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Thread: Connectedness

  1. #1
    Super Member Showcase_22's Avatar
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    Connectedness

    Prove that any infinte set with co-finite topology is connected.
    Does this question even make sense?

    Let $\displaystyle (X,T)$ be an infinite space with co-finite topology.
    Let $\displaystyle (U,V)$ be a two point discrete space where $\displaystyle U=\{0,1\}$.

    Let $\displaystyle f: X \rightarrow U$ be any continuous function.

    I need to show that f is constant.

    My big problem is this:

    We know $\displaystyle f^{-1}(1)$ and $\displaystyle f^{-1}(0)$ are open.

    Hence $\displaystyle f^{-1}(0), \ f^{-1}(1) \in T$.

    However, $\displaystyle X-T$ is finite so there exist finitely many points $\displaystyle x_1, \ldots , x_n$ where$\displaystyle f(x_i) \neq 0$ or $\displaystyle 1$ where $\displaystyle i=1, \ldots , n$.

    By that reasoning, we have a finite number of points where the function is not defined. These points are in $\displaystyle X$ so the function is discontinuous at each $\displaystyle x_i$.

    Why is this right or wrong?
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  2. #2
    Senior Member Tinyboss's Avatar
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    Suppose an infinite set $\displaystyle X$ with cofinite topology were disconnected. Then there are open subsets $\displaystyle U,V$ such that $\displaystyle U\cap V=\varnothing$ and $\displaystyle U\cup V=X$. In particular, we have that both $\displaystyle U$ and $\displaystyle X\setminus U$ are open in the co-finite topology. Is that possible?
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  3. #3
    Super Member Showcase_22's Avatar
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    No! that's not possible! =D

    If a set $\displaystyle U$ is open, it's complement should be closed! (That I do know!)

    I was misinterpreting what a co-finite topology was. I didn't realise that $\displaystyle X \setminus U$ was also open.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    No! that's not possible! =D

    If a set $\displaystyle U$ is open, it's complement should be closed! (That I do know!)

    I was misinterpreting what a co-finite topology was. I didn't realise that $\displaystyle X \setminus U$ was also open.
    The cofinite topology is a neat one in the sense that it cannot support two disjoint open sets. Thus it is also, in fact, a $\displaystyle T_1$ space which is not Hausdorff. Also, it is clear that since every totally disconnected space is Hausdorff that a cofinite space is also not totally disconnected (this was immediately obvious before actually doing the question was my point)

    A more interesting question you may want to ask yourself: is a cofinite space every locally connected?
    Last edited by Drexel28; Mar 15th 2010 at 09:19 AM.
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