351: Finite Intersection of Open Dense Subsets of Topological Space Is Open Dense

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A description/proof of that finite intersection of open dense subsets of topological space is open dense

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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 dense subset of topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, the intersection of any finite number of open dense subsets is open dense.

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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 finite number of open dense subsets, $\{U_i|i\in I\}$ where $I$ is any finite indices set, the intersection, ${\cap }_{i\in I}{U}_i$, is open dense.

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2: Proof

${\cap }_{i\in I}{U}_i$ is open.

Let us suppose that ${\cap }_{i\in I}{U}_i$ was not dense. Then, there would be a point, $p\in T\setminus \overline{{\cap }_{i\in I}{U}_i}$. There would be an open neighborhood, $U_p\subseteq T$, of $p$, such that $U_p\cap {\cap }_{i\in I}{U}_i=\mathrm{\varnothing }$, because $p$ would not be any accumulation point of ${\cap }_{i\in I}{U}_i$. Let us suppose that $I=\{1,2,...,n\}$ without loss of generality. There would be a point, $p_1\in U_p\cap U_1$, because $p$ would be a point of $U_1$ or an accumulation point of $U_1$. As $U_1$ would be open, $U_p\cap U_1$ would be an open neighborhood of $p_1$, and as $p_1$ would be a point of $U_2$ or an accumulation point of $U_2$, there would be a point, $p_2\in U_p\cap U_1\cap U_2$, and $U_p\cap U_1\cap U_2$ would be an open neighborhood of $p_2$, and so on. After all, there would be a point, $p_n\in U_p\cap {\cap }_{i\in I}{U}_i$, a contradiction.

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References

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