358: Open Set Complement of Measure 0 Subset Is Dense

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A description/proof of that open set complement of measure 0 subset is dense

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Topics

About: metric space
About: measure
About: map

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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 metric space.
• The reader knows a definition of measure.
• The reader knows a definition of denseness.

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Target Context

• The reader will have a description and a proof of the proposition that for any open set on any metric space, the complement of any measure 0 subset with respect to the open set is dense on the open set.

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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 metric space, M, any open set, $U\subseteq M$, and any measure 0 subset, $S\subseteq U$, the complement, $S^c=U\setminus S$, of $S$ with respect to $U$ is dense on $U$.

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2: Proof

Centered at any point, $p\in U$, there is any small-enough open ball, $B_{p-ϵ}\subseteq U$. $S^c\cap B_{p-ϵ}=(U\setminus S)\cap B_{p-ϵ}=U\cap B_{p-ϵ}\setminus S\cap B_{p-ϵ}$. $m(S^c\cap B_{p-ϵ})=m(U\cap B_{p-ϵ})-m(S\cap B_{p-ϵ})=m(B_{p-ϵ})-0$. As $m(B_{p-ϵ})>0$, $m(S^c\cap B_{p-ϵ})>0$, which means that $S^c\cap B_{p-ϵ}$ is not empty, which means that for any point, $p\in U$, and any small-enough open ball, $B_{p-ϵ}\subseteq U$, there is a point of $S^c$ inside the open ball, which means that $S^c$ is dense.

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References

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