description/proof of that for continuous map and net with directed index set that converges to point on domain, image of net converges to image of point and if codomain is Hausdorff, convergence of image of net is image of point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of net with directed index set.
- The reader knows a definition of continuous map.
- The reader admits the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(D\): \(\in \{\text{ the directed index sets }\}\)
\(N\): \(: D \to T_1\), \(\in \{\text{ the nets with } D\}\)
\(t\): \(\in T_1\), \(\in \{\text{ the convergences of } N\}\)
\(f (t)\): \(\in T_2\)
//
Statements:
\(f (t) \in \{\text{ the convergences of } f \circ N\}\)
\(\land\)
(
\(T_2 \in \{\text{ the Hausdorff topological spaces }\}\)
\(\implies\)
\(lim f \circ N = f (t)\)
)
//
2: Note
"the convergence of net" makes sense only when there is the unique convergence of the net.
So, 'the net converges to a point' is different from 'the convergence of the net is the point': the latter means that there is the unique convergence of the net.
The 1st-half of this proposition says that \(f \circ N\) converges to \(f (t)\); the 2nd-half of this proposition says that \(f \circ N\) has the unique convergence and the convergence is \(f (t)\).
3: Proof
Whole Strategy: Step 1: see that \(f \circ N\) is a net with \(D\); Step 2: take any neighborhood of \(f (t)\), \(N_{f (t)}\); Step 3: take an open neighborhood of \(t\), \(U_t\), such that \(f (U_t) \subseteq N_{f (t)}\); Step 4: take a \(j_0 \in D\) such that for each \(j \in D\) such that \(j_0 \le j\), \(N (j) \in U_t\); Step 5: see that \(f \circ N (j) \in N_{f (t)}\); Step 6: suppose that \(T_2\) is Hausdorff, and see that the convergence of \(f \circ N\) is unique and is \(f (t)\).
Step 1:
\(f \circ N\) is \(: D \to T_1 \to T_2\), and so, is a net with \(D\).
Step 2:
Let us take any neighborhood of \(f (t)\), \(N_{f (t)} \subseteq T_2\).
Step 3:
As \(f\) is continuous, there is an open neighborhood of \(t\), \(U_t \subseteq T_1\), such that \(f (U_t) \subseteq N_{f (t)}\).
Step 4:
As \(N\) converges to \(t\), there is a \(j_0 \in D\) such that for each \(j \in D\) such that \(j_0 \le j\), \(N (j) \in U_t\).
Step 5:
For each \(j \in D\) such that \(j_0 \le j\), \(f \circ N (j) \in f (U_t) \subseteq N_{f (t)}\), which means that \(f \circ N\) converges to \(f (t)\).
But that does not necessarily mean that \(f (t)\) is the unique convergence. So, it is not warranted to talk about "the convergence of \(f \circ N\)".
Step 6:
Let us suppose that \(T_2\) is Hausdorff.
By the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most, there is at most 1 convergence, but there is indeed a convergence, \(f (t)\), and so, \(f (t)\) is the unique convergence of \(f \circ N\).
So, it is warranted to denote \(lim f \circ N = f (t)\).
