description/proof of that topological space is Hausdorff iff its diagonal is closed
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of closed subset of topological space.
- The reader admits the local criterion for openness.
- The reader admits the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is Hausdorff iff its diagonal is closed.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(T \times T\): \(= \text{ the product topological space }\)
\(S\): \(= \{(t, t) \in T \times T \vert t \in T\}\)
//
Statements:
\(T \in \{\text{ the Hausdorff topological spaces }\}\)
\(\iff\)
\(S \in \{\text{ the closed subsets of } T \times T\}\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T\) is Hausdorff; Step 2: see that \(T \times T \setminus S\) is open using the local criterion for openness; Step 3: suppose that \(S\) is closed; Step 4: see that \(T\) is Hausdorff using the local criterion for openness.
Step 1:
Let us suppose that \(T\) is Hausdorff.
Step 2:
Let us see that \(T \times T \setminus S \subseteq T \times T\) is open.
Let \((t_1, t_2) \in T \times T \setminus S\) be any.
\(t_1 \neq t_2\).
As \(T\) is Hausdorff, there are an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T\), and an open neighborhood of \(t_2\), \(U_{t_2} \subseteq T\), such that \(U_{t_1} \cap U_{t_2} = \emptyset\).
\(U_{t_1} \times U_{t_2} \subseteq T \times T\) is an open neighborhood of \((t_1, t_2)\).
\(U_{t_1} \times U_{t_2} \subseteq T \times T \setminus S\), because for each \((t'_1, t'_2) \in U_{t_1} \times U_{t_2}\), \(t'_1 \neq t'_2\), because \(U_{t_1} \cap U_{t_2} = \emptyset\).
So, by the local criterion for openness, \((T \times T) \setminus S\) is open.
So, \(S\) is closed.
Step 3:
Let us suppose that \(S\) is closed.
Step 4:
\((T \times T) \setminus S \subseteq T \times T\) is open.
Let \(t_1, t_2 \in T\) be any such that \(t_1 \neq t_2\).
\((t_1, t_2) \in (T \times T) \setminus S\).
There is an open neighborhood of \((t_1, t_2)\), \(U_{(t_1, t_2)} \subseteq T \times T\), such that \(U_{(t_1, t_2)} \subseteq (T \times T) \setminus S\), by the local criterion for openness.
There are an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T\), and an open neighborhood of \(t_2\), \(U_{t_2} \subseteq T\), such that \(U_{t_1} \times U_{t_2} \subseteq U_{(t_1, t_2)}\), by the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
\(U_{t_1} \cap U_{t_2} = \emptyset\), because for each \((t'_1, t'_2) \in U_{t_1} \times U_{t_2}\), \((t'_1, t'_2) \in U_{(t_1, t_2)} \subseteq (T \times T) \setminus S\), which implies that \(t'_1 \neq t'_2\), so, for each \(t'_1 \in U_{t_1}\), \(t'_1 \notin U_{t_2}\).
So, \(T\) is Hausdorff.
