description/proof of that topological space is compact iff each net with directed index set into space has convergent subnet
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact topological space.
- The reader knows a definition of subnet of net with directed index set.
- The reader knows a definition of convergence of net with directed index set.
- The reader admits the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.
- The reader admits the proposition that any net with any directed index set has a universal subnet.
- The reader admits the proposition that any topological space is compact if and only if each universal net with directed index set into the space is convergent.
- The reader admits the proposition that for any net with directed index set eventually in any subset of any topological space that converges on the space, the convergence is on the closure of the subset.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is compact if and only if each net with directed index set into the space has a convergent subnet.
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:
\(T \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(\forall f: D \to T \in \{\text{ the nets with directed index sets into } T\} (\exists f': D' \to D \in \{\text{ the final maps }\} (f \circ f' \in \{\text{ the convergent nets }\}))\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T\) is compact; Step 2: take a universal subnet of \(f\), and see that the subnet is convergent; Step 3: suppose that each net has a convergent subnet; Step 4: see that for each collection of closed subsets for which each finite intersection of members is not empty, the intersection of the collection is not empty; Step 5: apply the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.
Step 1:
Let us suppose that \(T\) is compact.
Step 2:
Let \(f: D \to T\) be any net.
\(f\) has a universal subnet, \(f \circ f': D' \to T\), where \(f': D' \to D\) is a final map, by the proposition that any net with any directed index set has a universal subnet.
\(f \circ f'\) is convergent, by the proposition that any topological space is compact if and only if each universal net with directed index set into the space is convergent.
That means that \(f\) has a convergent subnet.
Step 3:
Let us suppose that each net has a convergent subnet.
Step 4:
We are going to see that for each collection of some closed subsets of \(T\) for which each finite intersection of members is not empty, the intersection of the collection is not empty.
Let \(\{C_j \vert j \in J\}\) be any such collection.
Let us augment it with all the finite intersections of members, with the result \(\{C_{j'} \vert j' \in J'\}\).
\(\{C_{j'} \vert j' \in J'\}\) is a collection of some closed subsets of \(T\) for which each finite intersection of members is not empty, because each added member is closed, as a finite intersection of closed subsets, and each finite intersection of members is a finite intersection of original members.
Let \(\{C_{j'} \vert j' \in J'\}\) have the relation (a partial ordering) such that \(C_{j'} \le C_{l'}\) if and only if \(C_{l'} \subseteq C_{j'}\).
\(\{C_{j'} \vert j' \in J'\}\) with the relation is a directed set, because 1) \(C_{j'} \leq C_{j'}\) for each \(C_{j'} \in \{C_{j'} \vert j' \in J'\}\): \(C_{j'} \subseteq C_{j'}\); 2) if \(C_{j'} \leq C_{l'}\) and \(C_{l'} \leq C_{m'}\), \(C_{j'} \leq C_{m'}\): \(C_{m'} \subseteq C_{l'} \subseteq C_{j'}\); 3) for each pair, \(C_{j'}, C_{l'} \in \{C_{j'} \vert j' \in J'\}\), there is a \(C_{m'} \in \{C_{j'} \vert j' \in J'\}\) such that \(C_{j'} \leq C_{m'}\) and \(C_{l'} \leq C_{m'}\): \(C_{m'} := C_{j'} \cap C_{l'} \in \{C_{j'} \vert j' \in J'\}\) and \(C_{m'} \subseteq C_{j'}, C_{l'}\).
Let us define a net with the directed index set into \(T\), \(f: \{C_{j'} \vert j' \in J'\} \to T\), that maps \(C_{j'}\) to a \(c_{j'} \in C_{j'}\), by the axiom of choice: \(C_{j'}\) is not empty, because it is a finite intersection of members of \(\{C_{j'} \vert j' \in J'\}\).
By the supposition, there is a convergent subnet, \(f \circ g: D \to T\), where \(g: D \to \{C_{j'} \vert j' \in J'\}\) is a final map.
Let \(t\) be a convergence of \(f \circ g\).
Let \(C_{j'} \in \{C_{j'} \vert j' \in J'\}\) be any.
There is a \(d \in D\) such that for each \(d' \in D\) such that \(d \le d'\), \(C_{j'} \le g (d')\), because \(g\) is final, which equals that \(g (d') \subseteq C_{j'}\).
\(f \circ g (d') \in g (d') \subseteq C_{j'}\), so, \(f \circ g\) is eventually in \(C_{j'}\).
\(t \in \overline{C_{j'}} = C_{j'}\), by the proposition that for any net with directed index set eventually in any subset of any topological space that converges on the space, the convergence is on the closure of the subset.
That implies that \(t \in \cap \{C_{j'} \vert j' \in J'\}\), which implies that \(\cap \{C_{j'} \vert j' \in J'\} \neq \emptyset\).
Step 5:
So, \(T\) is compact, by the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.
