A description/proof of that product of path-connected topological spaces is path-connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of path-connected topological space.
- The reader knows a definition of product topology.
- The reader admits the proposition that any map into the product of any topological spaces is continuous if and only if the projection of the map to each constituent topological space is continuous.
Target Context
- The reader will have a description and a proof of the proposition that the product of any path-connected topological spaces is path-connected.
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 product topological space, \(T = \times_{\alpha \in A} T_\alpha\) where \(A\) is any possibly uncountable indices set, such that each \(T_\alpha\) is path-connected, \(T\) is path-connected.
2: Proof
For any points, \(p_1, p_2 \in T\), \(p_{i-\alpha} \in T_\alpha\) where \(p_{i-\alpha}\) is the \(\alpha\) component of \(p_i\), which is really \(p_i (\alpha)\). As \(T_\alpha\) is path-connected, there is a path, \(\lambda_\alpha: [r_1, r_2] \rightarrow T_\alpha, \lambda_\alpha (r_i) = p_{i-\alpha}\). Let us define \(\lambda: [r_1, r_2] \rightarrow T\) such that \(\pi_\alpha \lambda (r) = \lambda_\alpha (r)\) where \(\pi_\alpha\) is the projection of \(T\) to \(T_\alpha\). \(\lambda\) is a path, by the proposition that any map into the product of any topological spaces is continuous if and only if the projection of the map to each constituent topological space is continuous. \(\lambda (r_1) = p_1\) and \(\lambda (r_2) = p_2\).
