Hallo. Back to set theory again. Hope you interested and not bored. Today we still play with natural numbers. Happy reading.
(x) Theorem :
If is a non-empty subset of a natural number , then there is a single natural number such that is a member of any member of set that different with .
Given a set of whose members are the set of nature and satisfies the Theorem (x) above. Will be proven . Clear that because the statement Theorem (x) is true if . Assume that then will be proven .
Take with , then there are 3 possibilities, that is , , or
Then according to the assumption, occur .
Then it is true that there is such that is a member of every set member which is different from . On the other hand, because of and then . As a result is a member of each member of different from .
Since is a set that has one member then it is true that there is such that is a member of each member of different from
From (i), (ii), and (iii) then which means that .
Now it will be proved for the unity of . Suppose that there are and with such that respectively and are members of each member of set which is different from and . Since is a member of every set which is different from then . On the other hand, for the same reason, we get . According to the Theorem (x) in previous post then and are equivalent to . There is a contradiction that means is unique.
Thanks for reading. See you in next post.