This is theorem..
and this is proof..
Imageshack - picture001c
sup özelliği = sup property
bağlantılıdır = connected
(this is a p->q, ~q->~p proof)
When I learned about supremums we were taught the convention that an empty set has supremum and a set unbounded above has supremum , so the set of real numbers has supremum , this is known as an affine extension.
Regardless, this question still doesn't make much sense.
Taking the supremum property to mean:
Every bounded subset of has a supremum in ,
the proof goes as follows.
Let a subset of be both open and closed.
Take and using our supremum property, take b as the supremum of real numbers such that
If , then b is a limit point of U and since U is closed,
Using the assumption that is also open we also must have that
Hence which contradicts b being a supremum. Therefore
Repeating this arguement for an infimum we have that or
Therefore is connected.
Hope this cleared some things up