Let and .
There can be no sequence with for all n
no sequence with for all n
Towards a contradiction suppose there is such a sequence with .
Since is non-decreasing and , we have
Since g is continous, , so that taking limits on both sides leads to the contradiction
Analogous in the second case.
Now consider . , because is closed.
for some .
Then and for all according to the definition of y.
So is nonempty such that there is a sequence with , which violates lemma 1.
So is nonempty such that there is a sequence with and which violates lemma 1.