A 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 the 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: Description
For any directed set, \(S_1\), any topological space, \(T\), any net with directed index set, \(f_1: S_1 \rightarrow T\), and any accumulation value of \(f_1\), \(p \in T\), there is a subnet of \(f_1\), \(f_1 \circ f_2: S_2 \rightarrow T\) where \(S_2\) is a directed set and \(f_2: S_2 \rightarrow S_1\), such that \(p\) is a convergence of \(f_1 \circ f_2\).
2: Proof
Let us define \(S_2\) as \(\{ (\alpha, N_p)\vert \alpha \in S_1, N_p \text{ is any neighborhood of } p \text{ such that } f_1 (\alpha) \in N_p\}\) with the relation such that \((\alpha_1, N_{p-1}) \leq (\alpha_2, N_{p-2})\) if and only if \(\alpha_1 \leq \alpha_2\) and \(N_{p-2} \subseteq N_{p-1}\), where there is at least 1 such \(N_p\) for each \(\alpha\), because there is a neighborhood of \(p\) and there is a neighborhood of \(f_1 (\alpha)\) and the union of the 2 neighborhoods is a one (\(N_p\) does not need to be connected). Note that all the possible pairs are included in \(S_2\).
\(S_2\) is indeed a directed set, because 1) \((\alpha_1, N_{p-1}) \leq (\alpha_1, N_{p-1})\); 2) if \((\alpha_1, N_{p-1}) \leq (\alpha_2, N_{p-2})\) and \((\alpha_2, N_{p-2}) \leq (\alpha_3, N_{p-3})\), \((\alpha_1, N_{p-1}) \leq (\alpha_3, N_{p-3})\); 3) for any \((\alpha_1, N_{p-1})\) and \((\alpha_2, N_{p-2})\), there is a \((\alpha_3, N_{p-3})\) such that \((\alpha_1, N_{p-1}) \leq (\alpha_3, N_{p-3})\) and \((\alpha_2, N_{p-2}) \leq (\alpha_3, N_{p-3})\), because by the definition of directed set, there is an \(\alpha_4\) such that \(\alpha_1 \leq \alpha_4\) and \(\alpha_2 \leq \alpha_4\) and as \(p\) is an accumulation value, there is an \(\alpha_3\) such that \(\alpha_4 \leq \alpha_3\) and \(f_1 (\alpha_3) \in N_{p-1} \cap N_{p-2} := N_{p-3}\), then \(\alpha_1 \leq \alpha_3\) and \(\alpha_2 \leq \alpha_3\) and \(N_{p-3} \subseteq N_{p-1}\) and \(N_{p-3} \subseteq N_{p-2}\).
Let us define \(f_2\) as \((\alpha, N_p) \mapsto \alpha\), which is final, because for any \(\alpha\), \(\alpha \leq f_2 ((\alpha, N_p)) = \alpha\).
\(p\) is a convergence of \(f_1 \circ f_2\), because for any neighborhood of \(p\), \(N_{p-0}\), there is a \((\alpha_0, N_{p-0})\) because \(p\) is an accumulation value of \(f_1\), and for every \((\alpha_0, N_{p-0}) \leq (\alpha, N_p)\), \(f_1 \circ f_2 ((\alpha, N_p)) = f_1 (\alpha) \in N_{p-0}\), because \(f_1 (\alpha) \in N_p \subseteq N_{p-0}\).
