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Math Help - Uncountable set with cofinite Topology

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    Uncountable set with cofinite Topology

    Can I have help:
    let X be Uncountable set with cofinite Topology. Show that every continuous function  X \to R^1 is constant.
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    MHF Contributor Drexel28's Avatar
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    Re: Uncountable set with cofinite Topology

    Quote Originally Posted by rqeeb View Post
    Can I have help:
    let X be Uncountable set with cofinite Topology. Show that every continuous function  X \to R^1 is constant.
    Here's the simplest possible way I can think to explain it. Suppose that f(x)\ne f(y) for some x,y\in X we can then separated the two by open sets (since \mathbb{R} is Hausdorff) say f(x)\in U,f(y)\in V and U\cap V=\varnothing. We have then that f^{-1}(U)\cap f^{-1}(V)=\varnothing and f^{-1}(U)\cap f^{-1}(V) are open and non-empty. That said, note that if f^{-1}(U)\cap f^{-1}(V)=\varnothing\implies \left(X-f^{-1}(U)\right)\cup (X-f^{-1}(V))=X, but we know that both X-f^{-1}(U),X-f^{-1}(V) are finite since the only open sets in X are X,\varnothing and the complement of finite sets, and so if X-f^{-1}(U) is infinite then X-f^{-1}(U)=X which implies that f^{-1}(U)=\varnothing contradictory to assumption. Thus, (X-f^{-1}(U))\cup (X-f^{-1}(V)) is the union of two finite sets, and so finite, but this is ludicrous since their union is X.


    Note more generally this proves that any map X\to Y where X is infinite and Y Hausdorff is constant.
    Last edited by Drexel28; September 28th 2011 at 03:35 PM.
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    Re: Uncountable set with cofinite Topology

    Quote Originally Posted by Drexel28 View Post
    Here's the simplest possible way I can think to explain it. Suppose that f(x)\ne f(y) for some x,y\in \mathbb{R} we can then separated the two by open sets (since \mathbb{R} is Hausdorff) say x\in U,y\in V and U\cap V=\varnothing.
    While this argument is basically correct there are some mistakes.
    For example \{x,y\}\subset\mathbb{X} not in \mathbb{R}
    Therefore, we want f(x)\in U~\&~f(y)\in V.
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    MHF Contributor Drexel28's Avatar
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    Re: Uncountable set with cofinite Topology

    Quote Originally Posted by Plato View Post
    While this argument is basically correct there are some mistakes.
    For example \{x,y\}\subset\mathbb{X} not in \mathbb{R}
    Therefore, we want f(x)\in U~\&~f(y)\in V.
    Thanks for catching that, I changed it.
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