Results 1 to 4 of 4

Math Help - Topology: normal space

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    12

    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?
    Last edited by sssitex; November 21st 2010 at 11:02 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    12
    I'm really sorry, I meant regular space. But thanks a lot for your help.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by sssitex View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A normal Moore space is completely normal
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 19th 2011, 12:52 PM
  2. Fort Space Topology
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: February 8th 2010, 09:46 AM
  3. Topology of a metric space
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: February 23rd 2009, 09:48 AM
  4. Topology Quotient Space
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 17th 2008, 01:42 PM
  5. Hausdorff Space- Topology
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 5th 2008, 05:42 PM

/mathhelpforum @mathhelpforum