Show if K contained in R is compact, then supK and inf K both exist and are elements of K.
The Attempt at a Solution
Ok we proved a theorem stating that if K is compact that means it is bounded and closed.
So if K is bounded that means that for every element k in K there exists an M in R so that,
which is equiv to:
meaning for all k in K
k>-M and k<M and therefore K is bounded above AND below.
According to the axiom of completeness, that means that the supK and the infK exists.
Alright, this is where I get stuck, I know that the supK and infK are the limit points of K, and since we know that K is closed, they would be contained in K, but I don't know how to show that supK and infK are the limit points. Any help would be great!