The Closure of Sets in Finite Topological Products

Recall from the The Interior of Sets in Finite Topological Products page that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces and $A_i \subseteq X_i$ for all $i \in \{1, 2, ..., n \}$ then the interior of the product of these sets is equal to the product of the interiors of these sets.

We will now look at a similar result for closures of sets.

In the following theorem we will show that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces and $A_i \subseteq X_i$ for all $i \in \{ 1, 2, ..., n \}$ then the closure of the product of these sets is equal to the product of the closures of these sets.

Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a collection of topological spaces and let $A_i \subseteq X_i$ for all $i \in \{1, 2, ..., n \}$. Then $\displaystyle{\overline{\left ( \prod_{i=1}^{n} A_i \right)} = \prod_{i=1}^{n} \overline{A_i}}$.

• Proof: $\Rightarrow$ Suppose that $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \overline{\prod_{i=1}^{n} A_i}}$. For each $i \in \{1, 2, ..., n \}$, let $U_i$ be an open neighbourhood of $x_i$.

• Then $\displaystyle{U = \prod_{i=1}^{n} U_i}$ is an open neighbourhood of $\mathbf{x}$, and so we have that:

(1)

\begin{align} \quad \left ( \prod_{i=1}^{n} A_i \right ) \cap U = \left ( \prod_{i=1}^{n} A_i \right ) \cap \left ( \prod_{i=1}^{n} U_i \right ) = \prod_{i=1}^{n} A_i \cap U_i \neq \emptyset \end{align}

• From above, this implies that $A_i \cap U_i \neq \emptyset$ for all $i \in \{1, 2, ..., n \}$. Therefore $x_i \in \overline{A_i}$ for all $i \in \{1, 2, ..., n \}$, i.e., $\displaystyle{\mathbf{x} \in \prod_{i=1}^{n} \overline{A_i}}$. Hence:

(2)

\begin{align} \quad \overline{\left ( \prod_{i=1}^{n} A_i \right)} \subseteq \prod_{i=1}^{n} \overline{A_i} \quad (*) \end{align}

• $\Leftarrow$ Suppose that $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \prod_{i=1}^{n} \overline{A_i}}$. Then $x_i \in \overline{A_i}$ for all $i \in \{ 1, 2, ..., n \}$.

• Let $\displaystyle{U = \prod_{i=1}^{n} U_i}$ be an open neighbourhood of $\mathbf{x}$ in $\displaystyle{\prod_{i=1}^{n} X_i}$. Then $U_i$ is an open neighbourhood of $x_i$ for all $i \in I$, and hence:

(3)

\begin{align} \quad A_i \cap U_i \neq \emptyset \end{align}

• Therefore:

(4)

\begin{align} \quad \prod_{i=1}^{n} A_i \cap U_i = \left( \prod_{i=1}^{n} A_i \right ) \cap \left ( \prod_{i=1}^{n} U_i \right ) = \left ( \prod_{i=1}^{n} A_i \right ) \cap U \neq \emptyset \end{align}

• This shows that $\displaystyle{\mathbf{x} \in \overline{\prod_{i=1}^{n} A_i}}$ and so:

(5)

\begin{align} \quad \overline{\left ( \prod_{i=1}^{n} A_i \right)} \supseteq \prod_{i=1}^{n} \overline{A_i} \quad (**) \end{align}

• From the inclusions in $(*)$ and $(**)$ we conclude that:

(6)

\begin{align} \quad \overline{\left ( \prod_{i=1}^{n} A_i \right)} = \prod_{i=1}^{n} \overline{A_i} \quad \blacksquare \end{align}