The Set of Accumulation Points in Finite Topological Products

# The Set of Accumulation Points in Finite Topological Products

On The Interior of Sets in Finite Topological Products and The Closure of Sets in Finite Topological Products pages we saw 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/closure of the product of these sets equals the product of the interiors/closures of these sets.

We will now see that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then the product of the set of accumulation points of sets is contained in the set of accumulation points of the product of sets.

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 $\displaystyle{\left ( \prod_{i=1}^{n} A_i \right )' \supseteq \prod_{i=1}^{n} A_i'}$. |

**Proof:**Let $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \prod_{i=1}^{n} A_i'}$. Then $x_i \in A_i'$ for all $i \in \{ 1, 2, ..., \}$.

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

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

- Taking the product from $i = 1$ to $n$ gives us that:

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

- Therefore:

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

- Hence $\mathbf{x}$ is an accumulation point of $\displaystyle{\prod_{i=1}^{n} A_i}$, i.e., $\displaystyle{\mathbf{x} \in \left ( \prod_{i=1}^{n} A_i \right )'}$, and so:

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