description/proof of that for lower-closed interval and nonempty subset, infimum of subset exists and equals infimum 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 lower-closed interval and any nonempty subset, the infimum of the subset exists and equals the infimum 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
\(J\): \(\in \{\text{ the lower-closed intervals of } \mathbb{R}\}\), with the lower end, \(r_1\)
\(S\): \(\subseteq J\) such that \(S \neq \emptyset\)
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Statements:
\(\exists Inf_J (S)\), where \(Inf_J (S)\) is the infimum of \(S \subseteq J\)
\(\land\)
\(Inf_J (S) = Inf_{\mathbb{R}} (S)\), where \(Inf_{\mathbb{R}} (S)\) is the infimum of \(S \subseteq \mathbb{R}\)
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2: Note
\(Inf_J (S)\) and \(Inf_{\mathbb{R}} (S)\) are conceptually different: for example, \(J = [-1, 1)\) and \(S = \{0\}\), then \(Inf_J (S) = Max (Lb_J (S)) = Max ([-1 , 0])\) and \(Inf_{\mathbb{R}} (S) = Max (Lb_{\mathbb{R}} (S)) = Max ((- \infty, 0])\).
\(J\) needs to be lower-closed for this proposition: for example, \(J = (-1, 1)\) and \(S = (-1, 0)\), then, \(Inf_{\mathbb{R}} (S) = -1\), but \(Inf_J (S)\) does not exist.
3: Proof
Whole Strategy: Step 1: see that \(p := Inf_{\mathbb{R}} (S)\) exists; Step 2: see that \(r_1 \le p\); Step 3: see that \(Inf_J (S) = p\).
Step 1:
\(Inf_{\mathbb{R}} (S)\) 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 \(p := Inf_{\mathbb{R}} (S)\).
Step 2:
Let us see that \(r_1 \le p\).
Let us suppose that \(p \lt r_1\).
\(p \lt p + (r_1 - p) / 2 \lt r_1\).
For each \(s \in S\), \(p + (r_1 - p) / 2 \lt r_1 \le s\).
So, \(p + (r_1 - p) / 2 \in Lb_{\mathbb{R}} (S)\), where \(Lb_{\mathbb{R}} (S)\) would be the set of the lower bounds of \(S\) on \(\mathbb{R}\), a contradiction against that \(p\) was the maximum of \(Lb_{\mathbb{R}} (S)\).
So, \(r_1 \le p\).
Step 3:
For any \(s \in S \subseteq J\), \(r_1 \le p \le s\), which implies that \(p \in J\), because \(J\) is an interval.
For each \(s \in S\), \(p \le s\), which means that \(p \in Lb_J (S)\).
If \(p\) was not the maximum of \(Lb_J (S)\), there would be a \(u \in Lb_J (S)\) such that \(u \le p\) did not hold, but \(u \in Lb_{\mathbb{R}} (S)\), a contradiction against that \(p\) was the maximum of \(Lb_{\mathbb{R}} (S)\), so, \(p\) is the maximum of \(Lb_J (S)\).
That means that \(p = Inf_J (S)\).
