348: Topological Space Is Normal Iff for Closed Set and Its Containing Open Set There Is Closed-Set-Containing Open Set Whose ~

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A description/proof of that topological space is normal iff for closed set and its containing open set there is closed-set-containing open set whose ~

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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 normal topological space.
• The reader knows a definition of closed set.
• 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 the proposition that any topological space is normal if and only if for any closed set and its any containing open set, there is a closed-set-containing open set whose closure is contained in the former 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

Any topological space, $T$, is normal if and only if for any closed set, $C\subseteq T$, and any open set, $U_1\subseteq T$, such that $C\subseteq U_1$, there is an open set, $U_2\subseteq T$, such that $C\subseteq U_2\subseteq \overline{U_2}\subseteq U_1$ where $\overline{U_2}$ is the closure of $U_2$.

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2: Proof

Suppose that $T$ is normal. Suppose that there are C and $U_1$. $C_2:=T\setminus U_1$ is closed and C and $C_2$ are disjoint. There are some open sets, $U_2,U_3\subseteq T$, that are disjoint such that $C\subseteq U_2$ and $C_2\subseteq U_3$. $C_4:=T\setminus U_3$ is closed and $C_2=T\setminus U_1\subseteq U_3=T\setminus C_4$, so, $C_4\subseteq U_1$. As $U_2$ and $U_3$ are disjoint, $U_2\subseteq T\setminus U_3=C_4$. So, $U_2\subseteq C_4\subseteq U_1$, but as $\overline{U_2}$ is the intersection of all the closed sets that contain $U_2$, $U_2\subseteq \overline{U_2}\subseteq C_4$. So, $C\subseteq U_2\subseteq \overline{U_2}\subseteq U_1$.

Suppose that for C and $U_1$, there is $U_2$ such that $C\subseteq U_2\subseteq \overline{U_2}\subseteq U_1$. For any disjoint closed sets, $C_1,C_2\subseteq T$, $U_1:=T\setminus C_2$ is open and $C_1\subseteq U_1$. So, there is an open set, $U_2\subseteq T$, such that $C_1\subseteq U_2\subseteq \overline{U_2}\subseteq U_1$. As $\overline{U_2}$ is closed, $U_3:=T\setminus \overline{U_2}$ is open and $T\setminus U_3=\overline{U_2}\subseteq U_1=T\setminus C_2$, which implies $C_2\subseteq U_3$. As $U_3=T\setminus \overline{U_2}\subseteq T\setminus U_2$, $U_2$ and $U_3$ are disjoint.

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References

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