585: Boundary of Subset of Topological Space Is Set of Points of Each of Which Each Neighborhood Intersects Both Subset and Complement of Subset

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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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of boundary of subset of topological space.
• The reader knows a definition of neighborhood of point.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$\dot{S}$: $=\text{ the boundary of }S$
$\tilde{S}$: $=\{p\in T|\mathrm{\forall }{N}_p\in \{\text{ the neighborhoods of }p\text{ on }T\}(N_p\cap S\ne \mathrm{\varnothing }\wedge N_p\cap (T\setminus S)\ne \mathrm{\varnothing })\}$
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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 $\tilde{S}:=\{p\in T|\mathrm{\forall }{N}_p\in \{\text{ the neighborhoods of }p\text{ on }T\}(N_p\cap S\ne \mathrm{\varnothing }\wedge N_p\cap (T\setminus S)\ne \mathrm{\varnothing })\}$.

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3: Proof

For any $p\in \dot{S}$, for any neighborhood, $N_p\subseteq T$, of $p$, $N_p\cap S\ne \mathrm{\varnothing }$ and $N_p\cap (T\setminus S)\ne \mathrm{\varnothing }$, so, $p\in \tilde{S}$.

For any $p\in \tilde{S}$, $p\in \bar{S}$ and $p\in \overline{T\setminus S}$, so, $p\in \dot{S}$.

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References

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