description/proof of that for linearly-ordered set and \(2\) elements, if each element that is larger than 2nd element is larger than 1st element, 1st element is equal to or smaller than 2nd element
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of linearly-ordered set.
Target Context
- The reader will have a description and a proof of the proposition that for any linearly-ordered set and any \(2\) elements, if each element that is larger than the 2nd element is larger than the 1st element, the 1st element is equal to or smaller than the 2nd element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s_1\): \(\in S\)
\(s_2\): \(\in S\)
//
Statements:
\(\forall s \in S \text{ such that } s_2 \lt s (s_1 \lt s)\)
\(\implies\)
\(s_1 \le s_2\)
//
2: Note
Compare with the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number, which requires only \(r_1 \le r_2 + \epsilon\), while this proposition requires that \(s_1 \lt s\): \(s_1 \le s\) is not enough.
For example, let \(S = \mathbb{Z}\) with the canonical ordering, \(s_1 = 1\), and \(s_2 = 0\), then, for each \(s_2 \lt s\), \(s_1 \le s\), but "\(s_1 \le s_2\)" does not hold.
3: Proof
Whole Strategy: Step 1: suppose that \(s_2 \lt s_1\), and find a contradiction.
Step 1:
Let us suppose that \(s_2 \lt s_1\).
\(s_1 \lt s_1\), by the supposition, a contradiction.
So, \(s_1 \le s_2\).
