description/proof of that for map from metric space minus point into metric space, iff for each sequence on domain that converges to point, its image converges to codomain point, map converges w.r.t. point to codomain point
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of sequence on metric space.
- The reader knows a definition of convergence of map from topological space minus point into topological space with respect to point.
- 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.
- The reader admits the proposition that the topological space induced by any metric is Hausdorff.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any metric space minus any point into any metric space, if and only if for each sequence on the domain that converges to the point, its image converges to any same codomain point, the map converges with respect to the point to the codomain point.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(M_2\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(m_1\): \(\in M_1\), such that \(\{m_1\} \subseteq M_1 \notin \{\text{ the open subsets of } M_1\}\)
\(m_2\): \(\in M_2\)
\(f\): \(: M_1 \setminus \{m_1\} \to M_2\)
//
Statements:
\(\forall s: J \to M_1 \setminus \{m_1\} \in \{\text{ the sequences that converge to } m_1\} (f \circ s: J \to M_2 \in \{\text{ the sequences that converge to } m_2\})\)
\(\iff\)
\(lim_{m'_1 \to m_1} f (m'_1) = m_2\)
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2: Note
The reason why the case that \(\{m_1\} \subseteq M_1\) is open is excluded is that for that case, each point of \(M_2\) will be a convergence of \(f\) with respect to \(m_1\), as is mentioned in Note for the definition of convergence of map from topological space minus point into topological space with respect to point, and so, \(lim_{m'_1 \to m_1} f (m'_1)\) will not make sense in general and also there may not be any sequence that converges to \(m_1\), so, the condition can become vacuous.
The mechanism of this proposition is essentially the same with that of the proposition that for any map between any metric spaces and any domain point, if and only if for each sequence on the domain that converges to the point, its image converges to the image of the point, the map is continuous at the point: the difference is that for this proposition, as the point is not on the domain, "continuous at the point" does not make sense.
3: Proof
Whole Strategy: Step 1: suppose that \(f \circ s\) converges to \(m_2\) for each \(s\); Step 2: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\); Step 3: suppose that \(f\) converges to \(m_2\) with respect to \(m_1\); Step 4: see that \(f \circ s\) converges to \(m_2\) for each \(s\).
Step 1:
Let us suppose that for each \(s: J \to M_1\) that converges to \(m_1\), \(f \circ s: J \to M_2\) converges to \(m_2\).
Step 2:
Step 2 Strategy: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), as follows: Step 2-1: take the set of the infinite sequences that converge to \(m_1\), \(\{s_l \vert l \in L\}\), for each \(l \in L\), take \(N_l\) such that for each \(N_l \lt n\), \(f \circ s_l (n) \in B_{m_2, \epsilon}\), take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\})\), and take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), where \(L^` \subseteq L\); Step 2-2: see that \(0 \lt \delta\); Step 2-3: see that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\); Step 2-4: see that the convergence is unique.
Step 2-1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(\{s_l: J_l \to M_1 \setminus \{m_1\} \vert l \in L\}\) be the set of the infinite sequences that converge to \(m_1\), where \(L\) is a possibly uncountable index set and \(J_l \subseteq \mathbb{N}\).
\(L\) is not empty, because there is at least \(s_l: \mathbb{N} \to M_1 \setminus \{m_1\}\) such that \(s_l (n) \in B_{m_1, (1 / 2)^n} \setminus \{m_1\}\): as \(\{m_1\} \subseteq M_1\) is not open, there is such an \(s_l (n)\), because if \(B_{m_1, (1 / 2)^n} \setminus \{m_1\} = \emptyset\), \(B_{m_1, (1 / 2)^n} = \{m_1\}\), which would imply that \(\{m_1\}\) was open.
Let \(l \in L\) be any.
There is an \(N_l \subseteq \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_l \lt n\), \(f \circ s_l ({J_l}_n) \in B_{m_2, \epsilon}\), because \(f \circ s_l\) converges to \(m_2\).
Let \(L^` := \{l \in L \vert \{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\} \neq \emptyset\}\).
When \(L^` = \emptyset\), let \(\delta \in \mathbb{R}\) be any such that \(0 \lt \delta\). Then, \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because if there was an \(m'_1 \in B_{m_1, \delta} \setminus \{m_1\}\) such that \(f (m'_1) \notin B_{m_2, \epsilon}\), there would be an \(s_l: \mathbb{N} \to M_1 \setminus \{m_1\}\) such that \(s_l (0) = m'_1\) and \(s_l (n) \in B_{m_1, (1 / 2)^n} \setminus \{m_1\}\) for each \(0 \lt n\), which would converge to \(m_1\) and \(l \in L^`\): \(1 \le N_l\) and \(s_l (0) \notin B_{m_2, \epsilon}\), a contradiction.
Let us suppose that \(L^` \neq \emptyset\), hereafter.
Let \(l \in L^`\) be any.
Let us take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \in \mathbb{N} \text{ such that } n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\})\), which exists, because \(\{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\}\) is not empty.
Let us take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), which exists because \(\{\delta_l \vert l \in L^`\}\) is not empty and lower bounded by \(0\).
Step 2-2:
While \(0 \le \delta\), let us see that \(0 \lt \delta\).
Let us suppose that \(0 = \delta\).
Let us take \(s: \mathbb{N} \to M_1 \setminus \{m_1\}\) as follows.
There would be a \(\delta_{l_0} \lt (1 / 2)^0\), 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: \(\delta \lt (1 / 2)^0\).
That would means that \(f \circ s_{l_0} ({J_{l_0}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_0} ({J_{l_0}}_n), m_1) = \delta{l_0}\) for an \(n \le N_{l_0}\), and let \(s (0) = s_{l_0} ({J_{l_0}}_n)\), so, \(dist (s (0), m_1) = \delta{l_0} \lt (1 / 2)^0\) and \(f \circ s (0) \notin B_{m_2, \epsilon}\).
Then, there would be a \(\delta_{l_1} \lt (1 / 2)^1\), as before.
That would mean that \(f \circ s_{l_1} ({J_{l_1}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_1} ({J_{l_1}}_n), m_1) = \delta{l_1}\) for an \(n \le N_{l_1}\), and let \(s (1) = s_{l_1} ({J_{l_1}}_n)\), so, \(dist (s (1), m_1) = \delta{l_1} \lt (1 / 2)^1\) and \(f \circ s (1) \notin B_{m_2, \epsilon}\).
In general, there would be a \(\delta_{l_j} \lt (1 / 2)^j\), as before.
That would mean that \(f \circ s_{l_j} ({J_{l_j}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_j} ({J_{l_j}}_n), m_1) = \delta{l_j}\) for an \(n \le N_{l_j}\), and let \(s (j) = s_{l_j} ({J_{l_j}}_n)\), so, \(dist (s (j), m_1) = \delta{l_j} \lt (1 / 2)^j\) and \(f \circ s (j) \notin B_{m_2, \epsilon}\).
Thus \(s\) had been defined.
\(s\) converged to \(m_1\), because \(dist (s (j), m_1) \lt (1 / 2)^j\).
But \(f \circ s (j) \notin B_{m_2, \epsilon}\) for each \(j \in \mathbb{N}\), so, \(f \circ s\) would not converge to \(m_2\), a contradiction against the supposition.
So, \(0 \lt \delta\).
Step 2-3:
Then, \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because if there was a \(m'_1 \in B_{m_1, \delta} \setminus \{m_1\}\) such that \(f (m'_1) \notin B_{m_2, \epsilon}\), there would be an \(s_l\) such that \(s_l ({J_l}_{N_l}) = m'_1\), which would mean that \(\delta \le \delta_l \le dist (m_1, m'_1) \lt \delta\), a contradiction.
That means that \(f\) converges to \(m_2\) with respect to \(m_1\).
Step 2-4:
The convergence is unique, because \(M_2\) is Hausdorff, by the proposition that the topological space induced by any metric is Hausdorff, and Note for the definition of convergence of map from topological space minus point into topological space with respect to point applies.
So, \(lim_{m'_1 \to m_1} f (m'_1) = m_2\).
Step 3:
Let us suppose that \(f\) converges to \(m_2\) with respect to \(m_1\).
Step 4:
Let \(s: J \to M_1 \setminus \{m_1\}\) be any sequence that converges to \(m_1\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because \(f\) converges to \(m_2\) with respect to \(m_1\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(s (J_n) \in B_{m_1, \delta} \setminus \{m_1\}\), because \(s\) converges to \(m_1\).
Then, for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(f \circ s (J_n) \in B_{m_2, \epsilon}\).
That means that \(f \circ s\) converges to \(m_2\).
