prove that??

here is the solution:

for so so 1 is upper bound so

there is a supremum suppose 1 is not the smallest upper bound

suppose .

so by the limit definition for every epsilon there is M>0 so for every

x for which x>M

so

and it works for every x for which x>M

we take

for this it works too

where is the contradiction here??

we were talking about the supremum of not