# Topology: normal space

• November 21st 2010, 05:00 AM
sssitex
Topology: normal space
Can somebody help me solve this problem?

Let $(X_i,\tau_i)$ be regular for each $i \in I$. Then the product space $(X,\tau) = \prod_{i \in I} (X_i, \tau_i)$ is regular.

My biggest problem is that I don't know how closed sets look like in a product space. Is that an infinite product of closed sets? $\prod_{\alpha \in A}E_\alpha$ is closed for all $\alpha$. Or need only a few of $E_\alpha$ 's to be closed? What about the rest? Is it empty?
• November 21st 2010, 10:46 AM
FernandoRevilla
Consider $X=\mathbb{R}$ and $\mathcal{T}$ the topology on $X$ whose basic neighborhoods for $x$ are the intervals $[x,y)\;(y>x)$ .

Then, $X$ is normal however $X\times X$ it is not normal.

Regards.

Fernando Revilla
• November 21st 2010, 11:03 AM
sssitex
I'm really sorry, I meant regular space. But thanks a lot for your help.
• November 21st 2010, 11:49 AM
Drexel28
Quote:

Originally Posted by sssitex
Can somebody help me solve this problem?

Let $(X_i,\tau_i)$ be regular for each $i \in I$. Then the product space $(X,\tau) = \prod_{i \in I} (X_i, \tau_i)$ is regular.

My biggest problem is that I don't know how closed sets look like in a product space. Is that an infinite product of closed sets? $\prod_{\alpha \in A}E_\alpha$ is closed for all $\alpha$. Or need only a few of $E_\alpha$ 's to be closed? What about the rest? Is it empty?

Suppose that $\left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is a non-empty class of regular spaces and let $\displaystyle \prod_{\alpha\in\mathcal{A}}X_{\alpha}=X$ have the product topology. Recall that it suffices to prove that given $(x_\alpha)\in X$ and a neighborhood $N$ of it there exists a neighborhood $U$ of $(x_\alpha)$ such that $\overline{U}\subseteq N$.

So, let $\displaystyle \prod_{\alpha\in\mathcal{A}}N_{\alpha}$ be any basic neighborhood of $(x_\alpha)$ with $\alpha_1,\cdots,\alpha_n$ the finitely many indices for which $U_{\alpha_j}\ne X_{\alpha_j}$. Then, for each $j\in\{1,\cdots,n\}$ we may find some neighborhood $U_{\alpha_j}$ of $x_{\alpha_j}$ for which $x_{\alpha_j}\in U_{\alpha_j}$ and $\overline{U_{\alpha_j}}\subseteq N_{\alpha_j}$. Clearly $\displaystyle \prod_{\alpha\in\mathcal{A}}G_{\alpha}$ where $G_\alpha=\begin{cases}U_{\alpha_j} & \mbox{if}\quad \alpha=\alpha_1,\cdots,\alpha_n\\ X_{\alpha} & \mbox{if}\quad \alpha\ne \apha_1,\cdots,\alpha_n\end{cases}$ is a neighborhood of $(x_\alpha)$ and $\displaystyle \overline{\prod_{\alpha\in\mathcal{A}}G_\alpha}=\p rod_{\alpha\in\mathcal{A}}\overline{G_\alpha}\subs eteq\prod_{\alpha\in\mathcal{A}}N_\alpha$. By prior discussion the conclusion follow.s