Proving Pathwise Connectedness in Topological Spaces: A Counterexample

[e12514]

Homework Statement

Pathwise connectedness implies connectedness.
Show that the converse is false by considering the following subspace of R^2:
X = { (t, sin(1/t)) : t is nonzero } union { (0, t) : -1 <= t <= 1 }

Homework Equations

Definitions:

A topological space X is called pathwise connected if for any two distinct points x and y of X, there exists a continuous map f: [0,1] -> X with f(0) = x and f(1) = y.

A topological space X is disconnected if there exists two nonempty disjoint clopen subsets U, V of X such that (U union V) = X.

A topological space X is disconnected if it is not connected.

The Attempt at a Solution

So we have to show that
X = { (t, sin(1/t)) : t is nonzero } union { (0, t) : -1 <= t <= 1 }
is connected but is not pathwise connected.

To show X is not pathwise connected, choose x = (0,0) and say y = (1,sin1).
Then there is no continuous map f : [0,1] -> X with f(0) = (0,0) and f(1) = (1,sin1) because lim_(t -> 0) sin(1/t) does not exist so sin(1/t) is not continuous at t=0.
Therefore X is not pathwise connected.

To show X is connected, suppose X is disconnected (to try to get a contradiction),
so there exists two nonempty disjoint clopen subsets U, V of X such that (U union V) = X.
Now, let
A = { (t, sin(1/t)) : t > 0}
B = { (t, sin(1/t)) : t < 0}
C = { (0, t) : -1 <= t <= 1 }
then A,B,C are all connected sets and also X = A union B union C.
Take x in U, and y in V.
Then x and y cannot be both in A, or both in B, or both in C (A,B,C are all connected)
so without loss of generality assume either x is in A and y is in C, or x is in A and y is in B.

And then I'm stuck not knowing what to do next.
Can anyone give me some help?

 

[mighty2000]

Solution:

To show X is connected, suppose X is disconnected (to try to get a contradiction),
so there exists two nonempty disjoint clopen subsets U, V of X such that (U union V) = X.
Now, let
A = { (t, sin(1/t)) : t > 0}
B = { (t, sin(1/t)) : t < 0}
C = { (0, t) : -1 <= t <= 1 }
then A,B,C are all connected sets and also X = A union B union C.
Take x in U, and y in V.
Then x and y cannot be both in A, or both in B, or both in C (A,B,C are all connected)
so without loss of generality assume either x is in A and y is in C, or x is in A and y is in B.

If x is in A and y is in C, then there must exist a point z in B such that z is in U.
Since U is open and B is closed, there exists a small enough open ball B(z,r) contained in U.
But this ball must contain some points of A as well, contradicting the fact that U and A are disjoint.
Therefore, this case is not possible.

If x is in A and y is in B, then there must exist points z and w in C such that z is in U and w is in V.
Since C is compact, there exists a finite subcover {B(z_i,r_i)} of C.
Now, consider the set {B(z_i,r_i)} union {B(w,r)}.
This is a finite open cover of C, and since C is compact, there exists a finite subcover that covers all of C.
But this subcover must contain points of both U and V, contradicting the fact that they are disjoint.
Therefore, this case is also not possible.

Thus, our assumption that X is disconnected must be false, and therefore X is connected.