The Open and Closed Sets of Finite Topological Products

Recall from the Finite Topological Products of Topological Spaces page that if $X$ and $Y$ are both topological spaces then we defined the resulting topological product to be the topological space of the set $X \times Y$ whose topology is given by the following basis:

(1)

\begin{align} \quad \mathcal B = \left \{ U \times V : U \: \mathrm{is \: open \: in \:} X, \: V \: \mathrm{is \: open \: in \:} Y. \right \} \end{align}

More generally, if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then the resulting topological product is the set $\displaystyle{\prod_{i=1}^{n} X_i}$ whose topology is given by the basis:

(2)

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

Therefore, the open sets in a general finite product $\displaystyle{\prod_{i=1}^{n} X_i}$ are unions of elements from the basis given above. Elements in the basis above are in a sense, the simplest open sets in the topological product.

So, what exactly are the open sets in the basis above? Simply but, if $U_1, U_2, ..., U_n$ are open sets in $X_1, X_2, ..., X_n$ respectively, then $\displaystyle{\prod_{i=1}^{n} U_i = U_1 \times U_2 \times ... \times U_n}$ is a basis element of the product.

For example, consider the topological space $\mathbb{R}^n$. Then $\mathbb{R}^n$ is a finite topological product, namely, $\mathbb{R}^n = \displaystyle{\prod_{i=1}^{n} \mathbb{R}}$, and the following set is open in $\mathbb{R}^n$ with the product topology:

(3)

\begin{align} \quad U = \prod_{i=1}^{n} (i, i+1) \end{align}

In $n = 3$, then:

(4)

\begin{align} \quad U = (1, 2) \times (2, 3) \times (3, 4) \end{align}

So $U$ is an open cube in $\mathbb{R}^3$ as depicted below:

Hence the product of open sets in an open set in the topological product. The following proposition tells us that similarly, the product of closed sets is a closed set in the topological product.

Proposition 1: Let $\{ X_1, X_2, ..., X_n \}$ be a collection of topological spaces. If $C_i$ is a closed set in $X_i$ for all $i \in \{1, 2, ..., n \}$ then the product of these closed sets, $\displaystyle{\prod_{i=1}^{n} C_i}$, is closed in $\displaystyle{\prod_{i=1}^{n} X_i}$.

• Proof: If $C_i$ is a closed set in $X_i$ for all $i \in \{1, 2, ..., n \}$ then $X_i \setminus C_i$ is an open set in $X_i$ for all $i \in \{1, 2, ..., n \}$. So:

(5)

\begin{align} \quad \left ( \prod_{i=1}^{n} X_i \right ) \setminus \left ( \prod_{i=1}^{n} C_i \right ) = [(X_1 \setminus C_1) \times X_2 \times ... \times X_n] \cup [X_1 \times (X_2 \setminus C_2) \times ... \times X_n] \cup ... \cup [X_1 \times X_2 \times ... \times (X_n \setminus C_n)] \end{align}

• This shows that $\displaystyle{\left ( \prod_{i=1}^{n} X_i \right ) \setminus \left ( \prod_{i=1}^{n} C_i \right )}$ is a union of open sets in $\displaystyle{\prod_{i=1}^{n} X_i}$. Therefore $\displaystyle{\left ( \prod_{i=1}^{n} X_i \right ) \setminus \left ( \prod_{i=1}^{n} C_i \right )}$ is open in $\displaystyle{\prod_{i=1}^{n} X_i}$, so $\displaystyle{\prod_{i=1}^{n} C_i}$ is closed in $\displaystyle{\prod_{i=1}^{n} X_i}$. $\blacksquare$