Is $\{(x,y)|y=\sin(\frac{1}{x})\}\cup(0,1)$ connected on $R^2$?

Question

Source of definition: http://mathworld.wolfram.com/ConnectedSet.html
Definition: A connected set is a set that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set.

I don't think I can find an open set (in $R^2$ we consider the standard topology) contains $(0,1)$ that does not contain points on $y=\sin(\frac{1}{x})$ but I'm not completely sure. Does anyone know a formal prove or intuitive explanation?

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Yes, this is a classical example of connected space which is not path connected. See this thread.

Here's one more detailed proof.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 29 '14 at 16:07

Tomasz Kania

16,996

score 1 · Answer 2 · answered Oct 29 '14 at 16:20

This is a variant of the topologist's sine curve. I like another definition though:

The closure of the graph of $f:(0,1]\to\mathbb R$, $f(x)=\sin x$ in $\mathbb R^2$.

The graph of $f$ is clearly connected and it's not hard to prove that if you have a connected set $C$, then every set $C\subseteq B\subseteq\overline C$ is connected, which resolves your case too.

Is $\{(x,y)|y=\sin(\frac{1}{x})\}\cup(0,1)$ connected on $R^2$?

2 Answers2