description/proof of that topological subspace is locally closed iff it is intersection of closed subset and open subset of base space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally closed topological subspace.
- The reader knows a definition of subspace topology of subset of topological space.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- 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.
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\subseteq T'\), with the subspace topology of \(T'\)
//
Statements:
\(T \in \{\text{ the locally closed topological subspaces of } T'\}\)
\(\iff\)
\(\exists C' \in \{\text{ the closed subsets of } T'\}, \exists U' \in \{\text{ the open subsets of } T'\} (T = C' \cap U')\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(T\) is locally closed; Step 2: take \(C' = \overline{T}\) and \(U' := \cup_{t \in T} U'_t\), and see that \(T = C' \cap U'\); Step 3: suppose that \(T = C' \cap U'\); Step 4: take \(U'_t = U'\), and see that \(U'_t \cap T = U'_t \cap \overline{T}\).
Step 1:
Let us suppose that \(T\) is locally closed.
Step 2:
Let us take \(C' := \overline{T}\), closed on \(T'\).
Let us take \(U' := \cup_{t \in T} U'_t\), where \(U'_t \subseteq T'\) is any open neighborhood of \(t\) such that \(U'_t \cap T \subseteq U'_t\) is closed, open on \(T'\).
\(C' \cap U' = \overline{T} \cap \cup_{t \in T} U'_t = \cup_{t \in T} (\overline{T} \cap U'_t)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(= \cup_{t \in T} (T \cap U'_t)\), because on the one hand, \((T \cap U'_t) \subseteq (\overline{T} \cap U'_t) \subseteq \overline{T \cap U'_t}\), 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, \(T \cap U'_t = \overline{T \cap U'_t} \cap U'_t\), because as \(T \cap U'_t\) is closed on \(U'_t\), there is a closed subset, \(C'_t \subseteq T'\), such that \(T \cap U'_t = C'_t \cap U'_t\), but \(T \cap U'_t \subseteq C'_t\) and as \(C'_t\) is closed, \(\overline{T \cap U'_t} \subseteq C'_t\), and in fact, \(C'_t\) can be taken to be \(\overline{T \cap U'_t}\), so, \(T \cap U'_t = C'_t \cap U'_t = \overline{T \cap U'_t} \cap U'_t\), then \(T \cap U'_t \cap U'_t = T \cap U'_t \subseteq (\overline{T} \cap U'_t) \cap U'_t \subseteq \overline{T \cap U'_t} \cap U'_t\) while the 1st term and the last term equal, so, \((\overline{T} \cap U'_t) \cap U'_t = \overline{T} \cap U'_t\) equals the 1st term, \(T \cap U'_t\).
\(= T\), because \(\cup_{t \in T} (T \cap U'_t) \subseteq T\), because for each \(p \in \cup_{t \in T} (T \cap U'_t)\), \(p \in T \cap U'_t\) for an \(t \in T\), and \(p \in T \cap U'_t \subseteq T\); \(T \subseteq \cup_{t \in T} (T \cap U'_t)\), because for each \(t \in T\), \(t \in T \cap U'_t\), and \(t \in \cup_{t \in T} (T \cap U'_t)\).
So, \(C' \cap U' = T\).
Step 3:
Let us suppose that \(T = C' \cap U'\).
Step 4:
For each \(t \in T\), let us take \(U'_t = U'\), which is an open neighborhood of \(t\), because \(t \in T = C' \cap U' \subseteq U'\).
\(T = \overline{T} \cap U'\), because for each \(p \in T\), \(p \in \overline{T}\) and \(p \in T = C' \cap U' \subseteq U'\), so, \(p \in \overline{T} \cap U'\), so, \(T \subseteq \overline{T} \cap U'\); for each \(p \in \overline{T} \cap U'\), as \(T = C' \cap U' \subseteq C'\), \(\overline{T} \subseteq C'\), so, \(p \in C' \cap U' = T\), so, \(\overline{T} \cap U' \subseteq T\).
So, \(U' \cap T = U' \cap \overline{T} \cap U' = U' \cap \overline{T}\), closed on \(U'\), by 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.
So, \(T\) is locally closed.
