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.
Hello, luckyc1423!
Here's a rather primitive approach . . .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 $\displaystyle B$ 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: .$\displaystyle A = B$