description/proof of that for net with directed index set eventually in subset of topological space that converges on space, convergence is on closure of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of net with directed index set eventually in subset.
- The reader knows a definition of convergence of net with directed index set.
- The reader knows a definition of closure of subset of topological space.
- The reader admits a local characterization of closure: any point on any topological space is on the closure of any subset if and only if its every neighborhood intersects the subset.
Target Context
- The reader will have a description and a proof of 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.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
\(D\): \(\in \{\text{ the directed sets }\}\)
\(f\): \(: D \to T\), \(\in \{\text{ the nets eventually in } S\}\)
\(t\): \(\in T\)
//
Statements:
\(f\) converges to \(t\)
\(\implies\)
\(t \in \overline{S}\).
//
2: Note
While the proposition that for any net with directed index set on any subset of any topological space that converges on the space, the convergence is on the closure of the subset has been already proved, in fact, the net does not really need to be on the subset but to be eventually in the subset, according to this proposition.
3: Proof
Whole Strategy: Step 1: suppose that \(t \notin \overline{S}\) and find a contradiction.
Step 1:
Let us suppose that \(f\) converges to \(t\).
Let us suppose that \(t \notin \overline{S}\).
By a local characterization of closure: any point on any topological space is on the closure of any subset if and only if its every neighborhood intersects the subset, there would be a neighborhood of \(t\), \(N_t \subseteq T\), such that \(N_t \cap S = \emptyset\).
There would be a \(d \in D\), such that for each \(d' \in D\) such that \(d \le d'\), \(f (d') \in N_t\), because \(f\) converged to \(t\).
Then, for each \(d'' \in D\), there would be a \(d''' \in D\) such that \(d, d'' \le d'''\), and \(f (d''') \in N_t\), which would imply that \(f (d''') \notin S\), which would mean that \(f\) was not eventually in \(S\), a contradiction.
So, \(t \in \overline{S}\).
