description/proof of that for non-decreasing sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, convergence of sequence is supremum of range of sequence
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean metric space.
- The reader knows a definition of convergence of sequence on metric space.
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader admits the proposition that for any linearly-ordered set, any nonempty finite subset has the maximum and the minimum.
- 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 non-decreasing sequence on the \(1\)-dimensional Euclidean metric space with the canonical ordering, if the sequence is not upper-bounded, not the convergence of the sequence nor the supremum of the range of the sequence exists, and otherwise, the convergence of the sequence is the supremum of the range of the sequence.
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 \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\) with the canonical ordering
\(s\): \(: J \to \mathbb{R}\), such that for each \(j, j' \in J\) such that \(j \lt j'\), \(s (j) \le s (j')\)
//
Statements:
(
\(\nexists r \in \mathbb{R} (s \le r)\)
\(\implies\)
\(\nexists lim s \land \nexists Sup (Ran (s))\)
)
\(\land\)
(
\(\exists r \in \mathbb{R} (s \le r)\)
\(\implies\)
\(\exists lim s \land \exists Sup (Ran (s)) \land lim s = Sup (Ran (s))\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(s\) is not upper-bounded; Step 2 see that not \(lim s\) nor \(Sup (Ran (s))\) exists; Step 3: suppose that \(s\) is upper-bounded; Step 4: see that \(Sup (Ran (s))\) exists; Step 5: see that \(lim s = Sup (Ran (s))\).
Step 1:
Let us suppose that \(s\) is not upper-bounded.
Step 2:
Inevitably, \(J\) is infinite, because if \(\vert J \vert \in \mathbb{N} \setminus \{0\}\), \(s \le Max (Ran (s))\) where \(Max (Ran (s)) \in \mathbb{R}\) would exist, by the proposition that for any linearly-ordered set, any nonempty finite subset has the maximum and the minimum, and \(Max (Ran (s))\) would be an upper-bound, a contradiction.
\(lim s \in \mathbb{R}\) does not exist, because if \(lim s \in \mathbb{R}\) existed, there would be an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert lim s - s (J_n) \vert \lt 1\), so, \(s (J_n) \lt lim s + 1\), so, \(s \le Max (\{s (J_1), ..., s (J_N), lim s + 1\})\), so, \(Max (\{s (J_1), ..., s (J_N), lim s + 1\})\) would be an upper-bound, a contradiction.
\(Sup (Ran (s)) \in \mathbb{R}\) does not exist, because if \(Sup (Ran (s))\) existed, \(Sup (Ran (s)) = Min (Ub (Ran (s)))\) would be an upper-bound of \(s\), a contradiction.
Step 3:
Let us suppose that \(s\) is upper-bounded.
Step 4:
\(Sup (Ran (s))\) exists, by the well known property of \(\mathbb{R}\).
Step 5:
When \(\vert J \vert \in \mathbb{N} \setminus \{0\}\), \(Sup (Ran (s)) = s (J_{\vert J \vert})\), because \(s\) is non-decreasing, and \(lim s = s (J_{\vert J \vert})\), so, \(lim s = Sup (Ran (s))\).
Let us suppose that \(J\) is infinite, hereafter.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N} \setminus \{0\}\) such that \(Sup (Ran (s)) - \epsilon \lt s (J_N)\), 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.
For each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(Sup (Ran (s)) - \epsilon \lt s (J_N) \le s (J_n)\), because \(s\) is non-decreasing: \(J_N \lt J_n\) as \(N \lt n\).
On the other hand, \(s (J_n) \le Sup (Ran (s)) \lt Sup (Ran (s)) + \epsilon\), because \(Sup (Ran (s)) = Min (Ub (Ran (s)))\), so, \(Sup (Ran (s))\) is an upper-bound of \(s\).
So, \(Sup (Ran (s)) - \epsilon \lt s (J_n) \lt Sup (Ran (s)) + \epsilon\), which means that \(\vert s (J_n) - Sup (Ran (s)) \vert \lt \epsilon\).
So, \(lim s = Sup (Ran (s))\).
