A description/proof of that net to product topological space converges to point iff each projection after net converges to component of point
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Description 1
- 2: Proof 1
- 3: Description 2
- 4: Proof 2
- 5: Note
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition 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 possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space.
Target Context
- The reader will have a description and a proof of the proposition that any net to any product topological space converges to a point if and only if the projection to each constituent space after the net converges to the corresponding component 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: Description 1
For any product topological space, \(T = \times_{\alpha} T_\alpha\) where \(\alpha \in A\) is any possibly uncountable indices set, any net, \(n: D \rightarrow T\) where \(D\) is any directed set, converges to a point, \(p = (. . ., p_\alpha, . . .)\) (which is really a function, \(f\), such that \(f (\alpha) = p_\alpha\), which is \(lim \text{ }n = p\), if and only if each projection, \(\pi_\alpha: T \rightarrow T_\alpha\), after \(n\) converges to the corresponding component of \(p\), \(p_\alpha\), which is \(lim \text{ }\pi_\alpha \circ n = p_\alpha\).
2: Proof 1
Let us suppose that \(n\) converges to \(p\). For any neighborhood, \(N_p \subseteq T\), of \(p\), there is a \(d \in D\) such that for each \(d' \in D \text{ such that }d \leq d'\), \(n (d') \in N_p\). For any neighborhood, \(N_{p_\alpha} \subseteq T_\alpha\), of \(p_\alpha\), there is an open set, \(U_{p_\alpha} \subseteq N_{p_\alpha}\), around \(p_\alpha\). For any fixed \(\alpha\), let us take the neighborhood, \(N_p \subseteq T\), of \(p\), that is \(N_p = \times_{\beta \in A} U_\beta\) where \(U_\beta = U_{p_\alpha}\) when \(\beta = \alpha\) and \(U_\beta = T_\beta\) when \(\beta \neq \alpha\), which is indeed an open neighborhood of \(p\) by the definition of product topology. Then, \(n (d') \in N_p\), which implies that \(\pi_\alpha \circ n (d') \in U_{p_\alpha}\). So, \(\pi_\alpha \circ n\) converges to \(p_\alpha\).
Let us suppose that \(\pi_\alpha \circ n\) converges to \(p_\alpha\) for each \(\alpha \in A\). For each neighborhood, \(N_{p_\alpha} \subseteq T_\alpha\), of \(p_\alpha\), there is a \(d_\alpha \in D\) such that for each \(d' \in D \text{ such that } d_\alpha \leq d'\), \(n (d') \in N_{p_\alpha}\). For any neighborhood, \(N_p \subseteq T\), of \(p\), there is an open set, \(U_p \subseteq N_p\), around \(p\), where \(U_p = \cup_\beta \times_\alpha U_{\beta-\alpha}\) where \(\alpha \in A\) where \(\beta\) is a member of any possibly uncountable indices set, and for any fixed \(\beta\), only finite \(U_{\beta-\alpha}\)s are not \(T_\alpha\), by the definition of product topology. There is a \(d_{\beta-\alpha} \in D\) such that for each \(d' \in D \text{ such that } d_{\beta-\alpha} \leq d'\), \(\pi_{\alpha} \circ n (d') \in U_{\beta-\alpha}\). As \(D\) is directed, for any fixed \(\beta\), there is a \(d \in D\) such that \(d_{\beta-\alpha} \leq d\) for the finite number of \(U_{\beta-\alpha}\)s, then for each \(d' \in D \text{ such that } d \leq d'\), \(\pi_\alpha \circ n (d') \in U_{\beta-\alpha}\) for any \(\alpha\) for the fixed \(\beta\). So, \(n (d') \in \times_\alpha U_{\beta-\alpha} \subseteq U_p\). So, \(n\) converges to \(p\).
3: Description 2
For any product topological space, \(T = T_1 \times T_2 \times . . . \times T_n\), any net, \(n: D \rightarrow T\) where \(D\) is any directed set, converges to a point, \(p = (p_1, p_2, . . ., p_n)\), which is \(lim \text{ }n = p\), if and only if each projection, \(\pi_i: T \rightarrow T_i\), after \(n\) converges to the corresponding component of \(p\), \(p_i\), which is \(lim \text{ }\pi_i \circ n = p_i\).
4: Proof 2
By the proposition that any possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space, \(T\) is homeomorphic to \(T' = \times_i T_i\) where \(i \in \{1, 2, . . ., n\}\), and Description 1 applies to \(T'\).
If \(n\) converges to \(p\), the corresponding net, \(n': D \rightarrow T'\), converges to the corresponding point, \(p'\), because for any neighborhood, \(N_{p'} \subseteq T'\), of \(p'\), \(n\) is eventually in the corresponding neighborhood, \(N_p\), of \(p\), which implies that \(n'\) is eventually in \(N_{p'}\). By Description 1, \(\pi'_i \circ n'\) converges to \(p_i\) (\(p'_i\) and \(p_i\) are the same). Then, \(\pi_i \circ n\) converges to \(p_i\), because \(\pi'_i \circ n'\) and \(\pi_i \circ n\) are the same.
If \(\pi_i \circ n\) converges to \(p_i\) for each \(i\), \(\pi'_i \circ n'\) converges to \(p_i\). By Description 1, \(n'\) converges to \(p'\). Then \(n\) converges to \(p\), because for any neighborhood, \(N_p \subseteq T\), of \(p\), \(n'\) is eventually in the corresponding neighborhood, \(N_{p'}\), of \(p'\), which implies that \(n\) is eventually in \(N_p\).
5: Note
As is stated in the proposition that any possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space, \(T = T_1 \times T_2 \times . . . \times T_n\) and \(T' = \times_i T_i\) are not exactly the same, as any element of \(T\) is like a \(\langle p_1, p_2, . . ., p_n \rangle\) while any element of \(T'\) is a function, \(f: \{1, 2, . . ., n\} \rightarrow \cup_i T_i\), although some people may sloppily say that they are the same. While Description 2 seems obvious from the homeomorphism anyway, we have bothered to go more explicit.
