description/proof of that topological space is compact iff for each set of closed subsets whose intersection is empty, there is nonempty finite subset whose intersection is empty
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact topological space.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is compact if and only if for each possibly uncountable set of closed subsets whose intersection is empty, there is a nonempty finite subset whose intersection is empty.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
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Statements:
\(T \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(\forall \{C_j \in \{\text{ the closed subsets of } T\} \vert j \in J\} \text{ where } J \text{ is any possibly uncountable index set } \text{ such that } \cap_{j \in J} C_j = \emptyset (\exists J^` \subseteq J \text{ such that } \vert J \vert \in \mathbb{N} \setminus \{0\} (\cap_{j \in J^`} C_j = \emptyset))\)
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2: Proof
Whole Strategy: Step 1: suppose that there is a \(J^`\) for each such \(\{C_j \vert j \in J\}\); Step 2: see that for each open cover of \(T\), there is a finite subcover; Step 3: suppose that \(T\) is compact; Step 4: see that there is a \(J^`\) for each such \(\{C_j \vert j \in J\}\).
Step 1:
Let us suppose that for each such \(\{C_j \vert j \in J\}\), there is a \(J^`\).
Step 2:
Let \(\{U_j \vert j \in J\}\), where \(J\) is a possibly uncountable index set, be any open cover of \(T\).
Let us take \(\{T \setminus U_j \vert j \in J\}\), which is a set of some closed subsets.
\(\cap_{j \in J} (T \setminus U_j) = \emptyset\), because if there was a \(t \in \cap_{j \in J} (T \setminus U_j)\), \(t \in T \setminus U_j\) for each \(j \in J\), so, \(t \notin U_j\) for each \(j \in J\), so, \(t \notin \cup_{j \in J} U_j = T\), a contradiction.
By the supposition, there is a nonempty finite subset, \(J^` \subseteq J\), such that \(\cap_{j \in J^`} (T \setminus U_j) = \emptyset\).
\(\cup_{j \in J^`} U_j = T\), because if there was a \(t \in T \setminus \cup_{j \in J^`} U_j\), \(t \notin \cup_{j \in J^`} U_j\), so, \(t \notin U_j\) for each \(j \in J^`\), so, \(t \in T \setminus U_j\) for each \(j \in J^`\), so, \(t \in \cap_{j \in J^`} (T \setminus U_j) = \emptyset\), a contradiction.
So, \(\{U_j \vert j \in J^`\}\) is a finite subcover.
So, \(T\) is compact.
Step 3:
Let us suppose that \(T\) is compact.
Step 4:
\(\cup_{j \in J} (T \setminus C_j) = T\), because while \(\cup_{j \in J} (T \setminus C_j) \subseteq T\) is obvious, for each \(t \in T\), \(t \notin C_j\) for a \(j \in J\), because otherwise, \(t \in \cap_{j \in J} C_j = \emptyset\), a contradiction, so, \(t \in T \setminus C_j\) for a \(j \in J\), so, \(t \in \cup_{j \in J} (T \setminus C_j)\), so, \(T \subseteq \cup_{j \in J} (T \setminus C_j)\).
So, \(\{T \setminus C_j \vert j \in J\}\) is an open cover of \(T\).
There is a finite subcover, \(\{T \setminus C_j \vert j \in J^`\}\), because \(T\) is compact.
\(\cap_{j \in J^`} C_j = \emptyset\), because if there was a \(t \in \cap_{j \in J^`} C_j\), \(t \in C_j\) for each \(j \in J^`\), so, \(t \notin T \setminus C_j\) for each \(j \in J^`\), so, \(t \notin \cup_{j \in J^`} T \setminus C_j = T\), a contradiction.
