Suppose b = Sup S and b is not in S. prove b is an accumulation point.
if b is the supremum of S but not in S then b is in R/S.
Since b is the supremum then for all s S, s < b.
But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c S, but still less than b.
is this ok?
.
Since b=SupS we have by definition that:
1) for all ,x :
2) for all,ε>0 ,there exists a y belonging to S ,such that :
.................................................. .................................................. ..1
And since b does not belong to S we have :
.................................................. .......................................2
Hence from (1) and (2) we conclude ,that there exists a deleted neighbourhood round b containing yεS.
So b is an accumulation point.
Drexel28 ,your theorem and your proof are both wrong.
Your theorem should be:
Theorem: Let such that exists .And SupE does not belong to E . Then is a limit point of
And in your proof ,does not imply :
that there does not exists an such that .
But that:
for all ,x : or (SupE=x) or .
Which is equivalent to:
.
And that shows that:
Only if , then there does not exists an such that .
And since SupE does not belong to E we have:
and the contradiction follows.