# Topologist's sine curve is not path connected

• Feb 12th 2009, 07:39 AM
math8
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?
• Feb 12th 2009, 04:08 PM
aliceinwonderland
Quote:

Originally Posted by math8
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 + $\displaystyle \sigma$) belongs to A for a positive $\displaystyle \sigma$.

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.