361: Finite Union of Nowhere Dense Subsets of Topological Space Has Empty Interior

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A description/proof of that finite union of nowhere dense subsets of topological space has empty interior

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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
• 3: Note

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Starting Context

• The reader knows a definition of nowhere dense subset of topological space.
• The reader knows a definition of interior of subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, the union of any finite number of nowhere dense subsets has the empty interior.

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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, $T$, any finite number of nowhere dense subsets, $\{S_i|i\in I\}$ where $I$ is any finite indices set, and the union, $S={\cup }_{i\in I}{S}_i$, the interior of the union, $intS$, is empty, which is $intS=\mathrm{\varnothing }$.

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2: Proof

Let us suppose that $intS\ne \mathrm{\varnothing }$. There would be an open set, $U\subseteq S$. $\mathrm{\neg }U\subseteq \overline{S_1}$, because $S_1$ would be nowhere dense. There would be a point, $p_1\in U$ such that $p_1\notin \overline{S_1}$. As $p_1\notin S_1$ would not be any accumulation point of $S_1$, there would be an open neighborhood, $U_{p_1}\subseteq T$, such that $U_{p_1}\cap S_1=\mathrm{\varnothing }$. So, $U\cap U_{p_1}\subseteq S_2\cup S_3\cup ...\cup S_n$.

As $U\cap U_{p_1}$ would play the same role for $S_2\cup S_3\cup ...\cup S_n$ as $U$ would do for $S_1\cup S_2\cup ...\cup S_n$, there would be a neighborhood, $U\cap U_{p_1}\cap U_{p_2}\subseteq T$ such that $U\cap U_{p_1}\cap U_{p_2}\subseteq S_3\cup S_4\cup ...\cup S_n$, and so on. After all, $U\cap U_{p_1}\cap U_{p_2}\cap ...\cap U_{p_{n-1}}\subseteq S_n$, a contradiction against $S_n$'s being nowhere dense.

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3: Note

As is shown in the proof, $S_n$ does not really need to be nowhere dense: having the empty interior is enough, which is a weaker condition than being nowhere dense.

For an infinite union, the union may have a nonempty interior, but if $T$ is a complete metric space or a locally compact Hausdorff space, any countable union will have the empty interior, by the Baire category theorem.

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References

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