Path-connectedness of $S^2$ and $X=\mathbb{C}\setminus\{p_1,\ldots,p_n\}$

general-topologypath-connected

I've been given two examples of path-connected sets, but I am not sure how to prove their path-connectedness.

$S^2$: Let $x,y\in S^2$. Then, we can parameterise the points by $x=\left(\cos{\theta}\sin{\phi},\sin{\theta}\sin{\phi},\cos{\phi}\right)$ and $y=\left(\cos{\alpha}\sin{\beta},\sin{\alpha}\sin{\beta},\cos{\beta}\right)$ for some $\theta,\phi,\alpha,\beta\in\left[0,2\pi\right]$. Then, any $x,y\in S^2$ are connected by the path $\gamma(t)$, however I am not sure how to proceed from what I know so far here.

$X=\mathbb{C}\setminus\{p_1,\ldots,p_n\}$: This one I do not even know whether to use the general, polar or exponential form for proof.

Any advice will be greatly appreciated.

Best Answer

$S^2$:

Let $x,y \in S^2$ and let $$u : [0,1] \to \mathbb R^3, u(t) = ty + (1-t)x$$ be the linear path from $x$ to $y$. When do we have $u(t) = 0$? Clearly $u(0) = x \ne 0$, thus $u(t) = 0$ means that $y = -\frac{1-t}{t}x$ and thus $1 = \lVert y \rVert = \frac{1-t}{t}\lVert x \rVert = \frac{1-t}{t}$. This implies $t = \frac{1}{2}$ and $y = - x$.

  1. Therefore, if $y \ne -x$, then $$\gamma(t) = \frac{ty + (1-t)x}{\lVert ty + (1-t)x \Vert}$$ is a path in $S^2$ from $x$ to $y$.

  2. If $y = -x$, choose any $z \in S^2 \setminus \{x,-x\}$. Then $z \ne -x$ and $z \ne -y = x$, and 1. shows that $x,z$ and $y, z$ can be joined by paths in $S^2$. We conclude that also $x,y$ can be joined by a path in $S^2$.

$X = \mathbb{C}\setminus\{p_1,\ldots,p_n\}$:

Let $r = \frac{1}{2}\min\{ \lvert p_i - p_j \rvert : i, j \in \{1,\ldots,n\}, i \ne j \}$ . We have $r > 0$ and the $U_r(p_i) = \{z \in \mathbb C : \lvert z - p_i \rvert < r\}$ are pairwise disjoint. For two distinct points $x,y \in X$ let $u(t) = ty + (1-t)x$ be the linear path in $\mathbb C$ from $x$ to $y$. Assume that some of the $p_i$ lie on this path. Then $p_i = u(t_i)$ for unique $t_i \in (0,1)$. For sufficiently small $\epsilon_i > 0$ we have $0 < t_i - \epsilon_i < t_i + \epsilon_i < 1$ and $u([t_i - \epsilon_i, t_i + \epsilon_i]) \subset U_r(p_i)$. Clearly $u(t_i \pm \epsilon_i) \ne p_i$. Pick any $z_i \in U_r(p_i)$ which does not lie on the line through the two points $u(t_i \pm \epsilon_i)$ and replace $u \mid_{[t_i - \epsilon_i, t_i + \epsilon_i]}$ by the path going linearly on $[t_i - \epsilon_i,t_i]$ from $u(t_i - \epsilon_i)$ to $z_i$ and then linearly on $[t_i, t_i + \epsilon_i]$ from $z_i$ to $u(t_i + \epsilon_i)$. This produces a new path from $x$ to $y$ which does not contain any of the $p_i$.

Related Question