# Thread: X_{\alpha} is Hausdorff

1. ## X_{\alpha} is Hausdorff

If each $\displaystyle X_{\alpha}$ is a Hausdorff space, then $\displaystyle \prod X_{\alpha}$ is a Hausdorff space in both the box and product top.

How do I go about proving this?

I know since $\displaystyle X_{\alpha}$ is Hausdorff, if for all $\displaystyle x_1,x_2\in X_{\alpha}$ where $\displaystyle x_1\neq x_2$ there exists neighborhoods $\displaystyle U_1 \ \text{and} \ U_2$ of $\displaystyle x_1 \ \text{and} \ x_2$ that are disjoint.

2. ## Re: X_{\alpha} is Hausdorff

If $\displaystyle (x_\alpha)$ and $\displaystyle (y_\alpha)$ are distinct points in the product space, then they must differ in at least one coordinate. So there exists $\displaystyle \alpha_0$ say, such that $\displaystyle x_{\alpha_0} \ne y_{\alpha_0}$. Use the Hausdorff property in the $\displaystyle \alpha_0$-coordinate space to separate $\displaystyle x_{\alpha_0}$ and $\displaystyle y_{\alpha_0}$ by disjoint neighbourhoods. Then use those neighbourhoods to construct neighbourhoods in the product space that separate $\displaystyle (x_\alpha)$ and $\displaystyle (y_\alpha)$. The same construction will work for both the product and the box topologies.