Show that the unit sphere is connected I need to show that $\{(x,y,z)\in\mathbb{R}^{3}:x^2+y^2+z^2 = 1\}$ is connected. Intuitively I understand that it is path connected and, therefore, connected. However, I don't understand how I would define such a path mathematically. For example, if I wish to show that every point on the unit sphere connected to the north pole and, therefore, to every other point, how would I show this?  
 A: For a path between $x$ and the North Pole $\langle0,0,1\rangle$, assume that $x$ is not the South Pole. Consider $\phi:[0,1]\to S^2$ defined by the straight-line path between $x$ and the North Pole in $\Bbb R^3$ (in symbols, $\phi(t)=(1-t)x+t\langle0,0,1\rangle$). Then the projection of this path into the sphere:
$$t\mapsto\frac{\phi(t)}{\|\phi(t)\|}$$
is the desired path.
To connect the South Pole to the North Pole, I suppose you could connect the South Pole to some other point using the same trick, and then connect that point to the North Pole.
A: Each hemisphere in $S^2$ is homeomorphic to the closed unit disk $D^2$ and thus connected. (There's an obvious path from each $p\in D^2$ to the origin; alternatively, $D^2$ is the image of $[0, 1]^2$ under the continuous map $(r, \theta) \to (r\cos \theta, r\sin\theta)$.) Their intersection is nontrivial, so $S^2$ itself is connected.
A: If you know the stereographic projection, $p_N$ relatively to the north pole and $p_S$ relatively to the south pole, $p_N:S^2-N\rightarrow R^2$ is a diffeomorphism, thus $S^2-N$ is connected by arcs. You also have $p_S:S^2-S\rightarrow R^2$ is a diffeomorphism, thus there exists an arc between $N$ and any point of $S^2-S$.This shows that $S^2$ is connected by arcs.
A: The north pole is $(0,0,1)$ and the south pole is $(0,0,-1).$
For $p=(x,y,z)\in S^2$ and $r\in [0,1],$ let $$f_p(r)=((1-r)x,(1-r)y, z_{p,r})\in S^2$$ $$ \text {with }  z_{p,r}\geq 0\iff z\geq 0.$$ Then $f_p:[0,1]\to S^2$ is continuous with $f_p(0)=p, $ while  $f_p(1)=(0,0,1)$ if $z\geq 0, $ and $f_p(1)=(0,0,-1)$ if $z<0.$ So $f_p$ is a path from $p$ to the north pole or to the south pole.
For $s\in [0,1]$ let $h(s)=(\sin \pi s  ,0,\cos \pi s).$ Then $h:[0,1]\to S^2$ is continuous with $h(0)=(0,0,1)$ and $h(-1)=(0,0,-1).$ 
So for $p,p'\in S^2,$ the points $p$ and $p'$ are each path-connected  to one pole or the other, and the two poles are path-connected  to each other.
