description/proof of that for net with directed index set into finite-dimensional real or complex vectors space with canonical topology, convergence exists iff convergences of components 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 convergence of net with directed index set.
- The reader knows a definition of canonical topology for finite-dimensional real vectors space.
- The reader knows a definition of canonical topology for finite-dimensional complex vectors space.
- The reader admits the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
Target Context
- The reader will have a description and a proof of the proposition that for any net with any directed index set into any finite-dimensional real or complex vectors space with the canonical topology, the convergence exists if and only if the convergences of the components (with respect to any basis) 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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with the canonical topology
\(J\): \(\in \{\text{ the directed index sets }\}\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, .., b_d\}\)
\(s\): \(: J \to V, t \mapsto s^j (t) b_j\), \(\in \{\text{ the nets with } J\}\)
//
Statements:
(
\(\exists lim s\)
\(\iff\)
\(\forall j \in \{1, ..., d\} (\exists lim s^j)\)
)
\(\land\)
(
\(\exists lim s\)
\(\implies\)
\(lim s = (lim s^j) b_j\)
)
//
\(s^j: J \to F\) is regarded to be the net with the direct index set with \(F\) regarded as the Euclidean topological space or the complex Euclidean topological space.
2: Proof
Whole Strategy: Step 1: suppose that \(lim s\) exists and denote it as \(v = v^j b_j\); Step 2: see that \(lim s^j\) exists and equals \(v^j\); Step 3: suppose that \(lim s^j\) exists and denote it as \(v^j\); Step 4: see that \(lim s\) exists and equals \(v^j b_j\).
Step 0:
Let the topology of \(V\) be defined based on a basis, \(B = \{b_1, ..., b_d\}\), but in fact, the topology does not depend on the choice of the basis, as is mentioned in the definition of canonical topology for finite-dimensional real vectors space or the definition of canonical topology for finite-dimensional complex vectors space.
When \(F = \mathbb{R}\), there is the canonical homeomorphism, \(f: V \to \mathbb{R}^d\).
When \(F = \mathbb{C}\), there are the canonical homeomorphism, \(f': V \to \mathbb{C}^d\), the canonical homeomorphism, \(f'': \mathbb{R}^{2 d} \to \mathbb{C}^d\), and the canonical homeomorphism, \(f = f''^{-1} \circ f': V \to \mathbb{R}^{2 d}\).
When we talk of '\(\epsilon\)-'open ball' around \(v \in V\), \(B_{v, \epsilon} \subseteq V\)', \(B_{v, \epsilon} := f^{-1} (B_{f (v), \epsilon})\), where \(B_{f (v), \epsilon} \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\) is the \(\epsilon\)-'open ball' around \(f (v)\) on \(\mathbb{R}^d\) or \(\mathbb{R}^{2 d}\).
\(B_{v, \epsilon}\) is an open neighborhood of \(v\) on \(V\), because \(B_{f (v), \epsilon}\) is an open neighborhood of \(f (v)\) on \(\mathbb{R}^d\) or \(\mathbb{R}^{2 d}\) and \(f\) is a homeomorphism.
Note that \(V\), which is homeomorphic to \(\mathbb{R}^d\) or \(\mathbb{R}^{2 d}\), and \(F\), which is \(\mathbb{R}\) or \(\mathbb{C}\), are Hausdorff.
When a convergence of \(s\) or \(s^j\) exists, the convergence is unique, by the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
So, we do not need to worry about the uniqueness of convergences.
Step 1:
Let us suppose that \(lim s\) exists and denote it as \(v = v^j b_j\).
Step 2:
Let us see that \(lim s^j\) exists and equals \(v^j\).
Let us suppose that \(F = \mathbb{R}\).
Let \(N_{v^j} \subseteq \mathbb{R}\) be any neighborhood of \(v^j\).
There is an \(\epsilon\)-'open ball' around \(v^j\), \(B_{v^j, \epsilon} \subseteq \mathbb{R}\), such that \(B_{v^j, \epsilon} \subseteq N_{v^j}\).
For the \(\epsilon\)-'open ball' around \(v\), \(B_{v, \epsilon} \subseteq V\), there is a \(t_0 \in J\) such that for each \(t \in J\) such that \(t_0 \le t\), \(s (t) \in B_{v, \epsilon}\), because \(s\) converges to \(v\).
That means that \(f (s (t)) = (s^1 (t), ..., s^d (t)) \in f (B_{v, \epsilon}) = B_{f (v), \epsilon}\).
That means that \(\sum_{j \in \{1, ..., d\}} (s^j (t) - v^j)^2 \lt \epsilon^2\).
So, for each \(j \in \{1, ..., d\}\), \((s^j (t) - v^j)^2 \le \sum_{j \in \{1, ..., d\}} (s^j (t) - v^j)^2 \lt \epsilon^2\).
So, for each \(t \in J\) such that \(t_0 \le t\), \(s^j (t) \in B_{v^j, \epsilon} \subseteq N_{v^j}\), which means that \(lim s^j = v^j\).
So, \(lim f = v = v^j b_j = (lim s^j) b_j\).
Let us suppose that \(F = \mathbb{C}\).
Let \(f''': \mathbb{R}^2 \to \mathbb{C}\) be the canonical homeomorphism.
Let \(N_{v^j} \subseteq \mathbb{C}\) be any neighborhood of \(v^j\).
\(f'''^{-1} (N_{v^j}) \subseteq \mathbb{R}^2\) is a neighborhood of \(f'''^{-1} (v^j)\).
There is an \(\epsilon\)-'open ball' around \(f'''^{-1} (v^j)\), \(B_{f'''^{-1} (v^j), \epsilon} \subseteq \mathbb{R}^2\), such that \(B_{f'''^{-1} (v^j), \epsilon} \subseteq f'''^{-1} (N_{v^j})\).
\(f''' (B_{f'''^{-1} (v^j), \epsilon}) \subseteq N_{v^j}\).
For the \(\epsilon\)-'open ball' around \(v\), \(B_{v, \epsilon} \subseteq V\), there is a \(t_0 \in J\) such that for each \(t \in J\) such that \(t_0 \le t\), \(s (t) \in B_{v, \epsilon}\), because \(s\) converges to \(v\).
That means that \(f (s (t)) = (Re (s^1 (t)), Im (s^1 (t)), ..., Re (s^d (t)), Im (s^d (t))) \in f (B_{v, \epsilon}) = B_{f (v), \epsilon}\).
That means that \(\sum_{j \in \{1, ..., d\}} ((Re (s^j (t)) - Re (v^j))^2 + (Im (s^j (t)) - Im (v^j))^2) \lt \epsilon^2\).
So, for each \(j \in \{1, ..., d\}\), \((Re (s^j (t)) - Re (v^j))^2 + (Im (s^j (t)) - Im (v^j))^2 \le \sum_{j \in \{1, ..., d\}} ((Re (s^j (t)) - Re (v^j))^2 + (Im (s^j (t)) - Im (v^j))^2) \lt \epsilon^2\).
That means that \(f'''^{-1} (s^j (t)) \in B_{f'''^{-1} (v^j), \epsilon}\), so, for each \(t \in J\) such that \(t_0 \le t\), \(s^j (t) \in f''' (B_{f'''^{-1} (v^j), \epsilon}) \subseteq N_{v^j}\), which means that \(lim s^j = v^j\).
So, \(lim s = v = v^j b_j = (lim s^j) b_j\).
Step 3:
Let us suppose that for each \(j \in \{1, ..., d\}\), \(lim s^j\) exists and let us denote it as \(v^j\).
Step 4:
Let us see that \(lim s\) exists and equals \(v^j b_j\).
Let us define \(v := v^j b_j\).
Let \(N_v \subseteq V\) be any neighborhood of \(v\).
There is an \(\epsilon\)-'open ball' around \(v\), \(B_{v, \epsilon} \subseteq V\), such that \(B_{v, \epsilon} \subseteq N_v\): \(f (N_v) \subseteq \mathbb{R}^d\) is a neighborhood of \(f (v)\), there is a \(B_{f (v), \epsilon} \subseteq f (N_v)\), and \(B_{v, \epsilon} := f^{-1} (B_{f (v), \epsilon}) \subseteq N_v\).
Let us suppose that \(F = \mathbb{R}\).
For each \(j \in \{1, ..., d\}\), let \(B_{v^j, \epsilon / \sqrt{d}} \subseteq \mathbb{R}\) be the '\(\epsilon / \sqrt{d}\)'-'open ball' around \(v^j\).
There is a \(t_j \in J\), such that for each \(t \in J\) such that \(t_j \le t\), \(s^j (t) \in B_{v^j, \epsilon / \sqrt{d}}\), because \(s^j\) converges to \(v^j\).
Let \(t_0 \in J\) be any such that \(t_1, ..., t_d \le t_0\), which exists, because there is a \(t_{1, 2} \in J\) such that \(t_1, t_2 \le t_{1, 2}\), then, there is a \(t_{1, 2, 3} \in J\) such that \(t_{1, 2}, t_3 \le t_{1, 2, 3}\), which means that \(t_1, t_2, t_3 \le t_{1, 2, 3}\), ..., and then, there is a \(t_{1, ..., d} \in J\) such that \(t_{1, ..., d - 1}, t_d \le t_{1, ..., d}\), which means that \(t_1, ..., t_d \le t_{1, ..., d}\), and we can take \(t_0 := t_{1, ..., d}\).
For each \(j\), for each \(t \in J\) such that \(t_0 \le t\), \(s^j (t) \in B_{v^j, \epsilon / \sqrt{d}}\), which means that \((s^j (t) - v^j)^2 \lt \epsilon^2 / d\), so, \(\sum_{j \in \{1, ..., d\}} (s^j (t) - v^j)^2 \lt \sum_{j \in \{1, ..., d\}} \epsilon^2 / d = \epsilon^2\), which means that \(f (s (t)) \in B_{f (v), \epsilon}\).
So, for each \(t \in J\) such that \(t_0 \le t\), \(s (t) \in f^{-1} (B_{f (v), \epsilon}) = B_{v, \epsilon} \subseteq N_v\).
That means that \(s\) has the convergence with \(lim s = v\).
So, \(lim s = v = v^j b_j = (lim s^j) b_j\).
Let us suppose that \(F = \mathbb{C}\).
Let \(f''': \mathbb{R}^2 \to \mathbb{C}\) be the canonical homeomorphism.
For each \(j \in \{1, ..., d\}\), let \(B_{f'''^{-1} (v^j), \epsilon / \sqrt{d}} \subseteq \mathbb{R}^2\) be the '\(\epsilon / \sqrt{d}\)'-'open ball' around \(f'''^{-1} (v^j)\).
\(f''' (B_{f'''^{-1} (v^j), \epsilon / \sqrt{d}})\) is an open neighborhood of \(v^j\).
There is a \(t_j \in J\), such that for each \(t \in J\) such that \(t_j \le t\), \(s^j (t) \in f''' (B_{f'''^{-1} (v^j), \epsilon / \sqrt{d}})\), because \(s^j\) converges to \(v^j\).
Let \(t_0 \in J\) be any such that \(t_1, ..., t_d \le t_0\), which exists, as before.
For each \(j\), for each \(t \in J\) such that \(t_0 \le t\), \(s^j (t) \in f''' (B_{f'''^{-1} (v^j), \epsilon / \sqrt{d}})\), which means that \(f'''^{-1} (s^j (t)) \in B_{f'''^{-1} (v^j), \epsilon / \sqrt{d}}\), which means that \((Re (s^j (t)) - Re (v^j))^2 + (Im (s^j (t)) - Im (v^j))^2 \lt \epsilon^2 / d\), so, \(\sum_{j \in \{1, ..., d\}} ((Re (s^j (t)) - Re (v^j))^2 + (Im (s^j (t)) - Im (v^j))^2) \lt \sum_{j \in \{1, ..., d\}} \epsilon^2 / d = \epsilon^2\), which means that \(f (s (t)) \in B_{f (v), \epsilon}\).
So, for each \(t \in J\) such that \(t_0 \le t\), \(s (t) \in f^{-1} (B_{f (v), \epsilon}) = B_{v, \epsilon} \subseteq N_v\).
That means that \(s\) has the convergence with \(lim s = v\).
So, \(lim s = v = v^j b_j = (lim s^j) b_j\).
