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Problem #10

Question: 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 $\bigcap_{n=1}^\infty A_n$ consists of a single real number. 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. $\bigcap_{n=1}^\infty A_n$ consists of a single real number

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 $0 \leq x \leq \frac{1}{n+1}$, and every element of $[0, \frac{1}{n}]$ as $0 \leq x \leq \frac{1}{n}$.
$\frac{1}{n+1} < \frac{1}{n}$ 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 $n \rightarrow \infty$, $[0, \frac{1}{n}] = [0, 0]$, and $[0, 0]$ is a set with a single member $0$, so $\bigcap_{n=1}^\infty A_n$ consists of a single real number. $\blacksquare$