A description/proof of that product of finite number 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 finite number of 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 finite number of topological spaces, \(T_1, T_2, . . ., T_n\), \(T_1 \times T_2, \times . . ., \times T_n\) is connected.
2: Proof
1st, let us think of the case, \(n = 2\). Suppose that \(T_1 \times T_2\) was not connected. \(T_1 \times T_2 = U_1 \cup U_2\), \(U_1 \cap U_2 = \emptyset\) where \(U_i\) would be a non-empty open set on \(T_1 \times T_2\). By the definition of product topology, \(U_1 = \cup_{\alpha} U_{1-\alpha} \times U_{2-\alpha}\) where \(\{\alpha\}\) would be a possibly uncountable index set and \(U_{i-\alpha}\) would be non-empty open on \(T_i\); \(U_2 = \cup_{\beta} U_{1-\beta} \times U_{2-\beta}\) where \(\{\beta\}\) would be a possibly uncountable index set and \(U_{i-\beta}\) would be non-empty open on \(T_i\). For any point, \(p_1 \in T_1\), \(\{p_1\} \times T_2\) would have to be covered by \(U_1 \cup U_2\). Let us take the subset of \(\{\alpha\}\), \(\{\alpha'\} = \{\alpha' \in \{\alpha\}| p_1 \in U_{1-\alpha'}\}\) and the subset of \(\{\beta\}\), \(\{\beta'\} = \{\beta' \in \{\beta\}| p_1 \in U_{1-\beta'}\}\). \((\cup_{\alpha'} U_{2-\alpha'}) \cup (\cup_{\beta'} U_{2-\beta'}) = T_2\), because for each \(p_2 \in T_2\), \((p_1, p_2)\) would have to be in a \(U_{1-\alpha'} \times U_{2-\alpha'}\) or in a \(U_{1-\beta'} \times U_{2-\beta'}\). \(\cup_{\alpha'} U_{2-\alpha'}\) and \(\cup_{\beta'} U_{2-\beta'}\) would not share any point, because if \(p_2 \in \cup_{\alpha'} U_{2-\alpha'}, \cup_{\beta'} U_{2-\beta'}\), \((p_1, p_2) \in U_1, U_2\). \(\cup_{\alpha'} U_{2-\alpha'}\) or \(\cup_{\beta'} U_{2-\beta'}\) would be empty, because otherwise, \(T_2\) would not be connected, being a union of disjoint non-empty open sets. Suppose that \(\cup_{\alpha'} U_{2-\alpha'} = T_2\) without loss of generality. \(p_1 \in \cup_\alpha U_{1-\alpha}\), because otherwise, there would be no \((p_1, ?)\) on \(U_1 \cup U_2\), and \((p_1, ?) \notin U_2\), because otherwise, a \(p_1, p_2\) would be shared by \(U_1\) and \(U_2\). But not all the points of \(T_1\) would be on \(U_1\), because otherwise, \(U_2\) would be empty. So, while \((\cup_{\alpha} U_{1-\alpha}) \cup (\cup_{\beta} U_{1-\beta}) = T_1\), because each \(p_1 \in T_1\) would have to be in a \(U_{1-\alpha}\) or in a \(U_{1-\beta}\), any point would not be shared by \(\cup_{\alpha} U_{1-\alpha}\) and \(\cup_{\beta} U_{1-\beta}\), neither of which would be empty, so, \(T_1\) would not be connected, as being a union of disjoint non-empty open sets, a contradiction. So, \(T_1 \times T_2\) is connected.
\(T_1 \times T_2, \times . . ., \times T_n\) can be proved to be connected inductively, as \(T_1 \times T_2\) is connected, \(T_1 \times T_2 \times T_3 = (T_1 \times T_2) \times T_3\) is connected, and so on.
