Topologist's sine curve is not path connected

In summary, this conversation discusses a topological space T that is not path-connected. The example given shows that T, which is a union of two sets S and A with a specific topology, cannot be connected by a path. This is proven using a contradiction argument and the fact that the closure of a path must contain all of its limit points. The concept of closure and basic neighborhoods are also mentioned in the conversation.
  • #1
math8
160
0
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?
 
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  • #2
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.
 
  • #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}?
 
  • #4
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?
 
  • #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}?
 
  • #6
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.
 
  • #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])?
 
  • #8
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.
 
  • #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?
 
  • #10
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.
 
  • #11
Oh, that makes sense now :). Thanks a lot.
 

1. What is the Topologist's sine curve?

The Topologist's sine curve is a mathematical curve that was first described by the mathematician J. W. Alexander in 1924. It is an example of a connected set that is not path connected, meaning that there is no continuous path that connects all of its points.

2. How is the Topologist's sine curve constructed?

The Topologist's sine curve is constructed by taking the union of the graph of the function f(x) = sin(1/x) for 0 < x ≤ 1 and the line segment connecting the points (0,0) and (1,0). This results in a curve that looks like a sine wave with a "spike" at the origin.

3. What does it mean for a set to be path connected?

A set is considered to be path connected if there is a continuous path that connects any two points in the set. This means that you can draw a line from one point to another without ever leaving the set.

4. Why is the Topologist's sine curve not path connected?

The Topologist's sine curve is not path connected because there is no way to draw a continuous path from the "spike" at the origin to any other point on the curve. No matter how close you get to the origin, there will always be infinitely many points that are not connected by a path.

5. What is the significance of the Topologist's sine curve in mathematics?

The Topologist's sine curve is an important example in topology, the branch of mathematics that studies properties of geometric objects that are preserved under continuous deformations. It illustrates the difference between connectedness and path connectedness, and its properties have important implications in other areas of mathematics such as analysis and topology.

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