description/proof of that for metric space and subset, closure of subset is compact iff each sequence into subset has subsequence that converges in closure of subset iff closure of subset is complete and for each positive real number, there is set of finite open balls of number-radius that covers closure of subset
Topics
About: metric space
About: topological 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 closure of subset of topological space.
- The reader knows a definition of compact subset of topological space.
- The reader knows a definition of sequence.
- The reader knows a definition of subsequence of sequence.
- The reader knows a definition of metric subspace.
- The reader knows a definition of complete metric space.
- The reader knows a definition of open ball around point on metric space.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
- The reader admits the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point.
- The reader admits the proposition that for any metric space with the induced topology, any closed ball is closed and contains but not necessarily equal the closure of the open ball with the same radius.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the 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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(S\): \(\subseteq M\) such that \(S \neq \emptyset\)
\(\overline{S}\): as the metric subspace
//
Statements:
1) \(\overline{S} \in \{\text{ the compact subsets }\}\)
\(\iff\)
2) \(\forall s: \mathbb{N} \to S (\exists s^` \in \{\text{ the subsequences of } s\}, \exists m \in \overline{S} (s^` \text{ converges to } m))\)
\(\iff\)
3)
(
\(\overline{S} \in \{\text{ the complete metric spaces }\}\)
\(\land\)
\(\forall r \in \mathbb{R} \text{ such that } 0 \lt r (\exists \{s_1, ..., s_{n_r}\} \subseteq S (\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B_{s_j, r}))\)
)
//
2: Note
As an immediate corollary, \(S\) can be taken to be \(M\), then, \(\overline{S} = M\), so, \(M\) is compact if and only if each sequence into \(M\) has a subsequence that converges in \(M\) (which means that \(M\) is sequentially compact) if and only if \(M\) is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers \(M\).
"open balls" can be replace by "closed balls", because if there is a set of such open balls for each \(r\), the set of the closed balls of the same radius covers \(\overline{S}\), and if there is a set of such closed balls, for each \(r\), there are an \(r - \epsilon\) such that \(0 \lt r - \epsilon\) and a set of such closed balls with radius \(r - \epsilon\), then, the corresponding open balls of radius \(r\) covers \(\overline{S}\).
2: Proof
Whole Strategy: Step 1: suppose 2); Step 2: see that \(\overline{S}\) is complete; Step 3: see that \(\forall r \in \mathbb{R} \text{ such that } 0 \lt r (\exists \{s_1, ..., s_{n_r}\} \subseteq S (\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B_{s_j, r}))\); Step 4: suppose 3); Step 5: see that 1) holds; Step 6: suppose 1); Step 7: see that 2) holds.
Step 1:
Let us suppose 2).
Step 2:
Step 2 Strategy: Step 2-1: take any Cauchy sequence into \(\overline{S}\), \(f\); Step 2-2: take a sequence into \(S\), \(s\), such that \(dist (s_j, f_j) \lt 2^{-j}\); Step 2-3: see that there is a subsequence of \(s\), \(s^` = s \circ g\), that converges to \(m \in \overline{S}\) and \(f \circ g\) converges to \(m\).
Let us see that \(\overline{S}\) is complete.
Step 2-1:
Let \(f: \mathbb{N} \to \overline{S}\) be any Cauchy sequence.
Step 2-2:
Let us take a sequence, \(s: \mathbb{N} \to S\), such that for each \(j \in \mathbb{N}\), \(dist (s_j, f_j) \lt 2^{-j}\), which is possible, because as \(f_j \in \overline{S}\), \(B_{f_j, 2^{-j}} \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Step 2-3:
By 2), there is a subsequence of \(s\), \(s^`: \mathbb{N} \to \mathbb{N} \to S = s \circ g\), that converges to \(m \in \overline{S}\).
Let us see that \(f \circ g\) converges to \(m\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(2^{- g (n)} \lt \epsilon / 2\).
There is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(dist (m, s_{g (n)}) \lt \epsilon / 2\), because \(s^`\) converges to \(m\).
Then, for each \(n \in \mathbb{N}\) such that \(Max (\{N_1, N_2\}) \lt n\), \(dist (m, f_{g (n)}) \le dist (m, s_{g (n)}) + dist (s_{g (n)}, f_{g (n)}) \lt \epsilon / 2 + 2^{- g (n)} \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
So, \(f \circ g\) converges to \(m\).
By the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point, \(f\) converges to \(m\).
So, \(\overline{S}\) is complete.
Step 3:
Let us see that \(\forall r \in \mathbb{R} \text{ such that } 0 \lt r (\exists \{s_1, ..., s_{n_r}\} \subseteq S (\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B_{s_j, r}))\).
Let \(r \in \mathbb{R}\) be any such that \(0 \lt r\).
Let us suppose that there was no finite subset of \(S\), \(\{s_1, ..., s_{n_r}\} \subseteq S\), such that \(\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B'_{s_j, r}\), where \(B'_{s_j, r}\) was the closed ball around \(s_j\) of radius \(r\).
\(S\) would be an infinite set, because otherwise, taking \(S\) as the subset would cover \(\overline{S}\): for each \(m \in \overline{S}\), there would be an \(s_j \in B_{m, r} \cap S\), then, \(m \in B'_{s_j, r}\), a contradiction.
Note that \(\overline{S}\)'s being covered equals \(S\)'s being covered, because if \(\overline{S}\) was covered, \(S\) would be covered, and if \(S\) was covered, \(S \subseteq \cup_{j \in \{1, ..., n_r\}} B'_{s_j, r}\), but as \(\cup_{j \in \{1, ..., n_r\}} B'_{s_j, r}\) was closed, by the proposition that for any metric space with the induced topology, any closed ball is closed and contains but not necessarily equal the closure of the open ball with the same radius, it would contain \(\overline{S}\), because \(\overline{S}\) was the intersection of all the closed subsets that contained \(S\). So, as \(\overline{S}\) was not covered, \(S\) was not covered.
Let us take any \(s_0 \in S\).
Let us take any \(s_1 \in S \setminus B'_{s_0, r}\), which would be possible, because \(B'_{s_0, r}\) did not cover \(S\).
Inductively, when we had \(\{s_0, ..., s_{n - 1}\}\), there would be an \(s_n \in S \setminus \cup_{j \in \{0, ..., n - 1\}} B'_{s_j, r}\), because \(\{B'_{s_j, r} \vert j \in \{0, ..., n - 1\}\}\) did not cover \(S\).
So, we would have an infinite sequence, \((s_0, s_1, ...)\), into \(S\).
But that sequence would not have any convergent subsequence, because for each \(j \in \mathbb{N}\), for each \(l \in \mathbb{N}\) such that \(l \lt j\), \(r \lt dist (s_l, s_j)\), because \(s_j \notin B'_{s_l, r}\), which would mean that for each \(j, l \in \mathbb{N}\) such that \(j \neq l\), \(r \lt dist (s_l, s_j)\), because whether \(s_j \lt s_l\) or \(s_l \lt s_j\), \(r \lt dist (s_l, s_j) = dist (s_j, s_l)\).
That would be a contradiction against 2).
So, for each \(r\), there is a finite subset of \(S\), \(\{s_1, ..., s_{n_r}\} \subseteq S\), such that \(\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B'_{s_j, r}\).
Then, for each \(r\), there is an \(r - \epsilon \in \mathbb{R}\) such that \(0 \lt r - \epsilon\), and there is a finite subset of \(S\), \(\{s_1, ..., s_{n_{r - \epsilon}}\} \subseteq S\), such that \(\overline{S} \subseteq \cup_{j \in \{1, ..., n_{r - \epsilon}\}} B'_{s_j, r - \epsilon}\), and \(\overline{S} \subseteq \cup_{j \in \{1, ..., n_{r - \epsilon}\}} B_{s_j, r}\), because \(B'_{s_j, r - \epsilon} \subseteq B_{s_j, r}\).
So, \(\forall r \in \mathbb{R} \text{ such that } 0 \lt r (\exists \{s_1, ..., s_{n_r}\} \subseteq S (\overline{S} \subseteq \cup_{j \in \{1, ..., n_r\}} B_{s_j, r}))\).
Step 4:
Let us suppose 3).
Step 5:
Let us see that 1) holds.
Let \(\{U_j \vert j \in J\}\), where \(J\) is any index set, be any open cover of \(\overline{S}\).
Let us suppose that there was no finite subcover.
By 3), for each \(l \in \mathbb{N} \setminus \{0\}\), there would be a \(\{s_{l, 1}, ..., s_{l, n_l}\} \subseteq S\) such that \(\{B_{s_{l, j}, 2^{- l}} \vert j \in \{1, ..., n_l\}\}\) covers \(\overline{S}\).
So, especially, there would be a \(\{s_{1, 1}, ..., s_{1, n_1}\} \subseteq S\) such that \(\{B_{s_{1, j}, 2^{- 1}} \vert j \in \{1, ..., n_1\}\}\) covered \(\overline{S}\).
There would be a \(B_{s_{1, j}, 2^{- 1}}\) such that \(B_{s_{1, j}, 2^{- 1}} \cap \overline{S}\) was not covered by any finite \(U_j\) s, because otherwise, while \(\overline{S} = (\cup_{j \in \{1, ..., n_1\}} B_{s_{1, j}, 2^{- 1}}) \cap \overline{S} = \cup_{j \in \{1, ..., n_1\}} (B_{s_{1, j}, 2^{- 1}}) \cap \overline{S})\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(\overline{S}\) would be covered by some finite \(U_j\) s, a contradiction, and let \(p_1 := s_{1, j}\).
Let us suppose there was a \(B_{s_{n - 1, j}, 2^{- (n - 1)}}\) such that \(B_{s_{n - 1, j}, 2^{- (n - 1)}} \cap \overline{S}\) was not covered by any finite \(U_j\) s, with \(p_{n - 1} := s_{n - 1, j}\).
Then, while there would be some finite \(B_{s_{n, j}, 2^{- n}}\) s that intersected \(B_{p_{n - 1}, 2^{- (n - 1)}} \cap \overline{S}\), because otherwise, \(\{B_{s_{n, j}, 2^{- n}}\}\) would not cover \(\overline{S}\), there would be a \(B_{s_{n, j}, 2^{- n}}\) among them such that \(B_{s_{n, j}, 2^{- n}} \cap \overline{S}\) was not covered by any finite \(U_j\) s, because otherwise, \(B_{p_{n - 1}, 2^{- (n - 1)}} \cap \overline{S}\) would be covered by some finite \(U_j\) s, and let \(p_n := s_{n, j}\).
Thus, inductively, we got a sequence, \((p_1, p_2, ...)\), such that \(B_{p_n, 2^{- n}} \cap \overline{S}\) was not covered by any finite \(U_j\) s and \(B_{p_{n - 1}, 2^{- (n - 1)}} \cap B_{p_n, 2^{- n}} \cap \overline{S} \neq \emptyset\).
\(dist (p_{n - 1}, p_n) \lt 2^{- (n - 2)}\), because \(B_{p_{n - 1}, 2^{- (n - 1)}} \cap B_{p_n, 2^{- n}} \neq \emptyset\), which would mean that there was a \(p \in B_{p_{n - 1}, 2^{- (n - 1)}} \cap B_{p_n, 2^{- n}}\), and \(dist (p_{n - 1}, p_n) \le dist (p_{n - 1}, p) + dist (p, p_n) \lt 2^{- (n - 1)} + 2^{- n} \lt 2^{- (n - 1)} + 2^{- (n - 1)} = 2^{- (n - 2)}\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There would be an \(N \in \mathbb{N}\) such that \(2^{- (N - 2)} \lt \epsilon\).
For each \(a, b \in \mathbb{N}\) such that \(N \lt a, b\) and \(a \lt b\), \(dist (p_a, p_b) \le dist (p_a, p_{a + 1}) + ... + dist (p_{b - 1}, p_b) \lt 2^{- (a - 1)} + ... + 2^{- (b - 2)} \lt 2^{- (a - 1)} + 2^{- a} + ... = 2^{- (a - 1)} (1 + 1 / 2 + (1 / 2)^2 + ...) = 2^{- (a - 1)} 1 / (1 - 1 / 2) = 2^{- (a - 2)} \lt 2^{- (N - 2)} \lt \epsilon\).
So, \((p_1, p_2, ...)\) would be a Cauchy sequence into \(S\), so, a Cauchy sequence into \(\overline{S}\).
By 3), \(\overline{S}\) would be complete, and so, the sequence would converge to an \(m \in \overline{S}\).
\(m \in U_j\) for a \(j\), because \(\{U_j\}\) covered \(\overline{S}\).
There would be an \(r \in \mathbb{R}\) such that \(0 \lt r\) and \(B_{m, r} \subseteq U_j\), because \(U_j\) was an open neighborhood of \(m\).
As \((p_1, p_2, ...)\) converged to \(m\), there would be an \(n \in \mathbb{N}\) such that \(dist (m, p_n) \lt r / 2\) and \(2^{- n} \lt r / 2\).
Then, \(B_{p_n, 2^{- n}} \subseteq B_{m, r}\), because for each \(p \in B_{p_n, 2^{- n}}\), \(dist (m, p) \le dist (m, p_n) + dist (p_n, p) \lt r / 2 + 2^{- n} \lt r / 2 + r / 2 = r\).
So, \(B_{p_n, 2^{- n}} \subseteq B_{m, r} \subseteq U_j\), a contradiction against that \(B_{p_n, 2^{- n}} \cap \overline{S}\) was not covered by any finite \(U_j\) s.
So, \(\{U_j \vert j \in J\}\) has a finite subcover.
So, \(\overline{S}\) is compact.
Step 6:
Let us suppose 1).
Step 7:
Let us see that 2) holds.
\(\overline{S}\) is a compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
\(\overline{S}\) is countably compact, because any compact topological space is countably compact, obviously.
Let \(s\) be any sequence into \(S\).
\(s\) is a sequence into \(\overline{S}\).
There is a subsequence of \(s\) that converges to an \(m \in \overline{S}\), 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.
That means that 2) holds.
