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# product topology preserves the Hausdorff property

Theorem Suppose $\{X_{\alpha}\}_{{\alpha\in A}}$ is a collection of Hausdorff spaces. Then the generalized Cartesian product $\prod_{{\alpha\in A}}X_{\alpha}$ equipped with the product topology is a Hausdorff space.

*Proof.* Let $Y=\prod_{{\alpha\in A}}X_{\alpha}$, and
let $x,y$ be distinct points in $Y$. Then there is an index $\beta\in A$
such that $x(\beta)$ and $y(\beta)$ are distinct points in
the Hausdorff space $X_{\beta}$. It follows that there are open sets
$U$ and $V$ in $X_{\beta}$ such that $x(\beta)\in U$, $y(\beta)\in V$,
and $U\cap V=\emptyset$.
Let $\pi_{\beta}$ be the projection operator $Y\to X_{\beta}$ defined
here. By the definition of
the product topology, $\pi_{\beta}$ is continuous, so
$\pi_{\beta}^{{-1}}(U)$ and $\pi_{\beta}^{{-1}}(V)$ are open sets in $Y$. Also,
since the
preimage commutes with set operations,
we have that

$\displaystyle\pi_{\beta}^{{-1}}(U)\cap\pi_{\beta}^{{-1}}(V)$ | $\displaystyle=$ | $\displaystyle\pi_{\beta}^{{-1}}\big(U\cap V\big)$ | ||

$\displaystyle=$ | $\displaystyle\emptyset.$ |

Finally, since $x(\beta)\in U$, i.e., $\pi_{\beta}(x)\in U$, it follows that $x\in\pi_{\beta}^{{-1}}(U)$. Similarly, $y\in\pi_{\beta}^{{-1}}(V)$. We have shown that $U$ and $V$ are open disjoint neighborhoods of $x$ respectively $y$. In other words, $Y$ is a Hausdorff space. $\Box$

## Mathematics Subject Classification

54B10*no label found*54D10

*no label found*

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