description/proof of that for containing convergent map from open interval into power set of real numbers set with canonical ordering at boundary, supremum of containing convergence is convergence of map to supremums of subsets
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of containing convergence of map from open interval into power set of set at boundary.
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of convergence of map from topological space minus point into topological space with respect to point.
- The reader admits the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any containing convergent map from any open interval into the power set of the real numbers set with canonical ordering at boundary, the supremum of the containing convergence is the convergence of the map to the supremums of the subsets.
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:
\((r_1, r_2)\): \(\in \{\text{ the open intervals of } \mathbb{R}\}\), where \(r_1\) cannnot be \(- \infty\) and \(r_2\) cannot be \(\infty\)
\(\mathbb{R}\): with the canonical ordering
\(f\): \(: (r_1, r_2) \to Pow (\mathbb{R})\) such that \(\exists lim_{ci, r_1} f \lor \exists lim_{ci, r_2} f\)
\(f'\): \(: (r_1, r_2) \to \mathbb{R}, r \mapsto Sup (f (r))\), which may not exist
//
Statements:
(
\(\lnot \exists Sup (lim_{ci, r_1} f)\)
\(\implies\)
\(\lnot \exists lim_{r_1} f'\)
)
\(\land\)
(
\(\exists Sup (lim_{ci, r_1} f)\)
\(\implies\)
\(\exists lim_{r_1} f' \land Sup (lim_{ci, r_1} f) = lim_{r_1} f'\)
)
\(\land\)
(
\(\lnot \exists Sup (lim_{ci, r_2} f)\)
\(\implies\)
\(\lnot \exists lim_{r_2} f'\)
)
\(\land\)
(
\(\exists Sup (lim_{ci, r_2} f)\)
\(\implies\)
\(\exists lim_{r_2} f' \land Sup (lim_{ci, r_2} f) = lim_{r_2} f'\)
)
//
2: Note
The analogous proposition for a contained convergent map from an open interval into the power set of the real numbers set with canonical ordering at boundary does not hold in general.
For example, let \((r_1, r_2) = (0, 1)\) and \(f: (r_1, r_2) \to Pow (\mathbb{R}), r \mapsto (-1, 1) \cup (2 - r, 2)\). Then, \(lim_{ce, r_1} f = (-1, 1)\), because \((-1, 1) \subseteq (-1, 1) \cup (2 - r, 2)\) and for each \(r' \in (2 - 1, 2)\), there is an \({r_1}' \in (0, 1)\) such that \(r' \lt 2 - {r_1}'\), because \(0 \lt 2 - r' \lt 1\), and for each \(r \in (0, {r_1}')\), \(r' \notin (2 - r, 2)\), because \(r' \lt 2 - {r_1}' \lt 2 - r\). But \(Sup (lim_{ce, r_1} f) = lim_{r_1} f'\) does not hold, because \(f' (r) = Sup (f (r)) = 2\) for each \(r\), so, \(f'\) does not approach \(1 = Sup ((- 1, 1))\).
3: Proof
Whole Strategy: Step 1: suppose that \(Sup (lim_{ci, r_1} f)\) does not exist; Step 2: see that \(lim_{r_1} f'\) does not exist; Step 3: suppose that \(Sup (lim_{ci, r_1} f)\) exists; Step 4: see that \(Sup (lim_{ci, r_1} f) = lim_{r_1} f'\); Step 5: suppose that \(Sup (lim_{ci, r_2} f)\) does not exist; Step 6: see that \(lim_{r_2} f'\) does not exist; Step 7: suppose that \(Sup (lim_{ci, r_2} f)\) exists; Step 8: see that \(Sup (lim_{ci, r_2} f) = lim_{r_2} f'\).
Step 1:
Let us suppose that \(Sup (lim_{ci, r_1} f)\) does not exist.
Step 2:
That means that \(lim_{ci, r_1} f\) is not upper bounded.
That means that for each \(r' \in \mathbb{R}\), there is an \(r \in lim_{ci, r_1} f\) such that \(r' \lt r\).
Then, there is an \({r_1}' \in (r_1, r_2)\) such that for each \(r^` \in (r_1, {r_1}')\), \(r \in f (r^`)\), so, \(r' \lt r \le Sup (f (r^`))\), even if \(Sup (f (r^`))\) exists (otherwise, \(f'\) will not exist, and the claim will hold).
If there was a \(lim_{r_1} f' \in \mathbb{R}\), there would be an \({r_1}'' \in (r_1, r_2)\) such that for each \(r'' \in (r_1, {r_1}'')\), \(\vert lim_{r_1} f' - Sup (f (r'')) \vert \lt 1\), so, \(lim_{r_1} f' - 1 \lt Sup (f (r'')) \lt lim_{r_1} f' + 1\).
But taking \(r' = lim_{r_1} f' + 1\), for each \(r'' \in (r_1, {r_1}') \cap (r_1, {r_1}'')\), \(lim_{r_1} f' + 1 \lt Sup (f (r''))\), so, \(Sup (f (r'')) \lt lim_{r_1} f' + 1 \lt Sup (f (r'')))\), a contradiction.
So, there is no \(lim_{r_1} f' \in \mathbb{R}\).
Step 3:
Let us suppose that \(Sup (lim_{ci, r_1} f)\) exists.
Step 4:
That means that \(lim_{ci, r_1} f\) is upper bounded.
Each \(Sup (f (r))\) exists, because \(f (r) \subseteq lim_{ci, r_1} f\).
So, \(f'\) exists.
\(Sup (f (r)) \le Sup (lim_{ci, r_1} f)\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(r \in lim_{ci, r_1} f\) such that \(Sup (lim_{ci, r_1} f) - \epsilon \lt r\), 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.
But there is an \({r_1}' \in (r_1, r_2)\) such that for each \(r' \in (r_1, {r_1}')\), \(r \in f (r')\), so, \(Sup (lim_{ci, r_1} f) - \epsilon \lt r \le Sup (f (r'))\).
So, as \(f' (r') = Sup (f (r'))\), \(Sup (lim_{ci, r_1} f) - \epsilon \lt f' (r') \le Sup (lim_{ci, r_1} f)\) for each \(r' \in (r_1, {r_1}')\), which means that \(\vert f' (r') - Sup (lim_{ci, r_1} f) \vert \lt \epsilon\).
So, \(lim_{r_1} f' = Sup (lim_{ci, r_1} f)\).
Step 5:
Let us suppose that \(Sup (lim_{ci, r_2} f)\) does not exist.
Step 6:
That means that \(lim_{ci, r_2} f\) is not upper bounded.
That means that for each \(r' \in \mathbb{R}\), there is an \(r \in lim_{ci, r_2} f\) such that \(r' \lt r\).
Then, there is an \({r_2}' \in (r_1, r_2)\) such that for each \(r^` \in ({r_2}', r_2)\), \(r \in f (r^`)\), so, \(r' \lt r \le Sup (f (r^`))\), even if \(Sup (f (r^`))\) exists (otherwise, \(f'\) will not exist, and the claim will hold).
If there was a \(lim_{r_2} f' \in \mathbb{R}\), there would be an \({r_2}'' \in (r_1, r_2)\) such that for each \(r'' \in ({r_2}'', r_2)\), \(\vert lim_{r_2} f' - Sup (f (r'')) \vert \lt 1\), so, \(lim_{r_2} f' - 1 \lt Sup (f (r'')) \lt lim_{r_2} f' + 1\).
But taking \(r' = lim_{r_2} f' + 1\), for each \(r'' \in ({r_2}', r_2) \cap ({r_2}'', r_2)\), \(lim_{r_2} f' + 1 \lt Sup (f (r''))\), so, \(Sup (f (r'')) \lt lim_{r_2} f' + 1 \lt Sup (f (r'')))\), a contradiction.
So, there is no \(lim_{r_2} f' \in \mathbb{R}\).
Step 7:
Let us suppose that \(Sup (lim_{ci, r_2} f)\) exists.
Step 8:
That means that \(lim_{ci, r_2} f\) is upper bounded.
Each \(Sup (f (r))\) exists, because \(f (r) \subseteq lim_{ci, r_2} f\).
So, \(f'\) exists.
\(Sup (f (r)) \le Sup (lim_{ci, r_2} f)\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(r \in lim_{ci, r_2} f\) such that \(Sup (lim_{ci, r_2} f) - \epsilon \lt r\), 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.
But there is an \({r_2}' \in (r_1, r_2)\) such that for each \(r' \in ({r_2}', r_2)\), \(r \in f (r')\), so, \(Sup (lim_{ci, r_2} f) - \epsilon \lt r \le Sup (f (r'))\).
So, as \(f' (r') = Sup (f (r'))\), \(Sup (lim_{ci, r_2} f) - \epsilon \lt f' (r') \le Sup (lim_{ci, r_2} f)\) for each \(r' \in ({r_2}', r_2)\), which means that \(\vert f' (r') - Sup (lim_{ci, r_2} f) \vert \lt \epsilon\).
So, \(lim_{r_2} f' = Sup (lim_{ci, r_2} f)\).
