description/proof of that for upper-closed interval and nonempty subset, supremum of subset exists and equals supremum of subset on real numbers set
Topics
About: set
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any upper-closed interval and any nonempty subset, the supremum of the subset exists and equals the supremum of the subset on the real numbers set.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean set }\) with the canonical linear ordering
\(J\): \(\in \{\text{ the upper-closed intervals of } \mathbb{R}\}\), with the upper end, \(r_2\)
\(S\): \(\subseteq J\) such that \(S \neq \emptyset\)
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Statements:
\(\exists Sup_J (S)\), where \(Sup_J (S)\) is the supremum of \(S \subseteq J\)
\(\land\)
\(Sup_J (S) = Sup_{\mathbb{R}} (S)\), where \(Sup_{\mathbb{R}} (S)\) is the supremum of \(S \subseteq \mathbb{R}\)
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2: Note
\(Sup_J (S)\) and \(Sup_{\mathbb{R}} (S)\) are conceptually different: for example, \(J = (-1, 1]\) and \(S = \{0\}\), then \(Sup_J (S) = Min (Ub_J (S)) = Min ([0, 1])\) and \(Sup_{\mathbb{R}} (S) = Min (Ub_{\mathbb{R}} (S)) = Min ([0, \infty))\).
\(J\) needs to be upper-closed for this proposition: for example, \(J = (-1, 1)\) and \(S = (0, 1)\), then, \(Sup_{\mathbb{R}} (S) = 1\), but \(Sup_J (S)\) does not exist.
3: Proof
Whole Strategy: Step 1: see that \(p := Sup_{\mathbb{R}} (S)\) exists; Step 2: see that \(p \le r_2\); Step 3: see that \(Sup_J (S) = p\).
Step 1:
\(Sup_{\mathbb{R}} (S)\) exists, by the proposition that any nonempty upper-bounded subset of the real numbers set has the supremum and any nonempty lower-bounded subset of the real numbers set has the infimum.
Let \(p := Sup_{\mathbb{R}} (S)\).
Step 2:
Let us see that \(p \le r_2\).
Let us suppose that \(r_2 \lt p\).
\(r_2 \lt p - (p - r_2) / 2 \lt p\).
For each \(s \in S\), \(s \le r_2 \lt p - (p - r_2) / 2\).
So, \(p - (p - r_2) / 2 \in Ub_{\mathbb{R}} (S)\), where \(Ub_{\mathbb{R}} (S)\) would be the set of the upper bounds of \(S\) on \(\mathbb{R}\), a contradiction against that \(p\) was the minimum of \(Ub_{\mathbb{R}} (S)\).
So, \(p \le r_2\).
Step 3:
For any \(s \in S \subseteq J\), \(s \le p \le r_2\), which implies that \(p \in J\), because \(J\) is an interval.
For each \(s \in S\), \(s \le p\), which means that \(p \in Ub_J (S)\).
If \(p\) was not the minimum of \(Ub_J (S)\), there would be a \(u \in Ub_J (S)\) such that \(p \le u\) did not hold, but \(u \in Ub_{\mathbb{R}} (S)\), a contradiction against that \(p\) was the minimum of \(Ub_{\mathbb{R}} (S)\), so, \(p\) is the minimum of \(Ub_J (S)\).
That means that \(p = Sup_J (S)\).
