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Thread: Topology: normal space

  1. #1
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    Topology: normal space

    Can somebody help me solve this problem?

    Let $\displaystyle (X_i,\tau_i)$ be regular for each $\displaystyle i \in I$. Then the product space $\displaystyle (X,\tau) = \prod_{i \in I} (X_i, \tau_i)$ 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? $\displaystyle \prod_{\alpha \in A}E_\alpha$ is closed for all $\displaystyle \alpha$. Or need only a few of $\displaystyle E_\alpha$ 's to be closed? What about the rest? Is it empty?
    Last edited by sssitex; Nov 21st 2010 at 11:02 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider $\displaystyle X=\mathbb{R}$ and $\displaystyle \mathcal{T}$ the topology on $\displaystyle X$ whose basic neighborhoods for $\displaystyle x$ are the intervals $\displaystyle [x,y)\;(y>x)$ .

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

    Regards.

    Fernando Revilla
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  3. #3
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    I'm really sorry, I meant regular space. But thanks a lot for your help.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sssitex View Post
    Can somebody help me solve this problem?

    Let $\displaystyle (X_i,\tau_i)$ be regular for each $\displaystyle i \in I$. Then the product space $\displaystyle (X,\tau) = \prod_{i \in I} (X_i, \tau_i)$ 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? $\displaystyle \prod_{\alpha \in A}E_\alpha$ is closed for all $\displaystyle \alpha$. Or need only a few of $\displaystyle E_\alpha$ 's to be closed? What about the rest? Is it empty?
    Suppose that $\displaystyle \left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is a non-empty class of regular spaces and let $\displaystyle \displaystyle \prod_{\alpha\in\mathcal{A}}X_{\alpha}=X$ have the product topology. Recall that it suffices to prove that given $\displaystyle (x_\alpha)\in X$ and a neighborhood $\displaystyle N$ of it there exists a neighborhood $\displaystyle U$ of $\displaystyle (x_\alpha)$ such that $\displaystyle \overline{U}\subseteq N$.

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