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Natural Numbers – Theorem Part 1

Hallo. Back to set theory again. Hope you interested and not bored. Today we still play with natural numbers. Happy reading.

(x) Theorem :

If S is a non-empty subset of a natural number n, then there is a single natural number m\in S such that m is a member of any member of set S that different with m.

Proof

Given a set of N whose members are the set of nature and satisfies the Theorem (x) above. Will be proven N=\omega. Clear that 0\in N because the statement Theorem (x) is true if n=0. Assume that n\in N then will be proven n^{+}\in N.

Take S\subseteq n^{+}=n\cup\left\{ n\right\} with S\neq\emptyset, then there are 3 possibilities, that is S\subseteq n, S=R\cup\left\{ n\right\} ,\mbox{ for }\emptyset\neq R\subseteq n, or S=\left\{ n\right\}

(i) For S\subseteq n

Then according to the assumption, occur n^{+}\in N.

(ii) For S=R\cup\left\{ n\right\} ,\mbox{ for\,}\emptyset\neq R\subseteq n

Then it is true that there is m\in R such that m is a member of every set member R which is different from m. On the other hand, because of m\in R and R\subseteq n then m\in n. As a result m is a member of each member of S different from m.

(iii) For S=\left\{ n\right\}

Since S is a set that has one member then it is true that there is n such that n is a member of each member of S different from n

From (i), (ii), and (iii) then n^{+}\in N which means that N=\omega.

Now it will be proved for the unity of m. Suppose that there are m and m' with m\neq m' such that respectively m and m' are members of each member of set S which is different from m and m'. Since m is a member of every set S which is different from m then m\in m'. On the other hand, for the same reason, we get m'\in m. According to the Theorem (x) inĀ previous post then m\subseteq m' and m'\subseteq m are equivalent to m=m'. There is a contradiction that means m is unique.

Thanks for reading. See you in next post.