description/proof of that local characterization of closure: point is on closure of subset iff every neighborhood of point intersects subset
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of closure of subset.
Target Context
- The reader will have a description and a proof of a local characterization of closure: any point on any topological space is on the closure of any subset if and only if every neighborhood of the point intersects 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
\(p\): \(\in T\)
//
Statements:
\(p \in \overline{S}\)
\(\iff\)
\(\forall N_p \subseteq T, \in \{\text{ the neighborhoods of } p\} (N_p \cap S \neq \emptyset)\)
//
2: Natural Language Description
For any topological space, \(T\), and any subset, \(S \subseteq T\), any point, \(p \in T\), is on the closure of \(S\), \(\overline{S}\), which is \(p \in \overline{S}\), if and only if every neighborhood, \(N_p \subseteq T\), of \(p\), intersects \(S\), which is \(N_p \cap S \neq \emptyset\).
3: Proof
Let us suppose that \(p \notin \overline{S}\). There is a closed subset, \(C \subseteq T\), such that \(S \subseteq C\), that does not contain \(p\), by the definition of closure. \(p \in T \setminus C\) while \(T \setminus C\) is open, so, a neighborhood of \(p\), which does not intersect \(S\). So, if \(p \notin \overline{S}\), there is a neighborhood of \(p\) that does not intersect \(S\), so, as the logical contraposition, if there is no neighborhood of \(p\) that does not intersect \(S\), which means every neighborhood of \(p\) intersects \(S\), then \(p \in \overline{S}\).
Let us suppose that there is a neighborhood, \(N_p \subseteq T\), of \(p\), that does not intersect \(S\). There is an open neighborhood, \(U_p \subset T\), of \(p\) such that \(U_p \subseteq N_p\). \(U_p\) does not intersect \(S\). \(T \setminus U_p\) is closed, and \(S \subseteq T \setminus U_p\). So, \(p \notin \overline{S}\). So, if there is a neighborhood of \(p\) that does not intersect \(S\), \(p \notin \overline{S}\), so, as the logical contraposition, if \(p \in \overline{S}\), there is no neighborhood of \(p\) that does not intersect \(S\), which means that every neighborhood of \(p\) intersects \(S\).
