description/proof of that for topological space contained in ambient topological space, if space is ambient-space-wise locally topological subspace of ambient space, space is topological subspace of ambient space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- 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 open set on any open topological subspace is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space contained in any ambient topological space, if the space is ambient-space-wise locally topological subspace of the ambient space, the space is the topological subspace of the 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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological spaces }\}\), such that \(T \subseteq T'\)
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Statements:
\(\forall t \in T (\exists U'_t \in \{\text{ the open neighborhoods of } t \text{ on } T'\} (U'_t \cap T \in \{\text{ the open subsets of } T\} \land U'_t \cap T \subseteq T \text{ as the subspace of } T = U'_t \cap T \subseteq U'_t \text{ as the subspace of } U'_t))\)
\(\implies\)
\(T \in \{\text{ the topological subspaces of } T'\}\)
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2: Note
In this article, \(U'_t \cap T \subseteq T\) denotes \(U'_t \cap T\) with the topology as the subspace of \(T\); \(U'_t \cap T \subseteq U'_t\) denotes \(U'_t \cap T\) with the topology as the subspace of \(U'_t\). They are different in general, because \(U'_t \cap T \subseteq T\) is based on the topology of \(T\) while \(U'_t \cap T \subseteq U'_t\) is based on the topology of \(U'_t\), which is the topological subspace of \(T'\).
A useful reference example is a figure-8 contained in \(\mathbb{R}^2\), centered at the origin: the 2 curves from the origin approaches infinitely but not reach the origin.
It is not a topological subspace of \(\mathbb{R}^2\), because an open neighborhood of the origin as a curve is not the intersection of any open subset of \(\mathbb{R}^2\) and the figure-8: any open ball around the origin on \(\mathbb{R}^2\) inevitably contains the 2 ends of the figure-8 that approach the origin.
This proposition specifically says "ambient-space-wise", because "space-wise locally topological subspace of the ambient space" does not guarantee the conclusion: the open neighborhood of the origin is a topological subspace of \(\mathbb{R}^2\): as the open neighborhood of the origin excludes the 2 ends of the figure-8, it becomes the topological subspace, but that is not allowed for proving that \(T\) is a topological subspace of \(T'\).
3: Proof
Whole Strategy: Step 1: let \(U \subseteq T\) be any open subset of \(T\), and find an open subset, \(U' \subseteq T'\), of \(T'\) such that \(U = U' \cap T\); Step 2: let \(U' \subseteq T'\) be any open subset of \(T'\), and see that \(U' \cap T\) is an open subset of \(T\); Step 3: conclude the proposition.
Step 1:
Let \(U \subseteq T\) be any open subset of \(T\).
Let us find an open subset, \(U' \subseteq T'\), of \(T'\) such that \(U = U' \cap T\).
Note that \(U'_t\) is the one chosen for each \(t \in T\) according to the conditions of this proposition.
\(T \subseteq \cup_{t \in T} U'_t\). So, \(U = U \cap \cup_{t \in T} U'_t = \cup_{t \in T} (U \cap U'_t)\), 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.
\(U \cap U'_t = U \cap U'_t \cap T \subseteq U'_t \cap T\) is an open subset of \(U'_t \cap T \subseteq T\), because \(U\) is an open subset of \(T\).
By the supposition, \(U \cap U'_t\) is an open subset of \(U'_t \cap T \subseteq U'_t\). That means that there is an open subset, \(U''_t \subseteq U'_t\), of \(U'_t\), such that \(U \cap U'_t = U''_t \cap U'_t \cap T\). Here, \(U''_t\) is an open subset of \(T'\), by the proposition that any open set on any open topological subspace is open on the base space, and so, \(U''_t \cap U'_t\) is an open subset of \(T'\).
Now, \(U = \cup_{t \in T} (U \cap U'_t) = \cup_{t \in T} (U''_t \cap U'_t \cap T) = \cup_{t \in T} (U''_t \cap U'_t) \cap T\), 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. But \(U' := \cup_{t \in T} (U''_t \cap U'_t)\) is an open subset of \(T'\), and \(U = U' \cap T\).
Step 2:
Let \(U' \subseteq T'\) be any open subset of \(T'\).
Let us see that \(U' \cap T\) is an open subset of \(T\).
\(U' \cap T = U' \cap T \cap \cup_{t \in T} U'_t\), because \(T \subseteq \cup_{t \in T} U'_t\), \(= \cup_{t \in T} (U' \cap T \cap U'_t)\), 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, \(= \cup_{t \in T} ((U' \cap U'_t) \cap (U'_t \cap T))\).
As \(U' \cap U'_t\) is an open subset of \(U'_t\), \((U' \cap U'_t) \cap (U'_t \cap T)\) is an open subset of \(U'_t \cap T \subseteq U'_t\). By the supposition, \((U' \cap U'_t) \cap (U'_t \cap T)\) is an open subset of \(U'_t \cap T \subseteq T\).
As \(U'_t \cap T\) is an open subset of \(T\) by the supposition, \((U' \cap U'_t) \cap (U'_t \cap T)\) is an open subset of \(T\), by the proposition that any open set on any open topological subspace is open on the base space.
So, \(U' \cap T = \cup_{t \in T} ((U' \cap U'_t) \cap (U'_t \cap T))\) is open on \(T\).
Step 3:
Step 1 and Step 2 have shown that any subset of \(T\) is open if and only if it is the intersection of an open subset of \(T'\) and \(T\), which means that \(T\) is a topological subspace of \(T'\).
