Connectedness of $S^2$ I'm taking a real analysis course, as I've said before, and the professor has been teaching a lot of topology.  We don't have a textbook, so I have to use his notes which are very confusing.  One question we have is:
Prove that the sphere $S^2$ is connected using the definition of connectedness.
The definition of connected we have is:
A topological space $X$ is said to be $connected$ if every continuous map from $X$ to a discrete topological space is constant.
So, which discrete topological space should I be choosing to show that $S^2$ is connected?  And then how would I show that $every$ continuous function is constant, as opposed to just choosing some and demonstrating that they are constant?
 A: To disprove that every man is mortal, it is enough to exhibit one man, of whom you can prove that he is immortal.  But to prove that every man is mortal, it is not enough to consider just one man.  So where you ask "which discrete topological space should I be choosing", the presupposition is erroneous: you can't do it by choosing one.  If you want to show as space is not connected, then that might work.
Suppose $f:S^2\to X$ is continuous and $X$ is a discrete topological space.  Pick two points $a,b\in S^2$.  Let $g:[0,1]\to S^2$ be a continuous function such that $g(0)=a$ and $g(1)=b$ (you may need to write a proof that such a function $g$ exists; that's not hard).  The $f\circ g:[0,1]\to X$ is continuous.  Consider the inverse image of $f(a)$ under $f\circ g$, i.e. the set $(f\circ g)^{-1}(f(a))=\{x\in[0,1]: f(g(x))=f(a)\}$.  By continuity and discreteness, that is an open subset of $[0,1]$.  It has a supremum $c$ in $[0,1]$, i.e. $c=\sup\{x\in[0,1]: f(g(x))=f(a)\}$.
See if you can deduce that $f(g(c))=f(a)$.  The inverse-image of every open neighborhood of $f(a)$ under $f\circ g$ is an open subset of $[0,1]$ that contains $c$.  That means $c$ is a member of an open set whose supremum is $c$.  The only point $d$ in $[0,1]$ that is the supremum of an open subset of $[0,1]$ that contains $d$ as a member is $1$.  Hence $c=1$, and we have $f\circ g$ is constant on $[0,1]$.  Hence $f(a)=f(b)$.  This holds regardless of which two points $a,b\in S^2$ were chosen.  Therefore $g$ is constant.
