description/proof of that for map from topological space minus point into finite-dimensional real vectors space with canonical topology, convergence of map w.r.t. point exists if convergences of coefficients w.r.t. constant vectors w.r.t. point exist, and then, convergence is expressed with convergences
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of canonical topology for finite-dimensional real vectors 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 map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if the convergences of the coefficients with respect to any constant vectors with respect to the point exist, and then, the convergence is expressed with the convergences.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } \mathbb{R} \text{ vectors spaces }\}\), with the canonical topology
\(\{v_1, ..., v_n\}\): \(\subseteq V\)
\(t\): \(\in T\)
\(f\): \(: T \setminus \{t\} \to V, t' \mapsto f^j (t') v_j\)
//
Statements:
(
\(\forall j \in \{1, ..., n\} (\exists lim_{t' \to t} f^j (t'))\)
\(\implies\)
\(\exists lim_{t' \to t} f (t')\)
)
\(\implies\)
\(lim_{t' \to t} f (t') = lim_{t' \to t} f^j (t') v_j\)
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2: Note
The reverse does not necessarily hold: for example, when \(T = \mathbb{R}\), \(t = 0\), and \(f (t') = cos (1 / t') v - cos (1 / t') v\), \(f (t') = 0\), so, \(f\) converges with respect to \(0\), but \(cos (1 / t')\) does not converge with respect to \(0\).
Compare with the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences.
3: Proof
Whole Strategy: Step 1: suppose that \(lim_{t' \to t} f^j (t')\) exists and denote it as \(v^j\); Step 2: see that \(lim_{t' \to t} f (t')\) exists and equals \(v^j v_j\).
Step 0:
Note that \(V\) is Hausdorff.
Note that supposing the existence of \(lim_{t' \to t} f^j (t')\) implies that \(\{t\} \subseteq T\) is not open, because otherwise, the convergences would not be unique: refer to Note for the definition of convergence of map from topological space minus point into topological space with respect to point.
So, we do not need to worry about the uniqueness of convergences.
Step 1:
Let us suppose that for each \(j \in \{1, ..., n\}\), \(lim_{t' \to t} f^j (t')\) exists and denote it as \(v^j\).
Step 2:
Let \(B = \{b_1, .., b_d\}\) be any basis for \(V\).
\(v_j = {v_j}^l b_l\).
\(f (t') = f^j (t') v_j = f^j (t') {v_j}^l b_l\).
As \(f^j (t')\) converges with respect to \(t\), for each \(l \in \{1, ..., d\}\), \(f^j (t') {v_j}^l\) converges to \(v^j {v_j}^l\) with respect to \(t\).
By the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences, \(lim_{t' \to t} f (t')\) exists and equals \(v^j {v_j}^l b_l\).
\(v^j {v_j}^l b_l = v^j v_j\).
So, \(lim_{t' \to t} f (t') = v^j v_j = lim_{t' \to t} f^j (t') v_j\).
