- #1

Incand

- 334

- 47

## Homework Statement

Show that

##S = \{(x,y):0 < x \le 1, y=\sin (1/x)\} \cup \{(x,y): -1 \le x \le 0, y= 0\}##

is connected but not path connected in ##\mathbf R^2##.

## Homework Equations

[/B]

Let ##X## be a metric space.

A set ##E\subseteq X## is said to be connected if ##E## is not a union of two nonempty separated sets.

Two subsets ##A,B## of ##X## are said to be separated if ##\bar A \cap B = A \cap \bar B = \emptyset##

A set ##S## in ##\mathbf R^n## is said to be path connected if there for every pair of points ##a,b \in S## there is a continuous function ##f:[0,1] \to S## so that ##f(0) = a, f(1)=b##.

3. The Attempt at a Solution

3. The Attempt at a Solution

The way forward seems to be to assume ##S## is not connected, i.e. ##S=A\cup B## where ##A## and ##B## are seperated. Now I need to find a contradiction but I don't even know where to start, any hints?

As for proving it's not path connected we need to select two promising points ##a,b##. Clearly we need to select ##a \in [-1,0]## and ##b\in (0,1)## (or the opposite) or they're obviously path connected. Then I think it's easiest to assume that it is path connected and that there is a function ##f## satisfying the above and then get a contradiction again.

Again I'm quite clueless how to get started. As I remember it a vector valued function is continuous iff every coordinate is continuous. Initially I thought I only had to look at the ##y## coordinates in which case one can actually find such a continuous function by choosing ##\xi \in (0,b]## that satisfy ##f(\xi) = 0##

And then define

##g(t) = \begin{cases}

a+t \text{ for } t \in [0,-a]\\

b -\frac{(b-\xi)(1-t)}{1+a} \text{ for } t > -a

\end{cases}##

and use that as our ##x## coordinate. Luckily in this case ##x([0,1]## isn't continuous so at least I didn't prove the impossible.

Last edited: