description/proof of that for non-increasing sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, convergence of sequence is infimum 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 infimum 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 infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
Target Context
- The reader will have a description and a proof of the proposition that for any non-increasing sequence on the \(1\)-dimensional Euclidean metric space with the canonical ordering, if the sequence is not lower-bounded, not the convergence of the sequence nor the infimum of the range of the sequence exists, and otherwise, the convergence of the sequence is the infimum 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:
(
\(\nexists r \in \mathbb{R} (r \le s)\)
\(\implies\)
\(\nexists lim s \land \nexists Inf (Ran (s))\)
)
\(\land\)
(
\(\exists r \in \mathbb{R} (r \le s)\)
\(\implies\)
\(\exists lim s \land \exists Inf (Ran (s)) \land lim s = Inf (Ran (s))\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(s\) is not lower-bounded; Step 2 see that not \(lim s\) nor \(Inf (Ran (s))\) exists; Step 3: suppose that \(s\) is lower-bounded; Step 4: see that \(Inf (Ran (s))\) exists; Step 5: see that \(lim s = Inf (Ran (s))\).
Step 1:
Let us suppose that \(s\) is not lower-bounded.
Step 2:
Inevitably, \(J\) is infinite, because if \(\vert J \vert \in \mathbb{N} \setminus \{0\}\), \(Min (Ran (s)) \le s\) where \(Min (Ran (s)) \in \mathbb{R}\) would exist, by the proposition that for any linearly-ordered set, any nonempty finite subset has the maximum and the minimum, and \(Min (Ran (s))\) would be an lower-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, \(lim s - 1 \lt s (J_n)\), so, \(Min (\{s (J_1), ..., s (J_N), lim s - 1\}) \le s\), so, \(Min (\{s (J_1), ..., s (J_N), lim s - 1\})\) would be an lower-bound, a contradiction.
\(Inf (Ran (s)) \in \mathbb{R}\) does not exist, because if \(Inf (Ran (s))\) existed, \(Inf (Ran (s)) = Max (Lb (Ran (s)))\) would be an lower-bound of \(s\), a contradiction.
Step 3:
Let us suppose that \(s\) is lower-bounded.
Step 4:
\(Inf (Ran (s))\) exists, by the well known property of \(\mathbb{R}\).
Step 5:
When \(\vert J \vert \in \mathbb{N} \setminus \{0\}\), \(Inf (Ran (s)) = s (J_{\vert J \vert})\), because \(s\) is non-increasing, and \(lim s = s (J_{\vert J \vert})\), so, \(lim s = Inf (Ran (s))\).
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 \(s (J_N) \lt Inf (Ran (s)) + \epsilon\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
For each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(s (J_n) \le s (J_N) \lt Inf (Ran (s)) + \epsilon\), because \(s\) is non-increasing: \(J_N \lt J_n\) as \(N \lt n\).
On the other hand, \(Inf (Ran (s)) - \epsilon \lt Inf (Ran (s)) \le s (J_n)\), because \(Inf (Ran (s)) = Max (Lb (Ran (s)))\), so, \(Inf (Ran (s))\) is an lower-bound of \(s\).
So, \(Inf (Ran (s)) - \epsilon \lt s (J_n) \lt Inf (Ran (s)) + \epsilon\), which means that \(\vert s (J_n) - Inf (Ran (s)) \vert \lt \epsilon\).
So, \(lim s = Inf (Ran (s))\).
