Results 1 to 2 of 2

Thread: K-topology on R

  1. #1
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228

    K-topology on R

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


    I have already shown that [0,1] is not compact in this topology. I just need to show that $\displaystyle \mathbb{R}_K$ is connected. Obviously the sets $\displaystyle (-\infty,0)\cup (1,\infty)$ are connected. All I need is to show that the range inbetween is also connected. Where should I start?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by putnam120 View Post
    Let $\displaystyle K=\left\{\frac{1}{n} : n\in\mathbb{N}\right\}$. Define $\displaystyle \mathbb{R}_K$ to be the topology on $\displaystyle \mathbb{R}$ generated by the topology $\displaystyle K\cup (a,b)-K (typo?)$.


    I have already shown that [0,1] is not compact in this topology. I just need to show that $\displaystyle \mathbb{R}_K$ is connected. Obviously the sets $\displaystyle (-\infty,0)\cup (1,\infty)$ 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 $\displaystyle \bar{Y}$ 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 $\displaystyle \mathbb{Re}$ are intervals.

    We already know that $\displaystyle (-\infty,0)\cup (1,\infty)$ is connected with respect to K-topology on $\displaystyle \mathbb{Re}$, which denoted as $\displaystyle \mathbb{Re}_K$. By lemma 1 & 2, we only remain to show that (0,1) is connected with respect to $\displaystyle \mathbb{Re}_K$. In standard topology on $\displaystyle \mathbb{Re}$, (0,1) is connected by Lemma 3. Since $\displaystyle \mathbb{Re}_K$ is finer than the standard topology on $\displaystyle \mathbb{Re}$, we need to check added open sets in (0,1) by $\displaystyle \mathbb{Re}_K$. Since a basis element has the form $\displaystyle (a,b)-K$, 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 $\displaystyle \mathbb{Re}_K$ by lemma 2 & 3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Order topology = discrete topology on a set?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 6th 2011, 11:19 AM
  2. a topology such that closed sets form a topology
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jun 14th 2011, 04:43 AM
  3. Show quotient topology on [0,1] = usual topology on [0,1]
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Nov 5th 2010, 04:44 PM
  4. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Dec 13th 2008, 02:19 PM
  5. discrete topology, product topology
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Dec 7th 2008, 01:01 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum