Dense Sets in Finite Topological Products

Consider a finite collection of topological spaces $\{ X_1, X_2, ..., X_n \}$. If $A_i \subseteq X_i$ are dense in $X_i$ for all $i \in \{1, 2, ..., n \}$, then what can we say about the product $\displaystyle{\prod_{i=1}^{n} A_i}$?

Conversely, if $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$, what can we say about each individual set $A_i$ in $X_i$?

The following theorem gives us the desired answer. $A_i \subseteq X_i$ is dense in $X_i$ for all $i \in \{1, 2, ..., n \}$ if and only if $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$.

Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of topological spaces and let $A_i \subseteq X_i$ for all $i \in \{ 1, 2, ..., n \}$. Then $A_i$ is dense in $X_i$ for all $i \in \{ 1, 2, ..., n \}$ if and only if $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$.

• Proof: $\Rightarrow$ Suppose that $A_i$ is dense in $X_i$ for all $i \in \{1, 2, ..., n \}$. Let $\displaystyle{U = \prod_{i=1}^{n} U_i}$ be any open set in $\displaystyle{\prod_{i=1}^{n} X_i}$. Then $U_i$ is open in $X_i$ for all $i \in \{ 1, 2, ..., n \}$. So since $A_i$ is dense in $X_i$ we have that for all $i$ that:

(1)

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

• So take $x_i \in A_i \cap U_i$. Then $\mathbf{x} = (x_1, x_2, ..., x_n)$ is such that:

(2)

\begin{align} \quad \mathbf{x} \in \left ( \prod_{i=1}^{n} A_i \right ) \cap \left ( \prod_{i=1}^{n} U_i \right ) \end{align}

• Thus $\displaystyle{\left ( \prod_{i=1}^{n} A_i \right ) \cap \left ( \prod_{i=1}^{n} U_i \right ) \neq \emptyset}$ for all open sets $\displaystyle{U = \prod_{i=1}^{n} U_i}$ in $\displaystyle{\prod_{i=1}^{n} X_i}$ which shows that $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$.

• $\Leftarrow$ Suppose that $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$.

• Consider the set $A_i$ and let $U_i$ be any open set in $X_i$. Let:

(3)

\begin{align} \quad U = X_1 \times X_2 \times ... \times X_{i-1} \times U_i \times X_{i+1}, \times ... \times X_n \end{align}

• Then $U$ is an open set in $\displaystyle{\prod_{i=1}^{n} X_i}$. Since $\displaystyle{\prod_{i=1}^{n} A_i}$ is dense in $\displaystyle{\prod_{i=1}^{n} X_i}$ we have that:

(4)

\begin{align} \quad \left ( \prod_{i=1}^{n} A_i \right ) \cap U & \neq \emptyset \\ \quad \left ( A_1 \times A_2 \times ... \times A_n \right ) \cap \left (X_1 \times X_2 \times ... \times X_{i-1} \times U_i \times X_{i+1} \times ... \times X_n \right ) & \neq \emptyset \\ \quad ( A_1 \cap X_1) \times (A_2 \cap X_2) \times ... \times (A_{i-1} \cap X_{i-1}) \times (A_i \cap U_i) \times (A_{i+1} \cap X_{i+1}) \times ... \times (A_n \cap X_n) & \neq \emptyset \\ \quad A_1 \times A_2 \times ... \times A_{i-1} \times (A_i \cap U_i) \times A_{i+1} \times ... \times A_n & \neq \emptyset \end{align}

• This shows that $A_i \cap U_i \neq \emptyset$.

• So for all $i \in \{ 1, 2, ..., n \}$ we have that $A_i$ is dense in $X_i$. $\blacksquare$