description/proof of that for locally connected topological space, open subspace is locally connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally connected topological space.
- The reader knows a definition of topological subspace.
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
- The reader admits the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any locally connected topological space, any open subspace is locally connected.
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:
\(T'\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(T\): \(\in \{\text{ the open subsets of } T'\}\) with the subspace topology
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Statements:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
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2: Note
As Proof shows, \(T\) needs to be open in order for this proposition to be applied.
3: Proof
Whole Strategy: Step 1: for each \(t \in T\), take any neighborhood of \(t\) on \(T\), \(N'_t\), and see that \(N'_t\) is a neighborhood of \(t\) on \(T'\); Step 2: take a connected neighborhood of \(t\) on \(T'\) contained in \(N'_t\), \(N_t\); Step 3: see that \(N_t\) is a connected neighborhood of \(t\) on \(T\).
Step 1:
Let \(t \in T\) be any.
Let \(N'_t \subseteq T\) be any neighborhood of \(t\).
\(N'_t\) is a neighborhood of \(t \in T'\) on \(T'\), because there is an open neighborhood of \(t\) on \(T\), \(U'_t \subseteq T\), such that \(U'_t \subseteq N'_t\), which is open also on \(T'\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Step 2:
There is a connected neighborhood of \(t\) on \(T'\), \(N_t \subseteq T'\), such that \(t \in N_t \subseteq N'_t\), by the definition of locally connected topological space.
Step 3:
\(N_t \subseteq N'_t \subseteq T\) is a neighborhood of \(T\), because there is an open neighborhood of \(t\) on \(T'\), \(U_t \subseteq T'\), such that \(U_t \subseteq N_t\), which is an open neighborhood of \(t\) also on \(T\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
While \(N_t\) is a connected subspace of \(T'\), \(N_t\) is a connected subspace of \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(N_t\) is a connected neighborhood of \(t\) on \(T\).
So, \(T\) is locally connected.
