This question may be really dumb, but I am wondering if $\mathbb{S}^{1}\setminus\{(1,0),(-1,0)\}$ is connected or not..
I have this confusion since we know that for sphere $\mathbb{S}^{n}$, $n>1$, the set $$C:=\{(x_{1},\cdots, x_{n+1})\in\mathbb{S}^{n}:x_{n+1}\neq \pm 1\}$$ is connected, but it seems that we don't talk about if $\mathbb{S}^{1}$ also has such a property.
My guess is that $\mathbb{S}^{1}\setminus\{(1,0),(-1,0)\}$ is disconnected, since it can be broken into two disjoint open sets, namely $$A=\{(\cos\alpha,\sin\alpha):\alpha\in (0,\pi)\}\ \text{and}\ B=\{(\cos\alpha,\sin\alpha):\alpha\in (\pi,2\pi)\}.$$
Is my argument correct? If this is indeed disconnected, what happens when we move $\mathbb{S}^{1}$ to higher dimension so that $C$ is connected?
Thank you!
Your argument is correct. Instead of your comparison, you could make the comparison to $$C':=\{(x_{1},\cdots, x_{n+1})\in\mathbb{S}^{n}:\ x_n\neq0\}.$$ This similarly divides $\Bbb{S}^n$ into two disjoint open sets; the points with $x_n>0$ and the points with $x_n<0$.