description/proof of that for real numbers set with canonical ordering and subset, if supremum of minus subset exists, it is minus infimum of subset, and if infimum of minus subset exists, it is minus supremum of subset
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.
Target Context
- The reader will have a description and a proof of the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of minus the subset exists, it is minus the infimum of the subset, and if the infimum of minus the subset exists, it is minus the supremum of the subset.
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:
\(\mathbb{R}\): \(= \text{ the real numbers set }\) with the canonical ordering
\(S\): \(\subseteq \mathbb{R}\)
\(- S\): \(= \{- s \in \mathbb{R} \vert s \in S\}\)
//
Statements:
(
\(\exists Sup (- S)\)
\(\implies\)
\(\exists Inf (S) \land Sup (- S) = - Inf (S)\)
)
\(\land\)
(
\(\exists Inf (- S)\)
\(\implies\)
\(\exists Sup (S) \land Inf (- S) = - Sup (S)\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(Sup (- S)\) exists; Step 2: see that \(- Sup (- S)\) satisfies the conditions to be the infimum of \(S\); Step 3: suppose that \(Inf (- S)\) exists; Step 4: see that \(- Inf (- S)\) satisfies the conditions to be the supremum of \(S\).
Step 1:
Let us suppose that \(Sup (- S)\) exists.
Step 2:
For each \(- s \in - S\), \(- s \le Sup (- S)\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(Sup (- S) - \epsilon \lt - s\), 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.
So, for each \(s \in S\), \(- s \le Sup (- S)\), so, \(- Sup (- S) \le s\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(Sup (- S) - \epsilon \lt - s\), so, there is an \(s \in S\) such that \(s \lt - Sup (- S) + \epsilon\), which implies that \(- Sup (- S) = Inf (S)\), 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.
So, \(Sup (- S) = - Inf (S)\).
Step 3:
Let us suppose that \(Inf (- S)\) exists.
Step 4:
For each \(- s \in - S\), \(Inf (- S) \le - s\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(- s \lt Inf (- S) + \epsilon\), 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.
So, for each \(s \in S\), \(Inf (- S) \le - s\), so, \(s \le - Inf (- S)\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(- s \lt Inf (- S) + \epsilon\), so, there is an \(s \in S\) such that \(- Inf (- S) - \epsilon \lt s\), which implies that \(- Inf (- S) = Sup (S)\), 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.
So, \(Inf (- S) = - Sup (S)\).
