305: Topological Subspace Is Locally Closed iff It Is Intersection of Closed Subset and Open Subset of Base Space

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A description/proof of that topological subspace is locally closed iff it is intersection of closed subset and open subset of base space

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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 locally closed topological space.
• The reader knows a definition of subspace topology.
• The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
• The reader admits the proposition that for any topological space, the intersection of the closure of any subset and any open set is contained in the closure of the intersection of the subset and the open set.

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Target Context

• The reader will have a description and a proof of the proposition that any topological subspace is locally closed if and only if it is the intersection of a closed subset and an open subset of the base space.

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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 subspace, $T_1\subseteq T$, is locally closed if and only if $T_1=C\cap U$ where $C$ and $U$ are a closed subset and an open set, respectively, of $T$.

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2: Proof

Let us suppose that $T_1=C\cap U$. $T_1=\overline{T_1}\cap U$, because obviously $\overline{T_1}\cap U\subseteq C\cap U$, and for any point, $p\in C\cap U$, $p\in T_1$ and $p\in U$, so, $p\in \overline{T_1}\cap U$, so, $C\cap P\subseteq \overline{T_1}\cap U$. If $U\cap T_1$ is closed on $U$, $U$ can be taken to be an open neighborhood of each point, $p\in T_1$, required for $T_1$ to be locally closed. But $U\cap T_1$ is in fact closed on $U$, because $U\cap T_1=U\cap \overline{T_1}\cap U=\overline{T_1}\cap U$.

Let us suppose that $T_1$ is locally closed. Let us define $U:={\cup }_p{U}_p$ where $U_p\subseteq T$ is an open neighborhood of $p$ on $T$ required for $T_1$ to be locally closed, so, $U_p\cap T_1$ is closed on $U_p$. $U$ is open on $T$. Let us define $C=\overline{T_1}$, closed on $T$. Then, $C\cap U=\overline{T_1}\cap {\cup }_p{U}_p={\cup }_p(\overline{T_1}\cap U_p)={\cup }_p(T_1\cap U_p)=T_1$, where the penultimate equal is because on the one hand, $(T_1\cap U_p)\subseteq (\overline{T_1}\cap U_p)\subseteq \overline{T_1\cap U_p}$, by the proposition that for any topological space, the intersection of the closure of any subset and any open set is contained in the closure of the intersection of the subset and the open set, and on the other hand, $T_1\cap U_p=\overline{T_1\cap U_p}\cap U_p$, because as $T_1\cap U_p$ is closed on $U_p$, there is a closed subset, $C_p\subseteq T$, such that $T_1\cap U_p=C_p\cap U_p$, but $T_1\cap U_p\subseteq C_p$ and as $C_p$ is closed, $\overline{T_1\cap U_p}\subseteq C_p$, and in fact, $C_p$ can be taken to be $\overline{T_1\cap U_p}$, so, $T_1\cap U_p=C_p\cap U_p=\overline{T_1\cap U_p}\cap U_p$, then $T_1\cap U_p\cap U_p=T_1\cap U_p\subseteq (\overline{T_1}\cap U_p)\cap U_p\subseteq \overline{T_1\cap U_p}\cap U_p$ while the 1st term and the 3rd term equal, so, the 2nd term equals the 1st term, and the last equal is because $U$ covers $T_1$.

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References

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