Problem #9

Question: given an infinite collection $A_n, n = 1,2,...$ of intervals of the real line, their intersection is defined to be
$$\bigcap_{n=1}^\infty A_n = \{x|(\forall n)(x \in {A_n})\}$$
Give an example of a family of intervals $A_n, n = 1,2,...$, such that $A_{n+1} \subset A_n$ for all $n$ and $\bigcap_{n=1}^\infty A_n = \emptyset$. Prove that your example has the stated property.
Answer: $A_n = (0, \frac{1}{n}), n = 1,2,...$

Claim: the family of intervals $A_n = (0, \frac{1}{n}), n = 1,2,...$ has the next properties:
1. $A_{n+1} \subset A_n$ for all $n$
2. $\bigcap_{n=1}^\infty A_n = \emptyset$

Proof:
1. Lets prove it directly. Let's take an arbitrary $n$, then interval $A_n = (0, \frac{1}{n})$ and $A_{n+1} = (0, \frac{1}{n+1})$. Let's proof that $A_{n+1} \subset A_n$, i.e, $(0, \frac{1}{n+1}) \subset (0, \frac{1}{n})$. Every element of interval $(0, \frac{1}{n+1})$ we can write as $0 < x < \frac{1}{n+1}$, and every element of $(0, \frac{1}{n})$ as $0 < x < \frac{1}{n}$.
$\frac{1}{n+1} < \frac{1}{n}$ and thus we get that every element of $A_{n+1}$ is an element of $A_n$, therefore, by definition of subset, $A_{n+1}$ is a subset of $A_n$ for all $n$. $\blacksquare$
2. Lets prove it directly. To find an intersection $\bigcap_{n=1}^\infty A_n$ we will use the fact that $A_{n+1} \subset A_n$ and will try to find the last $A_n$ when $n \rightarrow \infty$, that will be an intersection of all $A_n$ because of $A_n \supset A_{n+1} \supset A_{n+2} ... \supset A_{n \rightarrow \infty}$ and the intersection definition. Let's find $A_n$ when $n \rightarrow \infty$. $\lim \limits_{n \rightarrow \infty} = 0$, so when $n \rightarrow \infty$, $(0, \frac{1}{n}) = (0, 0)$, but $(0, 0)$ is an empty set $\emptyset$, so $\bigcap_{n=1}^\infty A_n = A_{n \rightarrow \infty} = \emptyset$. $\blacksquare$