# The Hausdorff Property on Finite Topological Products

As well have already seen on the First Countability of Finite Topological Products, Second Countability of Finite Topological Products, and Separability of Finite Topological Products pages that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces that are all first countable / second countable / separable, then the resulting topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also respectively first countable / second countable /separable.

Now recall from the Hausdorff Topological Spaces page that a topological space $X$ is said to be Hausdorff if for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $x$ such that:

(1)We will now see that the Hausdorff property of a finite collection of topological spaces is also inherited onto the resulting finite topological product.

Theorem 1: Let $\{ X_1, X_2, …, X_n \}$ be a finite collection of Hausdorff topological spaces. If $X_i$ is separable for each $i \in I$ then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also Hausdorff. |

**Proof:**Let $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \prod_{i=1}^{n} X_i}$ be distinct points. Then $x_i, y_i \in X_i$ for all $i \in \{ 1, 2, ..., n \}$, and moreover, $x_i \neq y_i$ for at least some $i \in \{ 1, 2, ..., n \}$.

- Suppose that $j \in \{ 1, 2, ..., n \}$ is such that $x_j \neq y_j$. Then since these two points are distinct points in $X_j$ we have that there exists open neighbourhoods $U_j$ of $x_j$ and $V_j$ of $y_j$ in $X_j$ such that:

- So, consider the following open sets in $U$ and $V$ of $\displaystyle{\prod_{i=1}^{n} X_i}$:

- We claim that $U \cap V = \emptyset$. Suppose not, i.e., suppose that there exists a $\mathbf{z} = (z_1, z_2, ..., z_n) \in U \cap V$. Then this implies that $z_j \in U_j \cap V_j$ which contradicts $U_j \cap V_j = \emptyset$. Therefore $U \cap V = \emptyset$. So, for all distinct points $\mathbf{x}$ and $\mathbf{y}$ in $\displaystyle{\prod_{i=1}^{n} X_i}$ there exists open neighbourhoods $U$ of $\mathbf{x}$ and $V$ of $\mathbf{y}$. Therefore, $\displaystyle{\prod_{i=1}^{n} X_i}$ is Hausdorff. $\blacksquare$