSoroban
Hello, luckyc1423!

Quote:

Prove that the empty set is unique.

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

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

When are two sets $\displaystyle A$ and $\displaystyle B$**un**equal?

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: .$\displaystyle A = B$