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description/proof of that for (countably) compact metric space with induced topology, non-convergent sequence has more than \(1\) points to which subsequences converge
Topics
About:
metric space
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 (countably) compact metric space with the induced topology, any non-convergent sequence has some more than \(1\) points to which some subsequences converge.
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:
\(M\): \(\in \{\text{ the countably compact metric spaces }\}\), with the induced topology
\(s\): \(: J \to M\), \(\in \text{ the sequences }\)
//
Statements:
\(s \notin \{\text{ the convergent sequences }\}\)
\(\implies\)
\(\exists s^`, \widetilde{s^`} \in \{\text{ the convergent subsequences of } s\} (lim s^` \neq lim \widetilde{s^`})\)
//
If \(M\) is compact, \(M\) is countably compact, by the proposition that any metric space is compact if and only if it is countably compact, so, \(M\) can be required to be compact.
2: Note
There may not be some more than \(2\) points to which some subsequences converge.
For example, let \(M = [-1, 1]\) and \(s: \mathbb{N} \to [-1, 1], n \mapsto - 1 / 2 \text{ when } n \text{ is even }; \mapsto 1 / 2 \text{ when } n \text{ is odd }\), then, \(s\) is not convergent, and there are a subsequence that converges to \(- 1 / 2\) and a subsequence that converges to \(1 / 2\), but there is no other point to which a subsequence converges.
3: Proof
Whole Strategy: Step 1: see that \(M\) is compact; Step 2: see that \(M\) is sequentially compact; Step 3: take any convergent subsequence of \(s\), \(s^`\), such that \(lim s^` = m\); Step 4: for each \(\epsilon\), take a finite open cover of \(M\), \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert j \in J\}\), and see that for an \(\epsilon\), a \(B_{m_j, \epsilon / 2}\) contains some infinite points of \(s\), which determines the subsequence of \(s\), \(\widetilde{s^`}'\); Step 5: take a convergent subsequence of \(\widetilde{s^`}'\), \(\widetilde{s^`}\).
Step 1:
\(M\) is compact, by the proposition that any metric space is compact if and only if it is countably compact.
In fact, \(M\) can be required to be compact, because then, \(M\) is countably compact.
Step 2:
\(M\) is sequentially compact, by the proposition that any metric space with the induced topology is 1st-countable and the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.
Step 3:
There is a convergent subsequence of \(s\), \(s^`: J^` \to M\), with \(m := lim s^`\), because \(M\) is sequentially compact: refer to the proposition that for any metric space, if and only if each sequence on it has a convergent subsequence, each sequence on it from the natural numbers set has a convergent subsequence.
Step 4:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let us take the open cover of \(M\), \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert m_j \in M \setminus B_{m, \epsilon}\}\).
That is indeed an open cover, because for each \(m' \in M\), \(m' \in B_{m, \epsilon}\) or \(m' \in M \setminus B_{m, \epsilon}\), but when \(m' \in M \setminus B_{m, \epsilon}\), \(m' \in B_{m_j, \epsilon / 2}\) where \(m_j = m'\).
As \(M\) is compact, there is a finite subcover, \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert j \in J\}\), where \(J\) is a finite index set.
If for each \(\epsilon\), each element of \(\{B_{m_j, \epsilon / 2} \vert j \in J\}\) contained only some finite points of \(s\), there would be an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(s (J_n) \in B_{m, \epsilon}\), which would mean that \(dist (m, s (J_n)) \lt \epsilon\), which would mean that \(s\) converged to \(m\), a contradiction against that \(s\) was not convergent.
So, there is an \(\epsilon\) such that a \(B_{m_j, \epsilon / 2}\) contains some infinite points of \(s\).
That determines the subsequence of \(s\), \(\widetilde{s^`}': {J^`}' \to M = s \circ f'\), where \({J^`}' = \{j^` \in J \vert s (j^`) \in B_{m_j, \epsilon / 2}\}\) and \(f': {J^`}' \to J, j^` \to j^`\), which is indeed a subsequence of \(s\), because for each \(j^`_1, j^`_2 \in {J^`}'\) such that \(j^`_1 \lt j^`_2\), \(f' (j^`_1) = j^`_1 \lt j^`_2 = f' (j^`_2)\), and for each \(j \in J\), there is a \(j^` \in {J^`}'\) such that \(j \le j^` = f' (j^`)\), because \({J^`}'\) is infinite.
Step 5:
There is a convergent subsequence of \(\widetilde{s^`}'\), \(\widetilde{s^`}: J^` \to M = \widetilde{s^`}' \circ f\), with \(lim \widetilde{s^`} = \widetilde{m}\), because \(M\) is sequentially compact.
\(\widetilde{s^`} = \widetilde{s^`}' \circ f = s \circ f' \circ f\) is a subsequence of \(s\), because for each \(j^`_1, j^`_2 \in J^`\) such that \(j^`_1 \lt j^`_2\), \(f' \circ f (j^`_1) \lt f' \circ f (j^`_2)\), because \(f (j^`_1) \lt f (j^`_2)\), so, \(f' \circ f (j^`_1) \lt f' \circ f (j^`_2)\), and for each \(j \in J\), there is a \({j^`}' \in {J^`}'\) such that \(j \le f' ({j^`}')\) and there is a \(j^` \in J^`\) such that \({j^`}' \le f (j^`)\), so, \(j \le f' ({j^`}') \le f' \circ f (j^`)\).
\(dist (\widetilde{m}, m_j) \le \epsilon / 2\), because for each \(\epsilon' \in \mathbb{R}\) such that \(0 \lt \epsilon'\), there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(dist (\widetilde{m}, \widetilde{s^`} ({J^`}_n)) \lt \epsilon'\), because \(\widetilde{s^`}\) converges to \(\widetilde{m}\), and \(dist (\widetilde{m}, m_j) \le dist (\widetilde{m}, \widetilde{s^`} ({J^`}_n)) + dist (\widetilde{s^`} ({J^`}_n), m_j) \lt \epsilon' + \epsilon / 2\), so, \(dist (\widetilde{m}, m_j) \le \epsilon / 2\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
\(\epsilon \lt dist (m, m_j) \le dist (m, \widetilde{m}) + dist (\widetilde{m}, m_j) \le dist (m, \widetilde{m}) + \epsilon / 2\), so, \(\epsilon / 2 = \epsilon - \epsilon / 2 \lt dist (m, \widetilde{m})\).
So, \(m \neq \widetilde{m}\).
So, there are at least some \(2\) subsequences, \(s^`\) and \(\widetilde{s^`}\) such that \(lim s^` \neq lim \widetilde{s^`}\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for metric space, iff each sequence on it has convergent subsequence, each sequence on it from natural numbers set has convergent subsequence
The table of contents of this article
Main Body
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(2\) convergent sequences with same index set on \(1\)-dimensional Euclidean metric space, if for each index, there is equal or larger index s.t. 1st sequence element with 2nd index is equal to or smaller than 2nd sequence element with 1st index, convergence of 1st sequence is equal to or smaller than convergence of 2nd sequence
Topics
About:
metric space
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 \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 2nd index is equal to or smaller than the 2nd sequence element with the 1st index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd 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 \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s_1\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
\(s_2\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
//
Statements:
\(\exists lim s_1 \land \exists lim s_2 \land \forall j \in J (\exists j' \in J \text{ such that } j \le j' (s_1 (j') \le s_2 (j)))\)
\(\implies\)
\(lim s_1 \le lim s_2\)
//
2: Note
Also the proposition that for any \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 1st index is equal to or smaller than the 2nd sequence element with the 2nd index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd sequence holds.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: for each \(\epsilon\), take some \(N\) and \(N'\) such that for each \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\) and for each \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\), and take any \(N, N' \lt n\) and any \(J_n \le j'\) such that \(s_1 (j') \le s_2 (J_n)\), and see that \(lim s_1 - \epsilon \lt lim s_2\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim s_2 = s_2 (J_{\vert J \vert})\).
There is a \(j' \in J\) such that \(J_{\vert J \vert} \le j'\) and \(s_1 (j') \le s_2 (J_{\vert J \vert})\), by the supposition, but \(J_{\vert J \vert} \le j'\) means that \(j' = J_{\vert J \vert}\), so, \(s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert})\).
But \(lim s_1 = s_1 (J_{\vert J \vert})\).
So, \(lim s_1 = s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert}) = lim s_2\).
Let us suppose otherwise, hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\), so, \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
There is an \(N' \in \mathbb{N}\) such that for each \(n' \in \mathbb{N} \setminus \{0\}\) such that \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\), so, \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
Let \(n \in \mathbb{N} \setminus \{0\}\) be any such that \(N, N' \lt n\).
As especially \(N \lt n\), \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
There is a \(j' \in J\) such that \(J_n \le j'\) and \(s_1 (j') \le s_2 (J_n)\), by the supposition.
While \(j' = J_{n'}\), \(J_n \le j' = J_{n'}\) means that \(n \le n'\), so, \(N, N' \lt n \le n'\).
As especially, \(N' \lt n'\), \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
So, \(lim s_1 - \epsilon \lt s_1 (J_{n'}) - \epsilon / 2 \le s_2 (J_n) - \epsilon / 2 \lt lim s_2\).
So, \(lim s_1 \le lim s_2 + \epsilon\).
So, \(lim s_1 \le lim s_2\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(2\) convergent sequences with same index set on \(1\)-dimensional Euclidean metric space, if for each index, there is equal or larger index s.t. 1st sequence element with 1st index is equal to or smaller than 2nd sequence element with 2nd index, convergence of 1st sequence is equal to or smaller than convergence of 2nd sequence
Topics
About:
metric space
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 \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 1st index is equal to or smaller than the 2nd sequence element with the 2nd index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd 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 \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s_1\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
\(s_2\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
//
Statements:
\(\exists lim s_1 \land \exists lim s_2 \land \forall j' \in J (\exists j \in J \text{ such that } j' \le j (s_1 (j') \le s_2 (j)))\)
\(\implies\)
\(lim s_1 \le lim s_2\)
//
2: Note
Also the proposition that for any \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 2nd index is equal to or smaller than the 2nd sequence element with the 1st index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd sequence holds.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: for each \(\epsilon\), take some \(N'\) and \(N\) such that for each \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\) and for each \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\), and take any \(N', N \lt n'\) and any \(J_{n'} \le j\) such that \(s_1 (J_{n'}) \le s_2 (j)\), and see that \(lim s_1 - \epsilon \lt lim s_2\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim s_1 = s_1 (J_{\vert J \vert})\).
There is a \(j \in J\) such that \(J_{\vert J \vert} \le j\) and \(s_1 (J_{\vert J \vert}) \le s_2 (j)\), by the supposition, but \(J_{\vert J \vert} \le j\) means that \(j = J_{\vert J \vert}\), so, \(s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert})\).
But \(lim s_2 = s_2 (J_{\vert J \vert})\).
So, \(lim s_1 = s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert}) = lim s_2\).
Let us suppose otherwise, hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N' \in \mathbb{N}\) such that for each \(n' \in \mathbb{N} \setminus \{0\}\) such that \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\), so, \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\), so, \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
Let \(n' \in \mathbb{N} \setminus \{0\}\) be any such that \(N', N \lt n'\).
As especially \(N' \lt n'\), \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
There is a \(j \in J\) such that \(J_{n'} \le j\) and \(s_1 (J_{n'}) \le s_2 (j)\), by the supposition.
While \(j = J_n\), \(J_{n'} \le j = J_n\) means that \(n' \le n\), so, \(N', N \lt n' \le n\).
As especially, \(N \lt n\), \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
So, \(lim s_1 - \epsilon \lt s_1 (J_{n'}) - \epsilon / 2 \le s_2 (J_n) - \epsilon / 2 \lt lim s_2\).
So, \(lim s_1 \le lim s_2 + \epsilon\).
So, \(lim s_1 \le lim s_2\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for sequence on partially-ordered set and subsequence, if limit superior of sequence exists, limit superior of subsequence does not necessarily exist, but if it exists, it is equal to or smaller than limit superior of sequence
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 sequence on any partially-ordered set and any subsequence, if the limit superior of the sequence exists, the limit superior of the subsequence does not necessarily exist, but if it exists, it is equal to or smaller than the limit superior 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 \(J \neq \emptyset\)
\(S\): \(\in \{\text{ the partially-ordered sets }\}\) with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(J^`\): \(\subseteq \mathbb{N}\), such that \(J^` \neq \emptyset\)
\(s^`\): \(= s \circ f\), \(\in \{\text{ the subsequences of } s \text{ with } f: J^` \to J\}\)
//
Statements:
(
\(\exists lim sup s\)
\(\lnot \implies\)
\(\exists lim sup s^`\)
)
\(\land\)
(
\(\exists lim sup s \land \exists lim sup s^`\)
\(\implies\)
\(lim sup s^` \le lim sup s\)
)
//
2: Note
Compare with the proposition that for any convergent sequence on any metric space, its any subsequence converges to the convergence of the sequence.
\(lim sup s = lim sup s^`\) does not necessarily hold.
3: Proof
Whole Strategy: Step 1: see an example that \(lim sup s\) exists but \(lim sup s^`\) does not exist; Step 2: suppose that \(lim sup s\) and \(lim sup s^`\) exist; Step 3: apply 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 and the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some infimums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the infimum of the 1st subset is equal to or smaller than the infimum of the 2nd subset; Step 4: see an example that \(lim sup s = lim sup s^`\) does not hold.
Step 1:
Let us see an example that \(lim sup s\) exists but \(lim sup s^`\) does not exist.
Let \(b_0.b_1 b_2 ...\) be the decimal expression of \(\sqrt{2}\).
Let \(J = \mathbb{N}\), \(S = \mathbb{Q}\), \(s: J \to S, j \mapsto b_0.b_1 b_2 ... b_{j / 2} \text{ when } j \text{ is even }; \mapsto 2 \text{ when } j \text{ is odd }\), \(J^` = \mathbb{N}\), and \(s^` = s \circ f\) where \(f: J^` \to J, j^` \mapsto 2 j^`\).
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) = 2\) for each \(m \in \mathbb{N} \setminus \{0\}\), and \(Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = 2\).
But \(Sup (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\})\) does not exist for each \(m \in \mathbb{N} \setminus \{0\}\), because \(s^`\) approaches \(\sqrt{2}\) from below and does not have any supremum in \(\mathbb{Q}\) (would have the supremum in \(\mathbb{R}\)), so, \(lim sup s^` = Inf (\{Sup (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\}) \vert m \in \mathbb{N} \setminus \{0\}\})\) does not exist.
Step 2:
Let us suppose that \(lim sup s\) and \(lim sup s^`\) exist.
Step 3:
For each \(m \in \mathbb{N} \setminus \{0\}\), \(\{s^` (J^`_{n^`}) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le {n^`}\} \subseteq \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\), because \(s^` (J^`_{n^`}) = s \circ f (J^`_{n^`})\) but \(f (J^`_{n^`}) = J_n\) where \(n^` \le n\) (refer to Note for the definition of subsequence of sequence), so, \(m \le n\), so, \(s^` (J^`_{n^`}) = s \circ f (J^`_{n^`}) = s (J_n) \in \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\).
\(Sup (\{s^` (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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.
Then, \(lim sup s^` = Inf (\{Sup (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\}) \vert m \in \mathbb{N} \setminus \{0\}\}) \le Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = lim sup s\), by the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some infimums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the infimum of the 1st subset is equal to or smaller than the infimum of the 2nd subset.
Step 4:
Let us see an example that \(lim sup s = lim sup s^`\) does not hold.
Let \(J = \mathbb{N}\), \(S = \mathbb{R}\), \(s: J \to S, j \mapsto - 1 \text{ when } j \text{ is even }; \mapsto \text{ 1 } \text{ when } j \text{ is odd }\), \(J^` = \mathbb{N}\), and \(s^` = s \circ f\) where \(f: J^` \to J, j^` \mapsto 2 j^`\).
Then, \(lim sup s = 1\), but \(lim sup s^` = - 1\), and \(lim sup s^` \lt lim sup s\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for sequence on partially-ordered set and subsequence, if limit inferior of sequence exists, limit inferior of subsequence does not necessarily exist, but if it exists, it is equal to or larger than limit inferior of sequence
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 sequence on any partially-ordered set and any subsequence, if the limit inferior of the sequence exists, the limit inferior of the subsequence does not necessarily exist, but if it exists, it is equal to or larger than the limit inferior 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 \(J \neq \emptyset\)
\(S\): \(\in \{\text{ the partially-ordered sets }\}\) with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(J^`\): \(\subseteq \mathbb{N}\), such that \(J^` \neq \emptyset\)
\(s^`\): \(= s \circ f\), \(\in \{\text{ the subsequences of } s \text{ with } f: J^` \to J\}\)
//
Statements:
(
\(\exists lim inf s\)
\(\lnot \implies\)
\(\exists lim inf s^`\)
)
\(\land\)
(
\(\exists lim inf s \land \exists lim inf s^`\)
\(\implies\)
\(lim inf s \le lim inf s^`\)
)
//
2: Note
Compare with the proposition that for any convergent sequence on any metric space, its any subsequence converges to the convergence of the sequence.
\(lim inf s = lim inf s^`\) does not necessarily hold.
3: Proof
Whole Strategy: Step 1: see an example that \(lim inf s\) exists but \(lim inf s^`\) does not exist; Step 2: suppose that \(lim inf s\) and \(lim inf s^`\) exist; Step 3: apply 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 and the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some supremums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the supremum of the 1st subset is equal to or smaller than the supremum of the 2nd subset; Step 4: see an example that \(lim inf s = lim inf s^`\) does not hold.
Step 1:
Let us see an example that \(lim inf s\) exists but \(lim inf s^`\) does not exist.
Let \(b_0.b_1 b_2 ...\) be the decimal expression of \(\sqrt{2}\).
Let \(J = \mathbb{N}\), \(S = \mathbb{Q}\), \(s: J \to S, j \mapsto - b_0.b_1 b_2 ... b_{j / 2} \text{ when } j \text{ is even }; \mapsto - 2 \text{ when } j \text{ is odd }\), \(J^` = \mathbb{N}\), and \(s^` = s \circ f\) where \(f: J^` \to J, j^` \mapsto 2 j^`\).
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) = - 2\) for each \(m \in \mathbb{N} \setminus \{0\}\), and \(Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = -2\).
But \(Inf (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\})\) does not exist for each \(m \in \mathbb{N} \setminus \{0\}\), because \(s^`\) approaches \(- \sqrt{2}\) from above and does not have any infimum in \(\mathbb{Q}\) (would have the infimum in \(\mathbb{R}\)), so, \(lim inf s^` = Sup (\{Inf (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\}) \vert m \in \mathbb{N} \setminus \{0\}\})\) does not exist.
Step 2:
Let us suppose that \(lim inf s\) and \(lim inf s^`\) exist.
Step 3:
For each \(m \in \mathbb{N} \setminus \{0\}\), \(\{s^` (J^`_{n^`}) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le {n^`}\} \subseteq \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\), because \(s^` (J^`_{n^`}) = s \circ f (J^`_{n^`})\) but \(f (J^`_{n^`}) = J_n\) where \(n^` \le n\) (refer to Note for the definition of subsequence of sequence), so, \(m \le n\), so, \(s^` (J^`_{n^`}) = s \circ f (J^`_{n^`}) = s (J_n) \in \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\).
\(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le Inf (\{s^` (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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.
Then, \(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) \le Sup (\{Inf (\{s^` (J^`_{n^`}) \vert n^` \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n^`\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = lim inf s^`\), by the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some supremums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the supremum of the 1st subset is equal to or smaller than the supremum of the 2nd subset.
Step 4:
Let us see an example that \(lim inf s = lim inf s^`\) does not hold.
Let \(J = \mathbb{N}\), \(S = \mathbb{R}\), \(s: J \to S, j \mapsto 1 \text{ when } j \text{ is even }; \mapsto \text{ - 1 } \text{ when } j \text{ is odd }\), \(J^` = \mathbb{N}\), and \(s^` = s \circ f\) where \(f: J^` \to J, j^` \mapsto 2 j^`\).
Then, \(lim inf s = - 1\), but \(lim inf s^` = 1\), and \(lim inf s \lt lim inf s^`\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for partially-ordered set and \(2\) subsets with same index set which have supremums, if for each index, element of 1st subset is equal to or smaller than element of 2nd subset, supremum of 1st subset is equal to or smaller than supremum of 2nd subset
Topics
About:
set
The table of contents of this article
Starting Context
-
The reader knows a definition of supremum of subset of partially-ordered set.
-
The reader admits the proposition that for any partially-ordered set and any subset, if the minimum of the subset exists, the minimum is the infimum of the subset, and if the maximum of the subset exists, the maximum is the supremum of the subset.
-
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.
Target Context
-
The reader will have a description and a proof of the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some supremums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the supremum of the 1st subset is equal to or smaller than the supremum of the 2nd subset.
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:
\(S\): \(\in \{\text{ the partially-ordered sets }\}\) with any partial ordering, \(\lt\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S_1\): \(= \{s_{1, j} \in S \vert j \in J\}\)
\(S_2\): \(= \{s_{2, j} \in S \vert j \in J\}\)
//
Statements:
\(\exists Sup (S_1) \land \exists Sup (S_2) \land \forall j \in J (s_{1, j} \le s_{2, j})\)
\(\implies\)
\(Sup (S_1) \le Sup (S_2)\)
//
2: Proof
Whole Strategy: Step 1: see that \(Ub (S_2) \subseteq Ub (S_1)\); Step 2: see that \(Min (Ub (S_1)) \le Min (Ub (S_2))\).
Step 1:
\(Ub (S_2) \subseteq Ub (S_1)\), because for each \(s \in Ub (S_2)\), for each \(j \in J\), \(s_{2, j} \le s\), so, \(s_{1, j} \le s_{2, j} \le s\), so, \(s \in Ub (S_1)\).
Step 2:
\(Sup (S_1) = Min (Ub (S_1)) \le Min (Ub (S_2)) = Sup (S_2)\), by the proposition that for any partially-ordered set and any subset, if the minimum of the subset exists, the minimum is the infimum of the subset, and if the maximum of the subset exists, the maximum is the supremum of the subset and 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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
