description/proof of that boundary of subset of topological space is set of points of each of which each neighborhood intersects both subset and complement of 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 boundary of subset of topological space.
- The reader knows a definition of neighborhood of point.
Target Context
- The reader will have a description and a proof of the proposition that the boundary of any subset of any topological space is the set of the points of each of which each neighborhood intersects both the subset and the complement 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
\(\dot{S}\): \(= \text{ the boundary of } S\)
\(\tilde{S}\): \(= \{p \in T \vert \forall N_p \in \{\text{ the neighborhoods of } p \text{ on } T\} (N_p \cap S \neq \emptyset \land N_p \cap (T \setminus S) \neq \emptyset)\}\)
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Statements:
\(\dot{S} = \tilde{S}\).
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2: Natural Language Description
For any topological space, \(T\), and any subset, \(S \subseteq T\), the boundary of \(S\), \(\dot{S}\), equals \(\tilde{S} := \{p \in T \vert \forall N_p \in \{\text{ the neighborhoods of } p \text{ on } T\} (N_p \cap S \neq \emptyset \land N_p \cap (T \setminus S) \neq \emptyset)\}\).
3: Proof
For any \(p \in \dot{S}\), for any neighborhood, \(N_p \subseteq T\), of \(p\), \(N_p \cap S \neq \emptyset\) and \(N_p \cap (T \setminus S) \neq \emptyset\), so, \(p \in \tilde{S}\).
For any \(p \in \tilde{S}\), \(p \in \overline{S}\) and \(p \in \overline{T \setminus S}\), so, \(p \in \dot{S}\).
