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Thread: Show not Hausdorff

  1. #1
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    Show not Hausdorff

    Define an equivalence relation in X=[0,1]\times[0,1], by declaring (s_0,t_0)\equiv(s_1,t_1) if and only if t_0=t_1>0. Describe the quotient space X/\equiv and show that it is not a Hausdorff space.

    My solution:
    Let \pi:X\rightarrow X/\equiv be the projection map. The equivalence class \pi((s,0)) of (s,0), where s\in[0.1], is (s,0) since t=0.
    If (s_1,t_1)\equiv(s_2,t_2) then t_1=t_2. Then the equivalence class \pi((s,t)) of (s,t), where t\in(0,1], is \{(s_i,t)|s_i\in[0,1]\}.
    Thus the set X/\equiv is the union of \{(s,0)|s\in[0,1]\} and \{(s_i,t)|s_i\in[0,1]\} for all t.

    Can someone help me show that this space is not a Hausdorff space? Thanks
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  2. #2
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    Quote Originally Posted by bram kierkels View Post
    Define an equivalence relation in X=[0,1]\times[0,1], by declaring (s_0,t_0)\equiv(s_1,t_1) if and only if t_0=t_1>0. Describe the quotient space X/\equiv and show that it is not a Hausdorff space.

    My solution:
    Let \pi:X\rightarrow X/\equiv be the projection map. The equivalence class \pi((s,0)) of (s,0), where s\in[0.1], is (s,0) since t=0.
    If (s_1,t_1)\equiv(s_2,t_2) then t_1=t_2. Then the equivalence class \pi((s,t)) of (s,t), where t\in(0,1], is \{(s_i,t)|s_i\in[0,1]\}.
    Thus the set X/\equiv is the union of \{(s,0)|s\in[0,1]\} and \{(s_i,t)|s_i\in[0,1]\} for all t.

    Can someone help me show that this space is not a Hausdorff space?
    If a space is Hausdorff then a convergent sequence has a unique limit point. You could show that this space is non-Hausdorff by showing that the sequence (x_n) given by x_n = (0,1/n) converges to all the points (s,0)\;\,(s\in[0,1]).
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  3. #3
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    Quote Originally Posted by Opalg View Post
    If a space is Hausdorff then a convergent sequence has a unique limit point. You could show that this space is non-Hausdorff by showing that the sequence (x_n) given by x_n = (0,1/n) converges to all the points (s,0)\;\,(s\in[0,1]).

    Thanks, but why does this sequence converges to (s,0)? For every open neigborhood (s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1} there is no N such that x_i\in(s_i,0)\times(s_{i+1},0) for all  i>N..?
    Last edited by bram kierkels; Nov 2nd 2010 at 11:06 AM.
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  4. #4
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    Quote Originally Posted by bram kierkels View Post
    Thanks, but why does this sequence converges to (s,0)? For every open neigborhood (s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1} there is no N such that x_i\in(s_i,0)\times(s_{i+1},0) for all  i>N..?
    I don't understand what you mean by the "open neigborhood (s_i,0)\times(s_{i+1},0), s_i\leq s\leq s_{i+1}".

    The way I think of it is that the projection map  \pi is continuous. So if x_n\to x in X then \pi(x_n)\to\pi(x). Since (s,1/n)\to(s,0) in X (for each s\in[0,1]), it follows that \pi(s,1/n)\to\pi((s,0). But \pi(s,1/n) = \pi(0,1/n) = \pi(x_n). It follows that \pi(x_n)\to(s,0) in the quotient space.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bram kierkels View Post
    Define an equivalence relation in X=[0,1]\times[0,1], by declaring (s_0,t_0)\equiv(s_1,t_1) if and only if t_0=t_1>0. Describe the quotient space X/\equiv and show that it is not a Hausdorff space.

    My solution:
    Let \pi:X\rightarrow X/\equiv be the projection map. The equivalence class \pi((s,0)) of (s,0), where s\in[0.1], is (s,0) since t=0.
    If (s_1,t_1)\equiv(s_2,t_2) then t_1=t_2. Then the equivalence class \pi((s,t)) of (s,t), where t\in(0,1], is \{(s_i,t)|s_i\in[0,1]\}.
    Thus the set X/\equiv is the union of \{(s,0)|s\in[0,1]\} and \{(s_i,t)|s_i\in[0,1]\} for all t.

    Can someone help me show that this space is not a Hausdorff space? Thanks
    Another possible way is to notice the following theorem:
    Theorem: Let X be a compact Hausdorff space and q:X\to Y a quotient map. Then, Y is Hausdorff if and only if q is closed.
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