description/proof of that accumulation value of net with directed index set is convergence of subnet
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 subnet of net with directed index set.
- The reader knows a definition of accumulation value of net with directed index set.
Target Context
- The reader will have a description and a proof of the proposition that any accumulation value of any net with directed index set is a convergence of a subnet.
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:
\(D_2\): \(\in \{\text{ the directed sets }\}\)
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(f_2\): \(: D_2 \to T\)
\(t\): \(\in \{\text{ the accumulation values of } f_2\}\)
//
Statements:
\(\exists D_1 \in \{\text{ the directed sets }\}, \exists f_1: D_1 \to D_2 \in \{\text{ the final maps }\} (t \in \{\text{ the convergences of } f_2 \circ f_1\})\)
//
2: Proof
Whole Strategy: Step 1: define \(D_1 := \{(d, U_t) \in D_2 \times \{\text{ the open neighborhoods of } t \text{ on } T\} \vert f_2 (d) \in U_t\}\) with the relation such that \((d_1, U_{t, 1}) \le (d_2, U_{t, 2})\) if and only if \(d_1 \le d_2\) and \(U_{t, 2} \subseteq U_{t, 1}\), and see that \(D_1\) is a directed set; Step 2: define \(f_1: (d, U_t) \mapsto d\), and see that \(f_1\) is final; Step 3: see that \(f_2 \circ f_1\) converges to \(t\).
Step 1:
Let us define \(D_1 := \{(d, U_t) \in D_2 \times \{\text{ the open neighborhoods of } t \text{ on } T\} \vert f_2 (d) \in U_t\}\).
For each \(d \in D_2\), there is at least \(1\) such \(U_t\), because there are an open neighborhood of \(t\) and an open neighborhood of \(f_2 (d)\) and the union of the 2 neighborhoods is a \(U_t\): \(U_t\) does not need to be connected.
Let us define the relation on \(D_1\) such that \((d_1, U_{t, 1}) \le (d_2, U_{t, 2})\) if and only if \(d_1 \le d_2\) and \(U_{t, 2} \subseteq U_{t, 1}\).
\(D_1\) is indeed a directed set, because 1) \((d_1, U_{t, 1}) \leq (d_1, U_{t, 1})\); 2) if \((d_1, U_{t, 1}) \leq (d_2, U_{t, 2})\) and \((d_2, U_{t, 2}) \leq (d_3, U_{t, 3})\), \((d_1, U_{t, 1}) \leq (d_3, U_{t, 3})\), because \(d_1 \le d_2 \le d_3\) and \(U_{t, 3} \subseteq U_{t, 2} \subseteq U_{t, 1}\); 3) for any \((d_1, U_{t, 1})\) and \((d_2, U_{t, 2})\), there is a \((d_3, U_{t, 3})\) such that \((d_1, U_{t, 1}) \leq (d_3, U_{t, 3})\) and \((d_2, U_{t, 2}) \leq (d_3, U_{t, 3})\), because by the definition of directed set, there is an \(d_4\) such that \(d_1 \leq d_4\) and \(d_2 \leq d_4\) and as \(t\) is an accumulation value, there is an \(d_3\) such that \(d_4 \leq d_3\) and \(f_2 (d_3) \in U_{t, 1} \cap U_{t, 2} := U_{t, 3}\), then \(d_1 \leq d_3\) and \(d_2 \leq d_3\) and \(U_{t, 3} \subseteq U_{t, 1}\) and \(U_{t, 3} \subseteq U_{t, 2}\).
Step 2:
Let us define \(f_1\) as \(: (d, U_t) \mapsto d\), which is final, because for any \(d \in D_2\), there is a \((d, U_t) \in D_1\), which satisfies that for each \((d', U'_t) \in D_1\) such that \((d, U_t) \le (d', U'_t)\), \(d \le f_1 ((d', U'_t)) = d'\).
Step 3:
\(t\) is a convergence of \(f_2 \circ f_1\), because for any open neighborhood of \(t\), \(U_{t, 0}\), there is a \((d_0, U_{t, 0})\), because \(t\) is an accumulation value of \(f_2\), and for each \((d, U_t) \in D_1\) such that \((d_0, U_{t, 0}) \leq (d, U_t)\), \(f_2 \circ f_1 ((d, U_t)) = f_2 (d) \in U_{t, 0}\), because \(f_2 (d) \in U_t \subseteq U_{t, 0}\).
