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Math Help - K-topology on R

  1. #1
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    K-topology on R

    Let K=\left\{\frac{1}{n} : n\in\mathbb{N}\right\}. Define \mathbb{R}_K to be the topology on \mathbb{R} generated by the topology K\cup (a,b)-K.


    I have already shown that [0,1] is not compact in this topology. I just need to show that \mathbb{R}_K is connected. Obviously the sets (-\infty,0)\cup (1,\infty) are connected. All I need is to show that the range inbetween is also connected. Where should I start?
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  2. #2
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    Quote Originally Posted by putnam120 View Post
    Let K=\left\{\frac{1}{n} : n\in\mathbb{N}\right\}. Define \mathbb{R}_K to be the topology on \mathbb{R} generated by the topology K\cup (a,b)-K (typo?).


    I have already shown that [0,1] is not compact in this topology. I just need to show that \mathbb{R}_K is connected. Obviously the sets (-\infty,0)\cup (1,\infty) are connected. All I need is to show that the range inbetween is also connected. Where should I start?
    Lemma 1. If Y is a connected subspace of a space X, then \bar{Y} is connected.
    Lemma 2. The union of a collection of connected subspace of X that have a point in common is connected.
    Lemma 3. The connected subsets of \mathbb{Re} are intervals.

    We already know that (-\infty,0)\cup (1,\infty) is connected with respect to K-topology on \mathbb{Re}, which denoted as \mathbb{Re}_K. By lemma 1 & 2, we only remain to show that (0,1) is connected with respect to \mathbb{Re}_K. In standard topology on \mathbb{Re}, (0,1) is connected by Lemma 3. Since \mathbb{Re}_K is finer than the standard topology on \mathbb{Re}, we need to check added open sets in (0,1) by \mathbb{Re}_K. Since a basis element has the form (a,b)-K, added open sets are simply unions of intervals. Now the whole (0,1) can be described as a union of connected intervals having intersections. Thus (0,1) is connected in \mathbb{Re}_K by lemma 2 & 3.
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