189: Local Characterization of Closure: Point Is on Closure of Subset iff Every Neighborhood of Point Intersects Subset

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description/proof of that local characterization of closure: point is on closure of subset iff every neighborhood of point intersects 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 topological space.
• The reader knows a definition of closure of subset.

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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.

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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:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$p$: $\in T$
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Statements:
$p\in \bar{S}$
$⟺$
$\mathrm{\forall }{N}_p\subseteq T,\in \{\text{ the neighborhoods of }p\}(N_p\cap S\ne \mathrm{\varnothing })$
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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 $S$, $\bar{S}$, which is $p\in \bar{S}$, if and only if every neighborhood, $N_p\subseteq T$, of $p$, intersects $S$, which is $N_p\cap S\ne \mathrm{\varnothing }$.

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3: Proof

Let us suppose that $p\notin \bar{S}$. 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 $p\notin \bar{S}$, 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 $p\in \bar{S}$.

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, $p\notin \bar{S}$. So, if there is a neighborhood of $p$ that does not intersect $S$, $p\notin \bar{S}$, so, as the logical contraposition, if $p\in \bar{S}$, there is no neighborhood of $p$ that does not intersect $S$, which means that every neighborhood of $p$ intersects $S$.

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References

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