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Math Help - Topolgy

  1. #1
    MHF Contributor Amer's Avatar
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    Topolgy

    to prove that T_0-space is a productive property

    a product topology \left(\prod_{k=1}^{n} X_k , T\right) is a T_0 space \Longleftrightarrow for every k=1,...,n (X_k , T_k) is a T_0 space

    Proof:

    (\Rightarrow)

    let \left(\prod_{k=1}^{n} X_k\right) is a T(0) space , let 1\leq i \leq n then \left(X_i , T_i \right) is homeomorphic to a subspace Y_i of the product space \prod_{k=1}^{n} X_k

    my question is how we know that there exist a subspace Y_i which is homeomorphic to (X_i,T_i) ?/

    thanks very much
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Amer View Post
    Proof:

    (\Rightarrow)

    let \left(\prod_{k=1}^{n} X_k\right) is a T(0) space , let 1\leq i \leq n then \left(X_i , T_i \right) is homeomorphic to a subspace Y_i of the product space \prod_{k=1}^{n} X_k

    my question is how we know that there exist a subspace Y_i which is homeomorphic to (X_i,T_i) ?/
    Let (a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n)\in \prod_{\substack{k=1\\k\neq i}}^n X_k. Then Y_i=\{a_1\}\times \{a_2\}\times \cdots\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \times \{a_n\} (with the topology induced by the topology on \prod_{k=1}^{n} X_k) is homeomorphic to X_i because the map \pi:Y_i \to  X_i defined by \pi\left((a_1,\ldots,a_{i-1},x_i,a_{i+1},\ldots, a_n)\right)= x_i is an homeomorphism. Can you show it ?
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hello,

    Let (a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n)\in \prod_{\substack{k=1\\k\neq i}}^n X_k. Then Y_i=\{a_1\}\times \{a_2\}\times \cdots\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \times \{a_n\} (with the topology induced by the topology on \prod_{k=1}^{n} X_k) is homeomorphic to X_i because the map \pi:Y_i \to  X_i defined by \pi\left((a_1,\ldots,a_{i-1},x_i,a_{i+1},\ldots, a_n)\right)= x_i is an homeomorphism. Can you show it ?
    I think I can define the projection function from P_i:\left(\prod_{k=0}^{n}  X_k,T_p\right) \cong \left(X_i,T_i\right)
    this is one-one and onto and the projection is continuous and the inverse for the projection is continuous thus it is homeomorphism right

    Thanks
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Amer View Post
    I think I can define the projection function from P_i:\left(\prod_{k=0}^{n}  X_k,T_p\right) \cong \left(X_i,T_i\right)
    this is one-one and onto and the projection is continuous and the inverse for the projection is continuous thus it is homeomorphism right
    P_i is onto and continuous but it not necessarily one-to-one. For example let
    \begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.
    We have P_x((1,0))=P_x((1,2))=1 so P_x is not injective.

    However the restriction of P_i to the set Y_i=\{a_1\}\times\cdots\times\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots\times \{a_n\} that I defined in my previous post is an homeomorphism... it is the function I named \pi.
    Last edited by flyingsquirrel; August 8th 2009 at 03:31 AM. Reason: typo
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    P_i is onto and continuous but it not necessarily one-to-one. For example let
    \begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.
    We have P_x((1,0))=P_x((1,2))=1 so P_x is not injective.

    However the restriction of P_i to the set Y_i=\{a_1\}\times\cdots\times\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \{a_n\} that I defined in my previous post is an homeomorphism... it is the function I named \pi.
    Thanks I get it now
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