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Natural Numbers – Theorem Part 2

Hallo, infinite-set readers. Set theory is back.

Today we will discuss a theorem. Still in natural numbers. This theorem just explore from previous post. Hope you like this post.

Theorem

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

Proof

For proof of unity of m can be seen on the proof of Theorem on previous post. Next, given any non-empty set S whose members are natural numbers. Take any h\in S. If there are no members from S who is a member of h then h is a member of all natural numbers contained in S that different with h. Choose m=h then the proof is complete.

If any member of S is a member of h, defined set M is the set that collects these elements. Obtained M\subseteq h with h is a natural number. According to Theorem (previous post) then there is a natural number m\in M such that m is a member of every member of set M that is different from m. Note that for n\in S-h then h=n or h\in n. If h=n, since m\in M\subseteq h=n then m\in n. If h\in n then according to Theorem (in this link) apply that h\subseteq n. Obtained m\in M\subseteq h\subseteq n if and only if m\in n. It is proved that there exists with a single natural number m\in S susch that m is a member of each member of set S that different with m.

Next post, we will discuss about Partially Ordered Set, Simply Ordered Set, and Well Ordered Set. Please open your book about relation. We will use it to explore that discussion. Hope you still waiting about my post.

Thanks for reading.