Connectedness of Box Topological Products

Connectedness of Box Topological Products

Recall from the Connectedness of Finite Topological Products page that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of connected topological spaces then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also connected. We also noted that if instead $\{ X_i \}_{i \in I}$ is an arbitrary collection of connected topological spaces then the topological product $\displaystyle{\prod_{i \in I} X_i}$ is also connected, though, we did not prove this.

Recall from the Box Topological Products of Topological Spaces page that another somewhat natural topology to put on the Cartesian product $\displaystyle{\prod_{i \in I} X_i}$ is called the box topology. The box topology on this Cartesian product is the topology $\tau$ with the basis:

(1)
\begin{align} \quad \mathcal B = \left \{ \prod_{i \in I} U_i : U_i \subseteq X_i \: \mathrm{is \: open \: for \: all \:} i \in I \right \} \end{align}

We already know that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then the resulting topological product and box topological product of these spaces are the same. So if this finite collection of topological spaces are all connected then by the theorem referenced above, $\displaystyle{\prod_{i=1}^{n} X_i}$ will also be connected.

Interestingly enough, if instead $\{ X_i \}_{i \in I}$ is an infinite collection of connected topological spaces then the box topological product need not be connected!

For example, consider the topological space $\mathbb{R}$ with the usual topology. Let $\{ \mathbb{R}_i \}_{i=1}^{\infty}$ be an infinite collection of copies of this topological space. We will show that even though each $\mathbb{R}_i$ is connected, the box topological product $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ is not connected.

Define the sets $A$ and $B$ as follows:

(2)
\begin{align} \quad A = \left \{ (x_i)_{i=1}^{\infty} \in \prod_{i=1}^{\infty} \mathbb{R}_i : (x_i)_{i=1}^{\infty} \: \mathrm{is \: a \: bounded \: sequence.} \right \} \end{align}
(3)
\begin{align} \quad B = \left \{ (x_i)_{i=1}^{\infty} \in \prod_{i=1}^{\infty} \mathbb{R}_i : (x_i)_{i=1}^{\infty} \: \mathrm{is \: an \: unbounded \: sequence.} \right \} \end{align}

Then clearly $A, B \not \emptyset$, $A \cap B \neq \emptyset$, and $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i = A \cup B}$. We only need to show that the sets $A$ and $B$ are open to prove that $\{ A, B \}$ is a separation of $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ and so $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ is disconnected.

Let $(x_i)_{i=1}^{\infty} \in A$. Then consider the following open neighbourhood $U$ of $(x_i)_{i=1}^{\infty}$:

(4)
\begin{align} \quad U = (x_1 - 1, x_1 + 1) \times (x_2 - 1, x_2 + 1) \times ... \times (x_n - 1, x_n + 1) \times ... \end{align}

Then $(x_i)_{i=1}^{\infty} \in U$, and $U$ is open in $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ with the box topology. Furthermore, $U$ contains only bounded sequences, so $(x_i)_{i=1}^{\infty} in U \subset A$ which shows that $A = \mathrm{int}(A)$, so $A$ is open in $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$.

Similarly, let $(x_i)_{i=1}^{\infty} \in B$, and consider the following open neighbourhood $V$ of $(x_i)_{i=1}^{\infty}$:

(5)
\begin{align} \quad V = (x_1 - 1, x+1 + 1) \times (x_2 - 1, x_2 + 1) \times ... \times (x_n - 1, x_n + 1) \times ... \end{align}

Then $(x_i)_{i=1}^{\infty} \in V$ and $V$ is open in $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ with the box topology. Furthermore, $V$ contains only unbounded sequences, so $(x_i)_{i=1}^{\infty} \in V \subset B$ which shows that $B = \mathrm{int}(B)$, so $B$ is open in $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$. Thus $\{ A, B \}$ is a separation of $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ with the box topology, so $\displaystyle{\prod_{i=1}^{\infty} \mathbb{R}_i}$ is disconnected!

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