/bio/skills/blog/math

Question 10

Question 10

Give an example of a family of intervals An,n=1,2,...A_n,n = 1,2,..., such that An+1AnA_{n+1} \subset A_n for all nn and n=1An\cap_{n=1}^\infty A_n consists of a single real number. Prove that your example has the stated property.

Solution

The family of intervals defined by An=[0,1n]A_n = [0, \frac{1}{n}] has the property that An+1AnA_{n+1} \subset A_n for all nn and n=1An={0}\cap_{n=1}^\infty A_n = \{0\}.

Part 1

We will prove [0,1n+1][0,1n][0, \frac{1}{n + 1}] \subset [0, \frac{1}{n}] for all nn directly.

  1. Let n be any natural number and take the interval An=[0,1n]A_n = [0, \frac{1}{n}]

    This means xAn0x1n\forall x \in A_n 0 \leq x \leq \frac{1}{n}

  2. Taking the successor interval An+1A_{n + 1}, we have [0,1n+1][0, \frac{1}{n + 1}].

  3. Since xAn+1x \in A_{n + 1} means 0x1n+1 0 \leq x \leq \frac{1}{n + 1} and 0x1n+1<1n 0 \leq x \leq \frac{1}{n + 1} \lt \frac{1}{n}

    xAn+1x \in A_{n + 1} means xAnx \in A_n

  4. And so we conclude that An+1AnA_{n+1} \subset A_n.

Part 2

We will prove, by contradiction, that n=1[0,1n]={0}\cap_{n=1}^\infty [0, \frac{1}{n}] = \{0\}

  1. First we note that 0 is in every interval. For any n,0[0,1n]\forall n, 0 \in [0, \frac{1}{n}]

  2. Next we assume for the sake of contradiction that there is another number with this property. Then:

    x>0,n,x[0,1n]\exists x \gt 0, \forall n, x \in [0, \frac{1}{n}] (Note xx must be greater than zero)

  3. Finally we pick a natural number m>1xm \gt \frac{1}{x}.

    Since both xx and mm are positive, we can multiply both sides by xx and divide both sides by mm to get 1m<x\frac{1}{m} \lt x

  4. Now we have an interval [0,1m][0, \frac{1}{m}] whose maximum is less than x.

    So x∉[0,1m]x \not{\in} [0, \frac{1}{m}]

    But we just assumed that x>0,n,x[0,1/n]\exists x \gt 0, \forall n, x \in [0, 1/n].

  5. And so we have a contradiction: x>0x > 0 is both a member of every interval and there is an interval that does not contain xx.

Therefore, we conclude that x>0x >0 does not exist in all intervals.

And so we have proven both that for An=[0,1n]A_n = [0, \frac{1}{n}],

n,An+1An\forall n, A_{n+1} \subset A_n and n=1An={0}\cap_{n=1}^\infty A_n = \{0\}.

2025 Stefano De Vuono