Originally Posted by
pomp Taking the supremum property to mean:
Every bounded subset of $\displaystyle \mathbb{R}$ has a supremum in $\displaystyle \mathbb{R}$,
the proof goes as follows.
Let a subset $\displaystyle U$ of $\displaystyle \mathbb{R}$ be both open and closed.
Take $\displaystyle a \in U$ and using our supremum property, take b as the supremum of real numbers such that $\displaystyle [a,b) \subset U$
If $\displaystyle b<\infty$, then b is a limit point of U and since U is closed, $\displaystyle b \in U.$
Using the assumption that $\displaystyle U$ is also open we also must have that
$\displaystyle \exists$ $\displaystyle \epsilon > 0$ such that $\displaystyle (b-\epsilon, b+\epsilon) \subset U$
Hence $\displaystyle [a, b + \epsilon) \subset U$ which contradicts b being a supremum. Therefore $\displaystyle b = \infty.$
Repeating this arguement for an infimum we have that $\displaystyle U = \mathbb{R}$ or $\displaystyle U = \oslash $
Therefore $\displaystyle \mathbb{R}$ is connected.
Hope this cleared some things up
Pomp.