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· 3 min read
Dominic Ng

##Question: An=[0,1n)={x∈R∣0≤x<1n}A_n = [0, \frac{1}{n}) = \{ x \in \mathbb{R} | 0 \leq x < \frac{1}{n} \}

A=⋂n=1∞An=A1∩A2∩⋯A = \bigcap_{n=1}^{\infty} A_n = A_1 \cap A_2 \cap \cdots

ProveA=0Prove A = {0}

This problem demonstrates the behavior of infinite intersections of nested sets. The sets AnA_n are nested, meaning that An+1⊆AnA_{n+1} \subseteq A_n for all n∈Nn \in \mathbb{N}. The interval AnA_n becomes arbitrarily small as nn increases, but it always contains numbers greater than 0. However, there is no real number that is in every AnA_n, and thus the intersection of all the AnA_n is the empty set. This can be shown formally as follows:

Let's suppose, for contradiction, that there exists an x∈A=⋂n=1∞Anx \in A = \bigcap_{n=1}^{\infty} A_n. By the definition of intersection, this would imply that xx is in every set AnA_n. However, for every real number xx in the interval [0,1n)[0, \frac{1}{n}), there exists an nn such that xx is not in the interval [0,1n+1)[0, \frac{1}{n+1}) because 1n+1<1n\frac{1}{n+1} < \frac{1}{n} for all n∈Nn \in \mathbb{N}, and thus x≥1n+1x \geq \frac{1}{n+1}. This is a contradiction, implying that there does not exist such an xx in the intersection of all AnA_n, and thus A=∅A = \emptyset.

Suppose for contradiction:

x∈A=⋂n=1∞Anx \in A = \bigcap_{n=1}^{\infty} A_n

By definition of intersection, xx is in every set AnA_n. However, for every real number xx in the interval [0,1n)[0, \frac{1}{n}), there exists an nn such that xx is not in the interval [0,1n+1)[0, \frac{1}{n+1}) because

1n+1<1n\frac{1}{n+1} < \frac{1}{n}

for all n∈Nn \in \mathbb{N}, and thus x≥1n+1x \geq \frac{1}{n+1}. This is a contradiction, implying that there does not exist such an xx in the intersection of all AnA_n, and thus A=∅A = \emptyset.

Alright, let's simplify this discussion with an analogy. Imagine you have a sequence of shrinking boxes:

  1. In the first scenario, the box (let's call it Box 1) has a length of 1 foot.
  2. In the next scenario, the box (Box 2) has a length of 1/2 foot.
  3. In the third scenario, the box (Box 3) has a length of 1/3 foot.
  4. And so on... The box keeps shrinking.

Now, the question is, if you keep going like this, shrinking the box in each scenario, what do you have left at the end?

The mathematical notation you provided describes this scenario but with intervals (a type of set) instead of boxes. The notation An=[0,1n)A_n = [0, \frac{1}{n}) is just a fancy way of describing these shrinking boxes (or intervals). Here, nn represents the scenario number, and 1n\frac{1}{n} tells you the length of the box in that scenario.

The expression A=⋂n=1∞AnA = \bigcap_{n=1}^{\infty} A_n is asking what is common to all these boxes if you go through all the scenarios (from 1 to infinity). In other words, it's like asking: Is there a piece of space that remains covered by all the boxes, no matter how small they get?

In this case, the answer is no. There's no piece of space that remains covered by all the boxes in all scenarios because the boxes keep getting smaller and smaller, and eventually, they don't cover anything. So the intersection of all these shrinking intervals (or the common space covered by all the boxes) is empty. In mathematical terms, we say the intersection of all these sets (or intervals) is the empty set, denoted as A=∅A = \emptyset.

Now, regarding the inequality n+1n>1n\frac{n+1}{n} > \frac{1}{n}, this is just a mathematical way of showing that as you go from one scenario to the next (e.g., from Box 1 to Box 2, from Box 2 to Box 3, etc.), the box is indeed getting smaller.