187: For Disjoint Subset and Open Set, Closure of Subset and Open Set Are Disjoint

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A description/proof of that for disjoint subset and open set, closure of subset and open set are disjoint

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of closure of subset.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint.

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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: Description

For any topological space, $T$, and any disjoint subset, $S\subseteq T$, and open set, $U\subseteq T$, $S\cap U=\mathrm{\varnothing }$, the closure of $S$, $\bar{S}$, and $U$ are disjoint, which is $\bar{S}\cap U=\mathrm{\varnothing }$.

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2: Proof

Suppose that $\bar{S}$ and $U$ were not disjoint. There would be a point, $p\in \bar{S},U$. By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, $p$ would be on $S$ or would be an accumulation point of $S$, but the former is impossible, because $S$ and $U$ are disjoint, so, $p$ would be an accumulation point of $S$. As $p\in U$, there would be an open set around $p$, $U_p$, such that $p\in U_p\subset U$, but $U_p$ would not contain any point of $S$, so, $p$ would not be any accumulation point of $S$, a contradiction.

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References

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