A description/proof of that for Hausdorff topological space, net with directed index set can have only 1 convergence
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of convergence of net with directed index set.
Target Context
- The reader will have a description and a proof of the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
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: Description
For any directed set, \(S\), any Hausdorff topological space, \(T\), and any net with directed index set, \(f: S \rightarrow T\), \(f\) can have only 1 convergence at most.
2: Proof
Suppose \(f\) had 2 convergences, \(p_1, p_2 \in T\). For any neighborhood of \(p_j\) where \(j = 1 \text{ or } 2\), \(N_j\), there would be an index, \(i_j \in S\), such that \(f (i) \in N_j\) for every \(i \in S, i_j \leq i\). As \(T\) is Hausdorff, let us take \(N_1\) and \(N_2\) as disjoint. By the definition of directed set, there would be an index, \(i_3 \in S\), such that \(i_1 \leq i_3\) and \(i_2 \leq i_3\). For every \(i_3 \leq i\), \(f (i) \in N_1\) and \(f (i) \in N_2\), which is a contradiction, as \(N_1\) and \(N_2\) would be disjoint.
