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 iff convergences of component maps 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
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 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.
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
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, .., b_d\}\)
\(t\): \(\in T\)
\(f\): \(: T \setminus \{t\} \to V, t' \mapsto f^j (t') b_j\)
//
Statements:
(
\(\exists lim_{t' \to t} f (t')\)
\(\iff\)
\(\forall j \in \{1, ..., d\} (\exists lim_{t' \to t} f^j (t'))\)
)
\(\implies\)
\(lim_{t' \to t} f (t') = lim_{t' \to t} f^j (t') b_j\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(lim_{t' \to t} f (t')\) exists and denote it as \(v = v^j b_j\); Step 2: see that \(lim_{t' \to t} f^j (t')\) exists and equals \(v^j\); Step 3: suppose that \(lim_{t' \to t} f^j (t')\) exists and denote it as \(v^j\); Step 4: see that \(lim_{t' \to t} f (t')\) exists and equals \(v^j b_j\).
Step 0:
Note that \(V\) is Hausdorff.
Note that each of supposing the existence of \(lim_{t' \to t} f (t')\) and 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 \(lim_{t' \to t} f (t')\) exists and denote it as \(v = v^j b_j\).
Step 2:
Let us see that \(lim_{t' \to t} f^j (t')\) exists and equals \(v^j\).
For each neighborhood of \(v\), \(N_v \subseteq V\), there is an open ball, \(B_{v, \epsilon} \subseteq V\), such that \(B_{v, \epsilon} \subseteq N_v\), by the definition of canonical topology: \(B_{v, \epsilon}\) means that the corresponding subset of \(\mathbb{R}^d\) is the \(\epsilon\)-'open ball'.
There is a neighborhood of \(t\), \(U_t \subseteq T\), such that \(f (U_t \setminus \{t\}) \subseteq B_{v, \epsilon} \subseteq N_v\), because \(f\) converges with respect to \(t\).
That means that for each \(t' \in U_t \setminus \{t\}\), \(f (t') = f^j (t') b_j \in B_{v, \epsilon}\), which means that \(\sum_{j \in \{1, ..., d\}} (f^j (t') - v^j)^2 \lt \epsilon^2\).
So, for each \(j\), \((f^j (t') - v^j)^2 \lt \epsilon^2\).
That means that the map, \(f^j: T_1 \setminus \{t\} \to \mathbb{R}\), has the convergence with respect to \(t\) with \(lim_{t' \to t} f^j (t') = v^j\).
So, \(lim_{t' \to t} f (t') = v = v^j b_j = lim_{t' \to t} f^j (t') b_j\).
Step 3:
Let us suppose that for each \(j \in \{1, ..., d\}\), \(lim_{t' \to t} f^j (t')\) exists and denote it as \(v^j\).
Step 4:
Let us see that \(lim_{t' \to t} f (t')\) exists and equals \(v^j b_j\).
Let us define \(v := v^j b_j\).
For each neighborhood of \(v\), \(N_v \subseteq V\), there is an open ball, \(B_{v, \epsilon} \subseteq V\), such that \(B_{v, \epsilon} \subseteq N_v\).
For each \(j\), let \(B_{v^j, \epsilon / \sqrt{d}} \subseteq \mathbb{R}\) be the open ball around \(v^j\).
For each \(j\), there is a neighborhood of \(t\), \(U_{j, t} \subseteq T\), such that \(f^j (U_{j, t} \setminus \{t\}) \subseteq B_{v^j, \epsilon / \sqrt{d}}\), because \(f^j\) converges with respect to \(t\).
Let \(U_t := \cap_{j \in \{1, ..., d\}} U_{j, t} \subseteq T\), which is a neighborhood of \(t\).
For each \(j\), \(f^j (U_t \setminus \{t\}) \subseteq B_{v^j, \epsilon / \sqrt{d}}\), which means that for each \(t' \in U_t \setminus \{t\}\), \((f^j (t') - v^j)^2 \lt \epsilon^2 / d\).
Then, for each \(t' \in U_t \setminus \{t\}\), \(f (t') = f^j (t') b_j \in B_{v, \epsilon}\), because \(\sum_{j \in \{1, ..., d\}} (f^j (t') - v^j)^2 \lt \sum_{j \in \{1, ..., d\}} \epsilon^2 / d = \epsilon^2\).
That means that \(f (U_t \setminus \{t\}) \subseteq B_{v, \epsilon} \subseteq N_v\).
That means that \(f\) has the convergence with respect to \(t\) with \(lim_{t' \to t} f (t') = v\).
So, \(lim_{t' \to t} f (t') = v = v^j b_j = lim_{t' \to t} f^j (t') b_j\).
