description/proof of that for partially-ordered ring and subset, if supremum of subset exists, supremum of (subset plus element) exists and equals (supremum of subset) plus element, and if infimum of subset exists, infimum of (subset plus element) exists and equals (infimum of subset) plus element
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of partially-ordered ring.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the 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:
\(R\): \(\in \{\text{ the partially-ordered rings }\}\) with any partial ordering, \(\lt\)
\(S\): \(\subseteq R\)
\(r\): \(\in R\)
//
Statements:
(
\(\exists Sup (S)\)
\(\implies\)
\(\exists Sup (S + r) \land Sup (S + r) = Sup (S) + r\)
)
\(\land\)
(
\(\exists Inf (S)\)
\(\implies\)
\(\exists Inf (S + r) \land Inf (S + r) = Inf (S) + r\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(Sup (S)\) exists; Step 2: see that \(Ub (S + r) = Ub (S) + r\); Step 3: see that \(Min (Ub (S + r)) = Min (Ub (S)) + r\); Step 4: suppose that \(Inf (S)\) exists; Step 5: see that \(Lb (S + r) = Lb (S) + r\); Step 6: see that \(Max (Lb (S + r)) = Max (Lb (S)) + r\).
Step 1:
Let us suppose that \(Sup (S)\) exists.
Step 2:
Let us see that \(Ub (S + r) = Ub (S) + r\).
Let \(r' \in Ub (S + r)\) be any.
For each \(s \in S\), \(s + r \le r'\), so, \(s \le r' - r\), so, \(r' - r \in Ub (S)\), so, \(r' \in Ub (S) + r\).
So, \(Ub (S + r) \subseteq Ub (S) + r\).
Let \(r' \in Ub (S) + r\) be any.
\(r' = p + r\) for a \(p \in Ub (S)\), which means that for each \(s \in S\), \(s \le p\), so, \(s + r \le p + r = r'\), so, \(r' \in Ub (S + r)\).
So, \(Ub (S) + r \subseteq Ub (S + r)\).
So, \(Ub (S + r) = Ub (S) + r\).
Step 3:
Let us see that \(Min (Ub (S + r)) = Min (Ub (S)) + r\).
As \(Ub (S + r) = Ub (S) + r\), we see that \(Min (Ub (S) + r) = Min (Ub (S)) + r\) instead.
\(Min (Ub (S)) + r \in Ub (S) + r\), because \(Min (Ub (S)) \in Ub (S)\), which implies that \(Min (Ub (S)) + r \in Ub (S) + r\).
For each \(p \in Ub (S)\), \(Min (Ub (S)) \le p\), so, \(Min (Ub (S)) + r \le p + r\), which means that \(Min (Ub (S)) + r = Min (Ub (S) + r)\).
So, \(Sup (S + r) = Min (Ub (S + r)) = Min (Ub (S) + r)\) exists and equals \(Min (Ub (S)) + r = Sup (S) + r\).
Step 4:
Let us suppose that \(Inf (S)\) exists.
Step 5:
Let us see that \(Lb (S + r) = Lb (S) + r\).
Let \(r' \in Lb (S + r)\) be any.
For each \(s \in S\), \(r' \le s + r\), so, \(r' - r \le s\), so, \(r' - r \in Lb (S)\), so, \(r' \in Lb (S) + r\).
So, \(Lb (S + r) \subseteq Lb (S) + r\).
Let \(r' \in Lb (S) + r\) be any.
\(r' = p + r\) for a \(p \in Lb (S)\), which means that for each \(s \in S\), \(p \le s\), so, \(r' = p + r = \le s + r\), so, \(r' \in Lb (S + r)\).
So, \(Lb (S) + r \subseteq Lb (S + r)\).
So, \(Lb (S + r) = Lb (S) + r\).
Step 6:
Let us see that \(Max (Lb (S + r)) = Max (Lb (S)) + r\).
As \(Lb (S + r) = Lb (S) + r\), we see that \(Max (Lb (S) + r) = Max (Lb (S)) + r\) instead.
\(Max (Lb (S)) + r \in Lb (S) + r\), because \(Max (Lb (S)) \in Lb (S)\), which implies that \(Max (Lb (S)) + r \in Lb (S) + r\).
For each \(p \in Lb (S)\), \(p \le Max (Lb (S))\), so, \(p + r \le Max (Lb (S)) + r\), which means that \(Max (Lb (S)) + r = Max (Lb (S) + r)\).
So, \(Inf (S + r) = Max (Lb (S + r)) = Max (Lb (S) + r)\) exists and equals \(Max (Lb (S)) + r = Inf (S) + r\).
