description/proof of that compact subset of Hausdorff topological space 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 compact subset of topological space.
- The reader knows a definition of closed subset of topological space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that any compact subset of any Hausdorff topological space 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:
\(T\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
\(K\): \(\in \{\text{ the compact subsets of } T\}\)
//
Statements:
\(K\): \(\in \{\text{ the closed subsets of } T\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(T \setminus K\) is open by for taking for each \(t \in T \setminus K\), an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \subseteq T \setminus K\).
Step 1:
Let us see that \(T \setminus K\) is open.
Let \(t \in T \setminus K\) be any.
For each \(k \in K\), as \(t \neq k\), there are an open neighborhood of \(k\), \(U_k \subseteq T\), and an open neighborhood of \(t\), \(U_{t, k} \subseteq T\), such that \(U_k \cap U_{t, k} = \emptyset\), because \(T\) is Hausdorff.
\(K \subseteq \cup_{k \in K} U_k\), which means that \(\{U_k \vert k \in K\}\) is an open cover of \(K\).
There is a finite subset, \(\{k_j \in K \vert j \in J\}\), where \(J\) is a finite index set, such that \(K \subseteq \cup_{j \in J} U_{k_j}\), because \(K\) is compact.
Let \(U_t := \cap_{j \in J} U_{t, k_j} \subseteq T\), which is an open neighborhood of \(t\).
\(U_t \cap \cup_{j \in J} U_{k_j} = \emptyset\), because for each \(j \in J\), \(U_t \cap U_{k_j} \subseteq U_{t, k_j} \cap U_{k_j} = \emptyset\).
So, \(U_t \cap K \subseteq U_t \cap \cup_{j \in J} U_{k_j} = \emptyset\).
So, \(U_t \subseteq T \setminus K\).
So, by the local criterion for openness, \(T \setminus K \subseteq T\) is open, and \(K\) is closed.
