description/proof of that for non-decreasing and non-increasing sequences on real numbers set s.t. 1st sequence is equal to or smaller than 2nd sequence, each element of 1st sequence is equal to or smaller than any element of 2nd sequence, and supremum of 1st sequence is equal to or smaller than infimum of 2nd sequence
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader knows a definition of infimum of subset of partially-ordered set.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any non-decreasing sequence and any non-increasing sequence on the real numbers set with any same domain such that the 1st sequence is equal to or smaller than the 2nd sequence, each element of the 1st sequence is equal to or smaller than any element of the 2nd sequence, and the supremum of the 1st sequence is equal to or smaller than the infimum of 2nd sequence.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(\mathbb{R}\): with the canonical ordering, \(\lt\)
\(s\): \(\in \{\text{ the non-decreasing sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
\(s'\): \(\in \{\text{ the non-increasing sequences }\}\), such that \(Dom (s') = J\) and \(Ran (s') \subseteq \mathbb{R}\)
//
Statements:
\(\forall j \in J (s (j) \le s' (j))\)
\(\implies\)
(
\(\forall j, j' \in J (s (j) \le s' (j'))\)
\(\land\)
\(Sup (Ran (s)) \le Inf (Ran (s'))\)
)
//
2: Proof
Whole Strategy: Step 1: see that \(s (j) \le s' (j')\); Step 2: see that for each \(\epsilon\), \(Sup (Ran (s)) - \epsilon / 2 \lt s (j)\) and \(s' (j') \lt Inf (Ran (s')) + \epsilon / 2\).
Step 1:
Let \(j, j' \in J\) be any.
\(j \le j'\) or \(j' \lt j\).
When \(j \le j'\), \(s (j) \le s (j')\), because \(s\) is non-decreasing, \(\le s' (j')\), so, \(s (j) \le s' (j')\).
When \(j' \lt j\), \(s' (j) \le s' (j')\), because \(s'\) is non-increasing, so, \(s (j) \le s' (j) \le s' (j')\).
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(j \in J\) such that \(Sup (Ran (s)) - \epsilon / 2 \lt s (j)\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
There is a \(j' \in J\) such that \(s' (j) \lt Sup (Ran (s')) + \epsilon / 2\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
\(Sup (Ran (s)) - \epsilon / 2 \lt s (j) \le s' (j) \lt Sup (Ran (s')) + \epsilon / 2\), by Step 1.
So, \(Sup (Ran (s)) \lt Sup (Ran (s')) + \epsilon\), so, \(Sup (Ran (s)) \le Sup (Ran (s')) + \epsilon\).
So, \(Sup (Ran (s)) \le Sup (Ran (s'))\), by 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.
