description/proof of that for containing convergent sequence of subsets of real numbers set with canonical ordering, supremum of containing convergence is convergence of sequence of supremums of subsets
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of containing convergence of sequence of subsets of set.
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader knows a definition of convergence of sequence on metric space.
- 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 sequence of subsets of the real numbers set with the canonical ordering, if the supremum of the containing convergence does not exist, the convergence of the sequence of the supremums of the subsets does not exist, and otherwise, the supremum of the containing convergence is the convergence of the sequence of 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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(\mathbb{R}\): with the canonical ordering
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq Pow (\mathbb{R})\) and \(\exists lim_{ci} s\)
\(s'\): \(: J \to \mathbb{R}, j \mapsto Sup (s (j))\), which may not exist
//
Statements:
(
\(\lnot \exists Sup (lim_{ci} s)\)
\(\implies\)
\(\lnot \exists lim s'\)
)
\(\land\)
(
\(\exists Sup (lim_{ci} s)\)
\(\implies\)
\(\exists lim s' \land Sup (lim_{ci} s) = lim s'\)
)
//
2: Note
The analogous proposition for a contained convergent sequence of subsets of the real numbers set with the canonical ordering does not hold in general.
For example, let \(J = \mathbb{N}\) and \(s: J \to Pow (\mathbb{R}), n \mapsto (-1, 1) \cup \{2 - (1 / 2)^j \vert j \in \mathbb{N} \text{ such that } n \le j\}\). Then, \(lim_{ce} s = (-1, 1)\), because \((-1, 1) \subseteq (-1, 1) \cup \{2 - (1 / 2)^j \vert j \in \mathbb{N} \text{ such that } n \le j\}\) and \(2 - (1 / 2)^{j'} \notin \{2 - (1 / 2)^j \vert j \in \mathbb{N} \text{ such that } n \le j\}\) for each \(j' \lt n\). But \(Sup (lim_{ce} s) = lim s'\) does not hold, because \(s' (n) = Sup (s (n)) = 2\) for each \(n\), so, \(s'\) does not approach \(1 = Sup ((- 1, 1))\).
Compare with the proposition that for any containing convergent sequence of subsets of the real numbers set with the canonical ordering, if the infimum of the containing convergence does not exist, the convergence of the sequence of the infimums of the subsets does not exist, and otherwise, the infimum of the containing convergence is the convergence of the sequence of the infimums of the subsets.
So, for any containing convergent sequence of subsets of the real numbers set with the canonical ordering, the infimum version and the supremum version hold, while for a contained convergent sequence of subsets of the real numbers set with the canonical ordering, not the infimum version nor the supremum version hold.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: suppose that \(Sup (lim_{ci} s)\) does not exist; Step 3: see that \(lim s'\) does not exist; Step 4: suppose that \(Sup (lim_{ci} s)\) exists; Step 5: see that \(Sup (lim_{ci} s) = lim s'\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim_{ci} s = s (J_{\vert J \vert})\).
When \(Sup (lim_{ci} s) = Sup (s (J_{\vert J \vert}))\) does not exist, \(s'\) does not exist, so, \(lim s'\) does not exist.
When \(Sup (lim_{ci} s) = Sup (s (J_{\vert J \vert}))\) exists, each \(Sup (s (j))\) exists, because \(s (j) \subseteq s (J_{\vert J \vert})\), so, \(s'\) exists, and \(lim s' = Sup (s (J_{\vert J \vert}))\) exists, and \(Sup (lim_{ci} s) = Sup (s (J_{\vert J \vert})) = lim s'\).
Let us suppose otherwise, hereafter.
Step 2:
Let us suppose that \(Sup (lim_{ci} s)\) does not exist.
Step 3:
That means that \(lim_{ci} s\) is not upper bounded.
That means that for each \(r' \in \mathbb{R}\), there is an \(r \in lim_{ci} s\) such that \(r' \lt r\).
Then, there is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(r \in s (J_n)\), so, \(r' \lt r \le Sup (s (J_n))\), even if \(Sup (s (J_n))\) exists (otherwise, \(s'\) will not exist, and the claim will hold).
If there was a \(lim s' \in \mathbb{R}\), there would be an \(N' \in \mathbb{N}\) such that for each \(n' \in \mathbb{N} \setminus \{0\}\) such that \(N' \lt n'\), \(\vert lim s' - Sup (s (J_{n'})) \vert \lt 1\), so, \(lim s' - 1 \lt Sup (s (J_{n'})) \lt lim s' + 1\).
But taking \(r' = lim s' + 1\), for each \(N, N' \lt n'\), \(lim s' + 1 \lt Sup (s (J_{n'}))\), so, \(Sup (s (J_{n'})) \lt lim s' + 1 \lt Sup (s (J_{n'}))\), a contradiction.
So, there is no \(lim s' \in \mathbb{R}\).
Step 4:
Let us suppose that \(Sup (lim_{ci} s)\) exists.
Step 5:
That means that \(lim_{ci} s\) is upper bounded.
Each \(Sup (s (j))\) exists, because \(s (j) \subseteq lim_{ci} s\).
So, \(s'\) exists.
\(Sup (s (j)) \le Sup (lim_{ci} s)\), 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 \(r \in lim_{ci} s\) such that \(Sup (lim_{ci} s) - \epsilon \lt r\), 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 \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(n \lt N\), \(r \in s (J_n)\), so, \(Sup (lim_{ci} s) - \epsilon \lt r \le Sup (s (J_n))\).
So, as \(s' (J_n) = Sup (s (J_n))\), \(Sup (lim_{ci} s) - \epsilon \lt s' (J_n) \le Sup (lim_{ci} s)\) for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), which means that \(\vert s' (J_n) - Sup (lim_{ci} s) \vert \lt \epsilon\).
So, \(lim s' = Sup (lim_{ci} s)\).
