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Show that $S:= \{ x+iy \;|\; x = 0 \text{ or } x>0 , y = \sin(\frac{1}{x}) \} $ is connected, even though there are points in $S$ that cannot be connected by any curve in $S$

Attempt: Suppose $S$ is disconnected. By definition there exists two disjoint sets $A$ and $B$ such that $S$ is contained in $A \cup B$ and $S$ is not contained in neither $A$ nor $B$.

This is where I'm stuck. Somehow I have to arrive at a contradiction or show that no such sets $A, B$ exist but don't know how to proceed. Any help would be greatly appreciated.

Michael Tagle
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1 Answers1

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I think you are looking for Topologist's sine curve, and trying to prove that it is connected. Here is a way that may help you.

Define, $X\subset \mathbb{R}^2$ by, $X=\{(0,0)\}\cup B$ where $B=\{(x,y):0<x<1,y=\sin(\frac{\pi}{x})\}$.

Clearly, the set $B$ is the graph of the function, $f:(0,1]\to \mathbb{R}^2$ defined by, $f(x)=(x,\sin (\frac{\pi}{x}))$.

Again it is clear that $f$ is continuous and $(0,1]$ is connected so $B$ is connected, and hence $\overline{B}$ is connected. Also, $(0,0)$ is a limit point of $B$ and $B\subset X \subset \overline{B}$. Hence $X$ is connected.

Hope it works.

Sujit Bhattacharyya
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  • Thank you! The book I'm using doesn't have the propositions you used, but it does provide the definition of a disconnected set and then mentions that a set is called connected if it's not disconnected. This is why I tried proof by contradiction. – Michael Tagle Jan 28 '19 at 06:59
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    $B$ connected implies $\bar{B}$ is connected? Perhaps you meant to link a proof to this? – Michael Tagle Jan 28 '19 at 07:00
  • @michael thank you for correcting me. I have updated the link. – Sujit Bhattacharyya Jan 29 '19 at 02:40