description/proof of that topological space is compact iff each universal net with directed index set into space is convergent
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 universal 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 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 universal net with directed index set into the space is convergent.
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:
\(T \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(\forall f \in \{\text{ the universal nets with directed index sets into } T\} (f \in \{\text{ the convergent nets }\})\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T\) is compact; Step 2: suppose that \(f\) did not converge, and find a contradiction; Step 3: suppose that each universal net converges; 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 us suppose that a universal net, \(f: D \to T\), did not converge.
Let \(t \in T\) be any.
\(t\) would not be any convergence of \(f\), because \(f\) did not have any convergence.
There would be an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(f\) was not eventually in \(U_t\), because otherwise, \(f\) would be eventually in each open neighborhood of \(t\), which would mean that \(t\) was a convergence of \(f\), a contradiction.
\(f\) would be eventually in \(T \setminus U_t\), because \(f\) was universal.
So, there would be a \(d \in D\) such that for each \(d' \in D\) such that \(d \le d'\), \(f (d') \in T \setminus U_t\), so, \(f (d') \notin U_t\).
\(\{U_t \vert t \in T\}\) would be an open cover of \(T\).
There would be a finite subcover, \(\{U_{t_j} \vert j \in J\}\), because \(T\) was compact. Let \(J = \{1, ..., n\}\) without loss of generality.
For each \(j \in J\), there would be a \(d_j \in D\) such that for each \(d' \in D\) such that \(d_j \le d'\), \(f (d') \notin U_{t_j}\).
There would be a \(d \in D\) such that \(d_1, ..., d_n \le d\): there wold be a \(d_{1, 2} \in D\) such that \(d_1, d_2 \le d_{1, 2}\), there wold be a \(d_{1, 2, 3} \in D\) such that \(d_{1, 2}, d_3 \le d_{1, 2, 3}\), which would imply that \(d_1, d_2, d_3 \le d_{1, 2, 3}\), and so on, and \(d := d_{1, ..., n}\) would do.
Then, for each \(d' \in D\) such that \(d \le d'\), \(f (d') \notin U_{t_1}, ..., U_{t_n}\), so, \(f (d') \notin U_{t_1} \cup ... \cup U_{t_n} = T\), a contradiction against that \(f\) was into \(T\).
So, \(f\) converges.
Step 3:
Let us suppose that each universal net converges.
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 \(\{C_{j'} \vert j' \in J'\}\).
There is a universal subnet, \(f \circ g: D \to T\), where \(g: D \to \{C_{j'} \vert j' \in J'\}\) is a final map, by the proposition that any net with any directed index set has a universal subnet.
By the supposition, \(f \circ g\) converges to a \(t \in T\).
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.
