description/proof of that subspace of Hausdorff topological space is Hausdorff
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological subspace.
- The reader knows a definition of Hausdorff topological space.
Target Context
- The reader will have a description and a proof of the proposition that any subspace of any Hausdorff topological space is Hausdorff.
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:
\(T'\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
\(T\): \(\subseteq T'\) with the subspace topology
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Statements:
\(T \in \{\text{ the Hausdorff topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(t_1, t_2 \in T\) such that \(t_1 \neq t_2\) and some open neighborhoods of \(t_1, t_2\), \(U'_{t_1}, U'_{t_2} \subseteq T'\), such that \(U'_{t_1} \cap U'_{t_2} = \emptyset\); Step 2: see that \(U'_{t_1} \cap T, U'_{t_2} \cap T \subseteq T\) are some open neighborhoods of \(t_1\) and \(t_2\) and \((U'_{t_1} \cap T) \cap (U'_{t_2} \cap T) = \emptyset\).
Step 1:
Let \(t_1, t_2 \in T\) be any such that \(t_1 \neq t_2\).
There are some open neighborhood of \(t_1\) and \(t_2\), \(U'_{t_1}, U'_{t_2} \subseteq T'\), such that \(U'_{t_1} \cap U'_{t_2} = \emptyset\).
Step 2:
\(U'_{t_1} \cap T \subseteq T\) is an open neighborhood of \(t_1\), because it is open on \(T\) and \(t_1 \in U'_{t_1} \cap T\).
\(U'_{t_2} \cap T \subseteq T\) is an open neighborhood of \(t_2\), likewise.
\((U'_{t_1} \cap T) \cap (U'_{t_2} \cap T) = \emptyset\), because \((U'_{t_1} \cap T) \cap (U'_{t_2} \cap T) \subseteq U'_{t_1} \cap U'_{t_2} = \emptyset\).
So, \(T\) is Hausdorff.
