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Topologist's sine curve is not path connected

  1. Feb 11, 2009 #1
    This example is to show that a connected topological space need not be path-connected.

    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). 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 Closure of p([0,k]) contains all of A. 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?
     
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 11, 2009 #2

    Dick

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    Your notation is about as confusing as you could possibly make it. A path p(t) going from p(0)=(0,0) in T to p(1)=(1,sin(1)) in T must approach a definite limit as t->0. But it has to follow the curve (t,sin(1/t)) to get there. Of course the Closure(p)=A. Pick an e>0. There is a point on p(t) within a distance e of any point on A. It oscillates infinitely often between -1 and 1 as t->0.
     
  4. Feb 12, 2009 #3
    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.

    Here I chose the points x=(1/pi, sin(1/(1/pi)) )= (1/pi, 0) and y = (0,0) in T and I assume T is path-connected so there is a path p:[0,1]-->T connecting x and y.

    I don't see how my notations are confusing (unless the definitions I have are wrong).

    For some reason I'm not getting the proof. Is there a simple way to show A is contained in the Closure of p([0,k]) where k = Inf {t in [0,1] : p(t) in A}?
     
  5. Feb 12, 2009 #4

    Dick

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    Perhaps I'm just easily confused. The point is that p(t) must contain all of the points O={(t,sin(1/t)) for t in (0,1/pi)}. Can you prove that? Now can you show that for any point a in A, and any e>0 there is a point in O whose distance from a is less than e?
     
  6. Feb 12, 2009 #5
    what do you mean by p(t) must contain all of the points O={(t,sin(1/t)) for t in (0,1/pi)}, is p(t) a set? How does it relate with Closure of p([0,k]) where k = Inf {t in [0,1] : p(t) in A}?
     
  7. Feb 12, 2009 #6

    Dick

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    The image of p(t) is a set, yes. I am saying that p(t) must follow the curve (t,sin(1/t)). And all of the points in A are limit points of that image set.
     
  8. Feb 12, 2009 #7
    So does it mean that A is contained in the closure of Image of p (i.e. closure of p([0,1])), but does it imply that A C closure of p([o,k])?
     
  9. Feb 12, 2009 #8

    Dick

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    I wouldn't worry about p([0,k]). The point is that A is contained in the closure of p([0,e)) for any e>0. This is going to make it difficult for p to have a limit as t->0. The main ingredient is showing that the image of p must contain all of the points in O.
     
  10. Feb 12, 2009 #9
    I am trying to think this way.Let a be any point of A. Let B be a basic neighborhood of a. We show there exists x in p([0,k])intersection B. By drawing I can see that if I can show p(k) is in A, then it works but how can I write it analytically?
     
  11. Feb 12, 2009 #10

    Dick

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    Suppose B has radius e>0. Pick a smaller e0>0 such that there is one full cycle of sin(1/t) (i.e. the value passes through -1 and 1) between t=e0 and t=e. Like e0=1/((1/e)+2pi). Though you don't have to be that detailed. There are an infinite number of sine cycles less than any e>0. The curve must intersect B since the x coordinate of a is 0 and the y coordinate of a is between -1 and 1.
     
  12. Feb 12, 2009 #11
    Oh, that makes sense now :). Thanks a lot.
     
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