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Proof / Disclaimer With Example

I think my post is a bit long and a lot of hassle lately. Today I will make a basic article. This is for website refreshment as well. Happy reading.

Most people use the equivalent method to prove mathematical statements. Even to prove the error of a statement. Though there are other ways to prove error statement, ie with an example of denial. This is considered sufficient because a statement must be perfect. Also because the truth must be perfect in mathematics.

Suppose there is a statement “If n\in\mathbb{N} then n^{2}+n+1 is prime”. Suppose we want to deny the statement. We have a presumption that n^{2}+n+1 not always prime. Then simply indicated there is a number n=k such that k^{2}+k+1 is not prime. For this case, we take n=4 then n^{2}+n+1=4^{2}+4+1=21 is not prime. This example is sufficient to prove that the above statement is false.

Thanks for reading.

Basic of Zorn’s Lemma – Initial Segment Part 2

Hello, infinite-set reader. Set theory is back.

To read this post, you should read the previous post first. It deals with the definition of \ll that we will use. Happy reading.

Theorem :

Given poset \left(Q,\ll\right) with Q is the set of all well ordered subsets of poset \left(P,\leq\right) and \ll is a relation that defined as before. For any simply ordered subset of poset \left(Q,\ll\right) then always have the smallest upper bound.

Proof :

Given S simply ordered subset of \left(Q,\ll\right). Suppose S=\emptyset then sup\, S=\emptyset. Next, let’s say S=\left\{ u,v,\ldots\right\} is non-empty simply ordered subset of \left(Q,\ll\right). Consider :

    \[ w=\cup S=\cup\left\{ u,v,\ldots\right\} =\bigcup_{u\in S}u\]

It will be proved that \left(w,\leq\right) is a well ordered subset of \left(P,\leq\right). Take any s\subseteq w with s\neq\emptyset then between the set u,v,\ldots, there is at least one that the element is on s. Suppose the set is u. Since \left(u,\leq\right) is well ordered set then u\cap s has the smallest element. Suppose the smallest element is a. It will be proved that a is also the smallest element for \left(s,\leq\right).

Take any b\in s with b\neq a. Will be proven a<b. Because of b\in s then there is v\in S such that b\in v. On the other hand, since \left(S,\ll\right) is simply ordered set then u\ll v or v\ll u. If v\ll u then a<b and if u\ll v then it is not possible to apply b<a . It because, if b<a then b\in u which results in a contradiction with a is the smallest element of u\cap s. In other words, apply a\leq b for all b\in s. So for any s\subseteq w, s has the smallest element. We get \left(w,\leq\right) is well ordered set which resulted w\in Q. Thus, we get w=sup\, S cause for any u\in S apply u\ll w.

Based on the description above, for the next well ordered subset w of poset P on above is called the greatest ordered subset of P. So w is called the greatest ordered subset of P if w is not the initial segment of a well ordered subset of P.

Thanks for reading. May be useful.