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proof10 .pdf


Original filename: proof10.pdf
Author: vargh

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10. Give an example of a family of intervals An,n=1,2,…, such that
An+1⊂An for all n and ⋂ n=1∞An consists of a single real number. Prove
that your example has the stated property.
An example is for An to be in the interval (-a/n, a/n) where n = 1, 2, ..., a.
In this, the only real number in ⋂ n=1∞An is 0.
Proof by contradiction.
Suppose there is one nonzero element in ⋂

n=1



An called x.

If x > 0, the limit of the upper bound a/n must be 0. As there always will
exist an n0 where m > n0, |a/m - 0| < |x|, which becomes a/m < x.
Therefore, x cannot belong to the intersection because it is greater than
the upper bound.
The same math applies if x were to be negative: |-a/m - 0| < |x|, which
simplifies to a/m > -x (because a is always a positive real number, you
must multiply both sides of the inequality by -1).
0 is the only real solution, because it will always fall into the interval
(-a/m, a/m) when a is a positive real number and m is nonzero.


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