- #1
kingwinner
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Homework Statement
Nested interval theorem:
Suppose [an,bn] is a nested sequence of closed intervals, i.e. [an+1,bn+1] is contained in [an,bn] for all n≥1. Then the intersection of all these intervals is nonempty.
Proof:
an is an increasing sequence bounded above by b1 (or any bn), so sup an = a exists and a≤bm for all m.
So "a" is a lower bound of bm
Thus let b = inf bm, then we have a≤b (since any lower bound is ≤ its greatest lower bound).
Then [a,b] is contained in [an,bn] for all n since an≤a and b≤bn for all n (because a is an upper bound of an and b is a lower bound of bn)
Therefore, the intersection is nonempty.
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1) In this theorem, it says that "...the seqeunce of CLOSED intervals...". Is it enough to have closed intervals, or must the endpoints also be FINITE numbers? Can any of the endpoints be -∞ or +∞? (reacall that intervals like [a,∞) are classified as closed interval)
2) In the proof, I am confused and I really don't understand why "a≤bm for all m". Can somebody please explain this?
Homework Equations
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The Attempt at a Solution
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Thank you! :)