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Math Help - Hausdorff spaces and convergence

  1. #1
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    Hausdorff spaces and convergence

    Hi Everyone,

    I'm new here and I'm dealing with some topology questions.

    Suppose that (X,T) is a C1 space then the following statements are equivalent:
    (1) (X,T) is a hausdorff space
    (2) Every convergence sequence has a unique limit
    (3) (X,T) is a T1-space and every compact subset is closed.

    I know this is not a homework service but honestly I have no idea where to start. In fact, in topology I always described convergence using filters and in this question I have to prove something about sequences so I'm confused.

    Can someone give me a hint?

    Thanks in advance!
    Cheers Impo
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  2. #2
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    Re: Hausdorff spaces and convergence

    Please tell us what "Suppose that (X,T) is a C1 space" means.

    C1 is not a common notation in topology. Does it mean first countable?
    Last edited by Plato; January 5th 2013 at 05:16 AM.
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  3. #3
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    Re: Hausdorff spaces and convergence

    Quote Originally Posted by Plato View Post
    Please tell us what "Suppose that (X,T) is a C1 space" means.

    C1 is not a common notation in topology. Does it mean first countable?
    Indeed.
    Last edited by Plato; January 5th 2013 at 05:16 AM.
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    Re: Hausdorff spaces and convergence

    Quote Originally Posted by Impo View Post
    Suppose that (X,T) is a C1 space then the following statements are equivalent:
    (1) (X,T) is a hausdorff space
    (2) Every convergence sequence has a unique limit
    (3) (X,T) is a T1-space and every compact subset is closed.

    I know this is not a homework service
    Note that a Hausdorff space is a T_2 space. Note the capital H.

    If l is the limit of (x_n) then any open set l\in O mean all but a finite collect \{x_1,x_,\cdots,x_n\} most belong to O. If you have two limits then in a Hausdorff there are disjoint open sets that separate those limits. That is impossible.

    Now you post some work.
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    Re: Hausdorff spaces and convergence

    (2)=>(3)
    If I'm allowed to use the fact that X is Hausdorff then (3) is proved immediately, but I don't think I'm allowed to.

    (3)=>(1)
    Every T_1 space is a T_0 space, if I can show that X is regular then we're done because every T_3 space is Hausdorff. Let A \subset X be a compact subset and suppose  x \notin A then there can be found disjoint open neigbourhouds for A and x, but by (3) every compact subspace is closed thus A is closed therefore X is regular.
    Last edited by Impo; January 4th 2013 at 02:42 PM.
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  6. #6
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    Re: Hausdorff spaces and convergence

    Well you are supposed to prove IFF implications.  (1) \text{if and  only if} (2) and  (2) \text{if and only if} (3) proving these two things is akin to showing all three statements are equivalent. Plato already gave you  (1) \to (2) now prove  (2) \to (1)
    Last edited by jakncoke; January 4th 2013 at 11:35 PM.
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  7. #7
    Senior Member jakncoke's Avatar
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    Re: Hausdorff spaces and convergence

    So let me give you some help for  (2) \to (1)

    Now a (X,T) being a C1 space means that every point has a countable neighborhood basis. (Meaning a sequence of open neighborhoods such that for any open neighborhood U of a, there exists an element in the sequence which is contained in U.

    Now if a, b were points in the topology, such that  a \not = b . Since C1, a has a countable neighborhood basis, call it \bar{A} = \{ N^{a}_i | i \in \mathbb{N} \} where  N^{a}_i is an open neighborhood of a. So for any open neighborhood U of a, there exists an  i \in \mathbb{N} such that  N^{a}_i \subset U . Now since the intersection of any finite number of open sets is open.  N_1 \cap N_2 \cap ... \cap N_i \subset N_i \subset U . For any  i \in \mathbb{N} let  O^{a}_i = N^{a}_1 \cap ... \cap N^{a}_i Note that  O^{a}_j \subset N^{a}_i for all  j \geq i . Note that since  O^{a}_j \subset N^{a}_i \subset U for  j \geq i . This should remind of you the definition for topological convergence. That is, a sequence  x_n \to a if for any open neighborhood of a, we can find a  j \in \mathbb{N} such that  p \geq j implies  x_p \in U .

    Now just as we did for a, for b take  \bar{B} = \{N^{b}_i | i \in \mathbb{N} \} (its countable basis) and O^{b}_i =  N^{b}_1 \cap ... \cap N^{b}_i | i \in \mathbb{N} .
    Now for any natural number i,  O^{a}_i \cap O^{b}_i is an open set (containing a, and containing b), since it is the finite intersection of 2 open sets.

    Now if we assume a and b have no disjoint open sets containing a and b. Then for no  i \in \mathbb{N} is   O^{a}_i \cap O^{b}_i  = \{ \phi \}
    So if we define a sequence x_i where  x_i \in O^{a}_i \cap O^{b}_i for  i \in \mathbb{N} . Now i claim x_i \to a . So pick any open neighborhood of a, call it U. since there exists a  j \in \mathbb{N} such that  N^{a}_j \subset U and since we know  O^{a}_p \subet N^{a}_j \subet U for all  p \geq j . This shows that for any x_p with  p \geq j  x_p \in O^{a}_p and thus  x_p \in U . Thus  x_i \to a .

    Now since we assumed,  x_i \in O^{a}_i \cap O^{b}_i for  i \in \mathbb{N} , then x_i \in O^{b}_i . I claim  x_i \to b . Now again pick any open neighborhood of b, call if U. then there exists a  j \in \mathbb{N} such that  N^{b}_j \subset U which is then  O^{b}_p \subset N^{b}_j \subet U for all  p \geq j . so this shows that for all  p \geq j  x_p \in O^{b}_p \subset U , thus  x_i \to b .

    Since we assumed every convergent sequence has a unique limit,  a = b which is a contradiction. Thus there be an open set containing a which is disjoint with an open set containing b. Thus hausdrauff.
    Last edited by jakncoke; January 4th 2013 at 06:43 PM.
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    Re: Hausdorff spaces and convergence

    Hi jakncoke, thanks for your answer. I wanted to prove the statements are equivalent by proving (1)=>(2)=>(3)=>(1). I'm only missing (2)=>(3) for now.
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  9. #9
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    Re: Hausdorff spaces and convergence

    Quote Originally Posted by Impo View Post
    I wanted to prove the statements are equivalent by proving (1)=>(2)=>(3)=>(1). I'm only missing (2)=>(3) for now.
    I commend you for wanting to show equivalence that way. But as you have already noted:
    Quote Originally Posted by Impo View Post
    (2)=>(3)
    If I'm allowed to use the fact that X is Hausdorff then (3) is proved immediately, but I don't think I'm allowed to.
    I think that I would use some sort of argument such as jakncoke used in reply #7 to show the space is Hausdorff. Then as you note it is an easy to precede.
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