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Smallest/Largest Element – Lower/Upper Bound In Set Theory

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

In the previous post we’ve learned about partial ordered set (poset), ordered set, minimum element, and maximum element. In the previous post, I said that the unity existence of the maximum elements and minimal elements are also not guaranteed. You can reopen the post on this link. This is done in order to avoid chaos in reading the this post. Ok, let’s start the discussion.

Definition :

Given any poset \left(S,\leq\right).

(i) p\in S said the smallest element (minimum, first) of S if p\leq x for all x\in S.

(ii) g\in S said the largest element (maximum, last) of S if x\leq g for all x\in S.

Example :

I will take the same example as yesterday.

A=\left\{ \left(a\right)\right\}

B=\left\{ \left(a;b\right),\left(d;e\right)\right\}

C=\left\{ \left(\ldots;a_{2};a_{1};b_{1};b_{2};\ldots\right)\right\}

D=\left\{ \left(\ldots;a_{2};a_{1};b_{1};b_{2};\ldots;c_{2};c_{1}\right),\left(a;b;c\right)\right\}

In the example above, A has the smallest element as well as the largest element a. For B,\, C,\,\mbox{and}\, D they do not have the smallest element and the largest element.

As another example, poset \left\{ \left(a_{1};a_{2};a_{3};\ldots;b_{3};b_{2};b_{1}\right)\right\} has the smallest element a_{1} and has the largest element b_{1}. (more…)

Ordered Set

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

Last post we discuss about poset or partially ordered set. As per my promise, today we will learn about the ordered set. Also will be discuss about minimal element and maximal element. Happy reading.

Definition :

Given any set S. A subset E\subseteq S\times S is said to be order in set S if for any x,y,z\in S apply :

(i) \left(x,x\right)\notin E (irreflexive)

(ii) if \left(x,y\right)\in E and \left(y,z\right)\in E then \left(x,z\right)\in E (transitive)

Example :

E=\left\{ \left(a,b\right),\left(c,d\right)\right\} is an order in the set \left\{ a,b,c,d\right\}.

Furthermore any order E is denoted by <. Just like a poset, \left(S,<\right) it says an ordered set if < is an order in the set S.

Theorem :

If \left(S,<\right) is an ordered set then for any x,y\in S it’s valid statement that x<y then y\nless x.

Proof :

Given x<y. Suppose y<x. Since < transitive then x<x. Contradictions with < are irreflexive.

Theorem :

Given any poset \left(S,\leq\right). If for any x,y\in S, defined :

    \[ x<y\mbox{ if and only if }\left(x\leq y\right)\wedge\left(x\neq y\right)\]

then \left(S,<\right) is an ordered set.

Proof : (more…)