description/proof of that on \(T_1\) topological space, point is \(\omega\)-accumulation point of subset iff it is accumulation point of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of \(T_1\) topological space.
- The reader knows a definition of \(\omega\)-accumulation point of subset of topological space.
- The reader knows a definition of accumulation point of subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that on any \(T_1\) topological space, any point is an \(\omega\)-accumulation point of any subset if and only if the point is an accumulation point of the subset.
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:
\(T\): \(\in \{\text{ the } T_1 \text{ topological spaces }\}\)
\(S\): \(\subseteq T\)
\(t\): \(\in T\)
//
Statements:
\(t \in \{\text{ the } \omega \text{ -accumulation points of } S\}\)
\(\iff\)
\(t \in \{\text{ the accumulation points of } S\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(t\) is an accumulation point of \(S\); Step 2: take any open neighborhood of \(t\), \(U_t\), see that \(t_1 \in U_t \cap (S \setminus \{t\})\), take an open neighborhood of \(t\), \(U_{t, 1}\), such that \(t_1 \notin U_{t, 1}\), see that \(t_2 \in U_t \cap U_{t, 1} \cap (S \setminus \{t\})\), and so on; Step 3: suppose that \(t\) is an \(\omega\)-accumulation point of \(S\); Step 4: see that \(t\) is an accumulation point of \(S\).
Step 1:
Let us suppose that \(t\) is an accumulation point of \(S\).
Step 2:
Let \(U_t \subseteq T\) be any open neighborhood of \(t\).
There is an \(t_1 \in U_t \cap (S \setminus \{t\})\), because \(t\) is an accumulation point of \(S\).
\(t \neq t_1\).
So, there is an open neighborhood of \(t\), \(U_{t, 1} \subseteq T\), such that \(t_1 \notin U_{t, 1}\), because \(T\) is \(T_1\).
\(U_t \cap U_{t, 1}\) is an open neighborhood of \(t\), so, there is an \(t_2 \in U_t \cap U_{t, 1} \cap (S \setminus \{t\})\), because \(t\) is an accumulation point of \(S\).
\(t_1 \neq t_2\), because \(t_2 \in U_{t, 1}\) while \(t_1 \notin U_{t, 1}\).
\(\{t_1, t_2\} \subseteq U_t \cap (S \setminus \{t\})\).
\(t \neq t_2\) and there is an open neighborhood of \(t\), \(U_{t, 2} \subseteq T\), such that \(t_2 \notin U_{t, 2}\), because \(T\) is \(T_1\).
So far, we have gotten a distinct \(\{t_1, t_2\}\) and \(\{U_{t, 1}, U_{t, 2}\}\) such that \(\{t_1, t_2\} \subseteq U_t \cap (S \setminus \{t\})\) and \(t_1, t_2 \notin U_{t, 1} \cap U_{t, 2}\).
Let us suppose that there are a distinct \(\{t_1, ..., t_n\}\) and \(\{U_{t, 1}, ..., U_{t, n}\}\) such that \(\{t_1, ..., t_n\} \subseteq U_t \cap (S \setminus \{t\})\) and \(t_1, ..., t_n \notin U_{t, 1} \cap ... \cap U_{t, n}\).
\(U_t \cap U_{t, 1} \cap ... \cap U_{t, n}\) is an open neighborhood of \(t\), so, there is an \(t_{n + 1} \in U_t \cap U_{t, 1} \cap ... \cap U_{t, n} \cap (S \setminus \{t\})\), because \(t\) is an accumulation point of \(S\).
\(\{t_1, ..., t_{n + 1}\}\) is distinct, because \(t_{n + 1} \in U_{t, 1} \cap ... \cap U_{t, n}\) while \(t_j \notin U_{t, 1} \cap ... \cap U_{t, n}\) for each \(j \in \{1, ..., n\}\).
\(t \neq t_{n + 1}\) and there is an open neighborhood of \(t\), \(U_{t, n + 1} \subseteq T\), such that \(t_{n + 1} \notin U_{t, n + 1}\), because \(T\) is \(T_1\).
So, we have gotten a distinct \(\{t_1, ..., t_{n + 1}\}\) and \(\{U_{t, 1}, ..., U_{t, n + 1}\}\) such that \(\{t_1, ..., t_{n + 1}\} \subseteq U_t \cap (S \setminus \{t\})\) and \(t_1, ..., t_{n + 1} \notin U_{t, 1} \cap ... \cap U_{t, n + 1}\).
So, \(U_t \cap (S \setminus \{t\})\) has at least some countably infinite points.
So, \(t\) is an \(\omega\)-accumulation point of \(S\).
Step 3:
Let us suppose that \(t\) is an \(\omega\)-accumulation point of \(S\).
Step 4:
For each open neighborhood of \(t\), \(U_t \subseteq T\), \(U_t \cap (S \setminus \{t\}) \neq \emptyset\).
So, \(t\) is an accumulation point of \(S\).
