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
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of countably compact topological space.
- The reader knows a definition of compact topological space.
- The reader knows a definition of convergence of sequence on metric space.
- The reader knows a definition of subsequence of sequence.
- The reader admits the proposition that any metric space is compact if and only if it is countably compact.
- The reader admits the proposition that any metric space with the induced topology is 1st-countable.
- The reader admits the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.
- The reader admits 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.
- The reader admits 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.
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 \(lim s^` \neq lim \widetilde{s^`}\).
