Connectedness of Finite Topological Products

# Connectedness of Finite Topological Products

Recall that if $X$ and $Y$ are topological spaces and $X \times Y$ is the Cartesian product of $X$ and $Y$, then we defined the product topology on $X \times Y$ to be the topology whose basis is:

(1)
\begin{align} \quad \mathcal B = \{ U \times V \subseteq X \times Y : U \: \mathrm{is \: open \: in \:} X, \: V \mathrm{is \: open \: in \:} Y \} \end{align}

We define the topological product of $X$ and $Y$ to then be $X \times Y$ with the product topology.

More generally, if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces with $\displaystyle{X_1 \times X_2 \times ... \times X_n = \prod_{i=1}^{n} X_i}$ then the product topology on $\displaystyle{\prod_{i=1}^{n} X_i}$ has the basis:

(2)
\begin{align} \quad \mathcal B = \left \{ \prod_{i=1}^{n} U_i \subseteq \prod_{i=1}^{n} X_i : U_i \mathrm{is \: open \: in \:} X_i, \: \forall \: i \in \{ 1, 2, ..., n \} \right \} \end{align}

Suppose that we have a finite collection of connected topological spaces, $\{ X_1, X_2, ..., X_n \}$. Then is the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ connected too? The answer is yes!

 Lemma 1: If $X$ and $Y$ are topological spaces such that $X$ is connected. Let $b \in Y$ and give the singleton $\{ b \}$ the subspace topology from $Y$. Then the product $X \times \{ b \}$ is connected.
• Proof: Suppose not, i.e., suppose that $X \times \{ b \}$ is disconnected. Then there exists open sets $U_1 \times V_1, U_2 \times V_2 \subseteq X \times \{ b \}$ such that $(U_1 \times V_1), (U_2 \times V_2) \neq \emptyset$, $(U_1 \times V_1) \cap (U_2 \times V_2) = \emptyset$ and:
(3)
\begin{align} \quad X \times \{ b \} = (U_1 \times V_1) \cup (U_2 \times V_2) \end{align}
• Notice that $V_1, V_2 \neq \emptyset$ since then $U_1 \times V_1 = \emptyset$ and/or $U_2 \times V_2 = \emptyset$ which is a contradiction. The only other open set in $\{ b \}$ with the subspace topology is the whole set $\{ b \}$, so $V_1 = \{ b \}$ and $V_2 = \{ b \}$. So $U_1 \times V_1 = U_1 \times \{ b \}$ and $U_2 \times V_2 = U_2 \times \{ b \}$ and do not contain any points in common we must have that $U_1, U_2 \subset X$, $U_1, U_2 = \emptyset$, and $U_1 \cap U_2 = \emptyset$. Note that:
(4)
\begin{align} \quad X \times \{ b \} = (U_1 \times \{ b \}) \cup (U_2 \times \{ b \} = (U_1 \cup U_2) \times \{ b \} \end{align}
• This implies that $X = U_1 \cup U_2$. Therefore $\{ U_1, U_2 \}$ is a separation of $X$ which contradicts $X$ being connected. So the assumption that $X \times \{ b \}$ was disconnected is false. $\blacksquare$
 Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of connected topological spaces. Then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is connected.
• Proof: Let $X$ and $Y$ be any two connected topological spaces. Let $(a, b)$ be any point in $X \times Y$. Since $X$ is connected, $X \times \{ b \}$ is also connected by Lemma 1. Similarly, for all $x \in X$ we have that $\{ x \} \times Y$ is also connected. Define $T_x$ by:
(5)
\begin{align} \quad T_x = (X \times \{ b \}) \cup (\{ x \} \times Y) \end{align} (6)
\begin{align} \quad X \times Y = \bigcup_{x \in X} T_x = \bigcup_{x \in X} (X \times \{ b \}) \cup (\{ x \} \times Y) \end{align}
• Each $T_x$ contains the point $(a, b)$ since $(a, b) \in (X \times \{ b \}) \subset (X \times \{ b \}) \cup (\{ x \} \times Y) = T_x$, and by the theorem referenced above, $X \times Y$ is therefore connected.
• Inductively, we see that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then $\displaystyle{\prod_{i=1}^{n} X_i}$ is also connected. $\blacksquare$

Theorem 1 can be generalized as true for an arbitrary collection of topological spaces. We state the result below but we will not prove it.

 Theorem 2: Let $\{ X_i \}_{i \in I}$ be an arbitrary collection of connected topological spaces. Then the topological product $\displaystyle{\prod_{i \in I} X_i}$ is also connected.