Proof if $A\subseteq \mathbb R$ then $\sup A>\inf A$ if and only if $A$ has at least two different elements. Let A be a nonempty, bounded set of real numbers. Show, that sup A > inf A if and only if A
contains at least two different points.
I was trying to do a proof by contradiction and say that A doesn't have two different points, and that all the points are the same. I get stuck from there. Any ideas? 
Thanks
 A: Assume our set $A$ contains two numbers $a$ and $b$, without loss of generality we may assume $a<b$. Then $\sup(A)$ is an upper bound by definition and so $a<b\le \sup(A)$. Also $\inf(A)$ is a lower bound by definition, so $\inf(A) \le a < b \le \sup(A)$. Thus $\inf(A)<\sup(A)$ and I win!
Now assume $\inf(A)<\sup(A)$ and suppose to the contrary there is only one element $x_0$ in $A$. Then $\inf(A) = x_0 = \sup(A).$ This contradicts our assumption. Yikes!
The desired result follows.
A: Others (H_B) have already given the correct answer. I'm just adding some more detail here:
$\implies$ (proving forward direction)
To prove by contradiction, assume $\sup(A)\gt\inf(A)$ and suppose $\nexists$x,y$\in$A such that x$\lt$y (alternatively $\forall$x,y$\in$A, x=y).  This means $\left\vert{A}\right\vert$ = 1.  So, choose $x_o\in$A.  By definition, $\sup(A)\ge$$x_o$.  Also by definition, $\inf(A)\le$$x_o$.  But since $\left\vert{A}\right\vert$ = 1, we know $\sup(A)$=$x_o$ and similarly $\inf(A)$=$x_o$.    $\therefore$ $\inf(A)$=$x_o$=$\sup(A)$, which is a contradiction.

$\impliedby$ (proving reverse direction)
Choose x,y$\in$A such that x$\ne$y (two different points in A).  Without loss of generality, we may assume that x$\lt$y.  By definition, $\sup(A)\ge$x and $\sup(A)\ge$y.  Also by definition, $\inf(A)\le$x and $\inf(A)\le$y.  Since x$\lt$y, it can be seen that $\inf(A)\le$x$\lt$y$\le\sup(A)$.    $\therefore$ $\inf(A)$<$\sup(A)$.
