- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I want to determine the following sets:
I want to determine the following sets:
- Let
. This means
To be able to divide by we have to consider three cases: , , . Case 1:
We get which is true for every . .
Case 2:
We get
Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection .
Case 3:
We get
Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection . Therefore we get that the intersection contains only the element and so we have .
Is everything correct and complete? (Wondering)
- We have that
for all . So it follows that .
Let . Now we have to prove that there is a such that and , or not? But how can we show that? (Wondering)
- Let
. This means
We see that it must hold that . Now we want to solve for .
From the first inequality we have .
From the second inequality we have . From that we get , since is positiv and so the case is rejected, right?
So we have Since is unbounded we can find always a value for that is bigger than , or not?
Would that means that the intersection can't contain any element, and so ? (Wondering)