Results 1 to 2 of 2

Math Help - Topologist's sine curve is not path connected

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    98

    Topologist's sine curve is not path connected

    This example is to show that a connected topological space need not be path-connected.
    The definition I have for X being path-connected is that for any 2 points of X, there is a path connecting them. So if the continuous map p:[0,1]-->X is the path connecting x and y in X, we have p(0)=x; p(1)=y.

    S={ (t,sin(1/t)): 0 <t <= 1 }
    A={ (0,t): -1 <= t <= 1 }
    let T = S U A
    with the topology induced from R^2.

    I show T is not path-connected. Assume to the contrary that there exists a path p:[0,1]-->T with p(0)=(1/pi, 0) and p(1)=(0,0).
    (Here I chose the points x=(1/pi, sin(1/(1/pi)) )= (1/pi, 0) and y = (0,0) in T )
    Let k = Inf {t in [0,1] : p(t) in A}.
    Then p([0,k]) contains at most one point of A. I want to show A is contained in the Closure of p([0,k]) . So p([0,k]) is not closed, and therefore not compact. But p is continuous and [0,k] is compact, so p([0,k]) must be compact,which is a contradiction.

    How do I show Closure of p([0,k]) contains all of A?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by math8 View Post
    This example is to show that a connected topological space need not be path-connected.
    The definition I have for X being path-connected is that for any 2 points of X, there is a path connecting them. So if the continuous map p:[0,1]-->X is the path connecting x and y in X, we have p(0)=x; p(1)=y.

    S={ (t,sin(1/t)): 0 <t <= 1 }
    A={ (0,t): -1 <= t <= 1 }
    let T = S U A
    with the topology induced from R^2.

    I show T is not path-connected. Assume to the contrary that there exists a path p:[0,1]-->T with p(0)=(1/pi, 0) and p(1)=(0,0).
    (Here I chose the points x=(1/pi, sin(1/(1/pi)) )= (1/pi, 0) and y = (0,0) in T )
    Let k = Inf {t in [0,1] : p(t) in A}.
    Then p([0,k]) contains at most one point of A. I want to show A is contained in the Closure of p([0,k]) . So p([0,k]) is not closed, and therefore not compact. But p is continuous and [0,k] is compact, so p([0,k]) must be compact,which is a contradiction.

    How do I show Closure of p([0,k]) contains all of A?
    I don't fully understand your question, but it seems like p([0,k]) does not contain all of A. As you mentioned, p([0,k]) contains at most one point of A and A is a subset of a real line which is a Hausdorff space. So, we can choose two distinct points p(k) and x in A whose open sets are disjoint. That means the closure of p([0,k]) is not A.

    My suggestion is
    Instead of using k = Inf {t in [0,1] : p(t) in A}, define k as k = sup {t in [0,1] : p(t) in S}.
    Now, p(k) belongs to S and p(k + \sigma) belongs to A for a positive \sigma.

    Lemma1. In the topologist's sine curve T, any connected subset C containing a point x in S and a point y in A has a diameter greater than 2.

    Using lemma1, we can draw a contradiction that p is continuous, so S and A are not path connected.
    Last edited by aliceinwonderland; February 12th 2009 at 05:41 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Every path-connected metric space is connected
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 13th 2011, 07:31 AM
  2. Connected and Path Connected
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 14th 2010, 01:28 PM
  3. is this set path-connected
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 2nd 2008, 08:47 AM
  4. (path) connected
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 11th 2008, 06:38 PM
  5. Replies: 1
    Last Post: April 18th 2008, 08:19 AM

Search Tags


/mathhelpforum @mathhelpforum