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Math Help - Topological space

  1. #1
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    removed.
    Last edited by johndon; May 17th 2010 at 09:03 AM. Reason: Re-titled.
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  2. #2
    Junior Member nimon's Avatar
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    I'm not sure what you mean by a 'net', but I can show you A \Leftrightarrow C to get you started.

    First, if X is Hausdorff then, given any (x,y)\in (X\times X) \backslash \Delta we can find disjoint, open neighbourhoods U_{x},U_{y}\subseteq X of x and y respectively. Then (x,y)\in U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta (the last inclusion is due to the disjointness of the neighbourhoods: if  (z,z) \in U_{x}\times U_{y} then z\in U_{x} \cap U_{y} which is wrong by assumption). It follows that (X\times X)\backslash \Delta is a union of open sets in X \times X, so must be open, and therefore \Delta is closed.

    The converse is similar. If \Delta is closed then (X\times X)\backslash \Delta is open, and so for any (x,y)\in X\times X we can find an open set V\subseteq (X\times X) \backslash \Delta such that (x,y)\in V, and it follows that (x,y)\in U_{x}\times U_{y} \subseteq V for some U_{x},U_{y} open in X. Since U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta, they must be disjoint, and so X is Hausdorff.

    Hope this helps.
    Last edited by nimon; May 4th 2010 at 02:18 AM. Reason: Relabelling
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by johndon View Post
    Q(i) Let X be a topological space. Show that the following
    statements are equivalent.
    (A) X is a Hausdorff space.
    (B) Every net in X has at most one limit.
    I gave you a lot of help in your last thread, what do you think about this one?

    (C) The diagonal delta= {(x,x):x∈X }⊂ X x X is closed in X X
    nimon already did this.

    (ii) Let X,Y be topological spaces, A ⊂ X, and Ᾱ = X. Let f: X →Y ,
    g: X →Y be continuous functions, and let f(a)= g(a) ∀ a∈A . Show that if
    Y is a Hausdorff space, then f(x)= g(x) ∀ x∈X . [HINT: Use (B) or (C).]
    Hint:

    Spoiler:


    The set of those points is closed.

    Partial solution (only look at this after much, much thought. Even then, feel bad):
    Spoiler:

    Let \varphi,\psi:X\to Y be continuous and let A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}. Now, define another map \varphi\oplus\psi:X\to Y\times Y:x\mapsto(\varphi(x),\psi(x)), this is easily verifed to be continuous and A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}\left(\Delta_Y\right) from where the conclusion follows


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