A description/proof of that product of connected topological spaces is connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological space.
- The reader knows a definition of product topology.
Target Context
- The reader will have a description and a proof of the proposition that the product of any connected topological spaces is 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 connected, \(T\) is connected.
2: Proof
Let us suppose that \(T\) was not connected. \(T = U_1 \cup U_2, U_1 \cap U_2 = \emptyset\), \(U_1 \neq \emptyset\) and \(U_2 \neq \emptyset\) where \(U_i\) would be an open set. \(U_i = \cup_{\beta \in B_i} \times_{\alpha \in A} U_{i-\beta-\alpha}\) where \(B_i\) would be any possibly uncountable indices set, \(U_{i-\beta-\alpha}\) would be an open set such that each of only a finite number of \(U_{i-\beta-\alpha}\)s would not be \(T_\alpha\) for each \(i\) and \(\beta\), by the definition of product topology.
Let us take any point, \(p_{\underline\gamma} \in \pi_{\underline\gamma} T\), where \(\pi_{\underline\gamma}\) would be the projection of \(T\) to \(T_{\underline\gamma} := \times_{\alpha \in A \setminus \{\gamma\}} T_\alpha\). Let us define \(B_{i-\underline\gamma} := \{\beta \in B_i\vert p_{\underline\gamma} \in \times_{\alpha \in A \setminus \{\gamma\}} U_{i-\beta-\alpha}\}\) and \(U_{i-\gamma} := \cup_{\beta \in B_{i-\underline\gamma}} U_{i-\beta-\gamma}\). \(U_{1-\gamma} \cup U_{2-\gamma} = T_\gamma\), because for any \(p_\gamma \in T_\gamma\), the corresponding point, \(p \in T\), such that \(\pi_{\underline{\gamma}} p = p_{\underline{\gamma}}\) and \(\pi_\gamma p = p_\gamma\) would be in \(\times_{\alpha \in A} U_{i-\beta-\alpha}\) for a \(\beta \in B_{i-\underline{\gamma}}\), as \(p \in U_1\) or \(p \in U_2\). \(U_{1-\gamma} \cap U_{2-\gamma} = \emptyset\), because if \(p_\gamma \in U_{1-\gamma} \cap U_{2-\gamma}\), the corresponding point, \(p \in T\), such that \(\pi_{\underline{\gamma}} p = p_{\underline{\gamma}}\) and \(\pi_\gamma p = p_\gamma\) would be in \(U_1 \cap U_2\).
As \(T_\gamma\) is connected, \(U_{1-\gamma} = \emptyset\) or \(U_{2-\gamma} = \emptyset\), which would mean that for any fixed \(p_{\underline{\gamma}}\), all the \(p_\gamma \in T_\gamma\)s would be entirely in \(U_1\) or in \(U_2\). Let us suppose that \(U_{2-\gamma} = \emptyset\) and so, inevitably \(U_{1-\gamma} = T_\gamma\) without loss of generality. For any \(p_\gamma \in U_{1-\gamma}\), \(p_\gamma \in U_{1-\beta-\gamma}\) for a \(\beta \in B_{1-\underline{\gamma}}\), and the corresponding point, \(p \in T\), such that \(\pi_{\underline{\gamma}} p = p_{\underline{\gamma}}\) and \(\pi_\gamma p = p_\gamma\) would be in a \(\times_{\alpha \in A} U_{1-\beta-\alpha}\) for a \(\beta \in B_{1-\underline{\gamma}}\).
But there are only finite number of \(U_{1-\beta-\alpha}\)s such that \(U_{1-\beta-\alpha} \neq T_\alpha\) for the \(\beta\). Let us define \(A_\beta\) as the set of such \(\alpha\)s, \(\{\alpha_1, \alpha_2, . . ., \alpha_n\}\). Then, every point, \(p' \in T\), such that \(\pi_{\alpha \in A_\beta} p' = p_\alpha \in U_{1-\beta-\alpha}\) for every \(\alpha \in A_\beta\) would be in \(\times_{\alpha \in A} U_{1-\beta-\alpha} \subseteq U_1\). \(\alpha_1\) could be taken to be the \(\gamma\) of the above procedure, then, every point, \(p' \in T\), such that \(\pi_{\alpha \in A_\beta \setminus \{\alpha_1\}} p' = p_\alpha \in U_{1-\beta-\alpha}\) for every \(\alpha \in A_\beta \setminus \{\alpha_1\}\) would be in \(U_1\), because as the 1 \(p_{\alpha_1}\) is known to be in \(U_1\), the whole \(T_{\alpha_1}\) have to be in \(U_1\). Then, \(\alpha_2\) could be taken to be the \(\gamma\) of the above procedure, then, every point, \(p' \in T\), such that \(\pi_{\alpha \in A_\beta \setminus \{\alpha_1, \alpha_2\}} p' = p_\alpha \in U_{1-\beta-\alpha}\) for every \(\alpha \in A_\beta \setminus \{\alpha_1, \alpha_2\}\) would be in \(U_1\), likewise, and so on. After all, every point, \(p' \in T\) would be in \(U_1\), a contradiction.
