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

*

*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 $y$.


*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 $p_i$.
A: For $S^2$ you can either try to find the equation for the arc of the great circle that any two points on a sphere share (IE the straight line in spherical geometry),   or you can just do a simple two step path,  first vary $\theta$ to $\alpha$ by standard interpolation then $\phi$ to $\beta$.  (i.e. from $0$ to $\frac 1 2$) have your path $$f(t)=(\cos (\frac t 2\theta +\frac {1-t} 2\alpha).....)$$ while leaving the second parameter fixed at $\phi$, then for the second half of the path do the same thing but vary $\phi$ to $\beta$.
For the second one,  you could always either A:  Use the straight line path if the straight line doesn't hit any of the $p_i's$ and if it does,  go around them in a piecewise linear fashion,  or if you wanted to try to come up with a general formula you could try just doing a piecewise formula varying $r$ and $\theta$ from point 1 to point 2 in such a way that you never duplicate one of the finite number of bad points.
A: After spending a long time staring at the computer screen, I think I figured it out and thought I'd share my solution.
$S^2$:
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 $\gamma\left(t\right)=\left(\cos{\left(t\alpha+\left(1-t\right)\theta\right)}\sin{\left(t\phi+\left(1-t\right)\beta\right)}, \sin{\left(t\alpha+\left(1-t\right)\theta\right)}\sin{\left(t\phi+\left(1-t\right)\beta\right)}, \cos{\left(t\phi+\left(1-t\right)\beta\right)}\right)$
$X=\mathbb{C}\setminus\{p_1,...,p_n\}$:
Let $x,y\in\mathbb{C}$ with $x=x_1+ix_2$ and $y=y_1+iy_2$.
Between any two distinct points on $\mathbb{C}$, there is a unique line.
Let $l_i$ be the line on which $x$ and $p_i$ lie and $m_i$ be a line on which $y$ and $p_i$ lie. Then, we have $2n$ lines in $\mathbb{C}$
There are infinitely many lines through $x$, so pick some line $l$ which does not equal any of the line $l_i$. similarly, pick a line $m$ which does not equal any of the line $m_i$ which is not parallel to $l$.
Then, we have two lines $l$ and $m$ such that $x$ is in $l$ and $y$ is in $m$ and $p_i$ does not lie on either $l$ or $m$. Since $l$ and $m$ are not parallel, there is a point $z$ where the lines intersect.
Therefore, there is a path from $x$ to $y$ by first travelling along $l$ from $x$ to $z$ and then travelling along $m$ from $z$ to $y$.
