Question 10
Question 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.
Solution
The family of intervals defined by An=[0,n1] has the property that An+1⊂An for all n and ∩n=1∞An={0}.
Part 1
We will prove [0,n+11]⊂[0,n1] for all n directly.
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Let n be any natural number and take the interval An=[0,n1]
This means ∀x∈An0≤x≤n1
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Taking the successor interval An+1, we have [0,n+11].
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Since x∈An+1 means 0≤x≤n+11 and 0≤x≤n+11<n1
x∈An+1 means x∈An
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And so we conclude that An+1⊂An.
Part 2
We will prove, by contradiction, that ∩n=1∞[0,n1]={0}
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First we note that 0 is in every interval. For any ∀n,0∈[0,n1]
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Next we assume for the sake of contradiction that there is another number with this property. Then:
∃x>0,∀n,x∈[0,n1] (Note x must be greater than zero)
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Finally we pick a natural number m>x1.
Since both x and m are positive, we can multiply both sides by x and divide both sides by m to get m1<x
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Now we have an interval [0,m1] whose maximum is less than x.
So x∈[0,m1]
But we just assumed that ∃x>0,∀n,x∈[0,1/n].
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And so we have a contradiction: x>0 is both a member of every interval and there is an interval that does not contain x.
Therefore, we conclude that x>0 does not exist in all intervals.
And so we have proven both that for An=[0,n1],
∀n,An+1⊂An and ∩n=1∞An={0}.