Some question about path connectedness I think that's intuitively evident but I can't prove that the set $\mathbb{S}^n\setminus\{(0,\cdots, 1),(0,\cdots, -1)\}\; (n>1)$ is path connected. Does anyone have a formal argument to prove it?
Thanks
 A: Stereographic projection provides a homeomorphism between $S^n \setminus \{NP, SP\}$ and $\Bbb R^n \setminus \{0\}$. Because path-connectedness is a homeomorphism invariant, it suffices to show that this second space is path-connected. What better way to do this than to write down a path between any two points? 
Let $x, y \in \Bbb R^n$ be nonzero. If $x$ is not a scalar multiple of $y$, the path $f(t) = (1-t)x+ty$ is a path from $x$ to $y$ that does not pass through zero. If $x$ is a scalar multiple of $y$, we'll need to modify this path just a tiny bit. Pick a vector $z$ not on the line spanned by $y$. Why not just use the path that first goes from $x$ to $z$ and then to $y$ by a straight line? $z$ is intentionally chosen to be a scalar multiple of neither $x$ nor $y$, so this path will never go through zero.
So let's define such a path: 
$$f(t) = \left\{
     \begin{array}{lr}
       (1-2t)x + 2tz & 0 \leq t \leq 1/2\\
       (2-2t)z+(2t-1)y & 1/2 \leq t \leq 1
     \end{array}
   \right.$$
This starts at $x$, is at $z$ at $t=1/2$, and ends at $y$. (You can check that this is continuous.) So we've drawn a path between any two points; this shows that our space is path-connected.
Note that to do this, it was key that we had some $z$ that wasn't a multiple of $y$, or else we couldn't have avoided $0$. This is where the $n>1$ hypothesis came in!
