__Lemma 1:__
Let

and

.

There can be no sequence

with

for all n

and

no sequence

with

for all n

** Proof:**
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.

qed

Now consider

.

, because

is closed.

Assume

.

case 1:

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.

case 2:

for some

Then

.

So

is nonempty such that there is a sequence

with

and

which violates lemma 1.

Therefor

.

qed