Due Thursday Sep 9th, 6pm.
In the following, a sequence $(a_n)$ means $(a_n)_{n=0}^\infty$, unless otherwise specified. You can only use properties of real number proved in Tao's book, section 5.4.
1. Let $(a_n)$ be a sequence in $\Q$. Suppose there is a rational $0 < r < 1$, such that $|a_{n+1} - a_n| < r |a_n - a_{n-1}|$, prove that $(a_n)$ is a Cauchy sequence.
2. Let $a_0$ be any positive integer, and $a_{n+1} = 3 + \frac{1}{a_n}$. Prove that the $(a_n)$ is a Cauchy sequence. (Hint: use the previous problem)
3. Let $(a_n)$ be a Cauchy sequence in $\Q$, and let the sequence $(b_n)$ be defined such that $b_n = a_{2n}$. Prove that $(b_n)$ is equivalent to $(a_n)$.
4. If $(a_n)$ and $(b_n)$ are Cauchy sequences, prove that $(a_n b_n)$ is also a Cauchy sequence.
5. If $(a_n)$, $(c_n)$ and $(b_n)$ are Cauchy sequences, and $(a_n) \sim (c_n)$, prove that $(a_n b_n) \sim (c_n b_n)$.
(Problem 4 and 5 together proves Tao proposition 5.3.10, multiplication of real are well-defined).
We first note that, for $n \geq 0$, by repeated using the given inequality, we have $$ |a_{n+1} - a_n| < r^n |a_1 - a_0| $$ We also note that $|a_1-a_0|>(1/r) |a_2-a_1|=0$ by assumption.
Then, for any $n < m$, we have $$ |a_m - a_n| \leq |a_m - a_{m-1}| + |a_{m-1} - a_{m-2}| + \cdots + |a_{n+1} - a_n| \leq |a_1 - a_0| (r^n + \cdots r^{m-1}) \leq |a_1 - a_0| \frac {r^n}{1-r}. $$
For any $\epsilon>0$, we may take integer $N$ large enough, such that $$ |a_1 - a_0| \frac {r^N}{1-r} < \epsilon $$ Thus, for all $n,m > N$, we have $$ |a_m - a_n| \leq |a_1 - a_0| \frac {r^n}{1-r} \leq |a_1 - a_0| \frac {r^N}{1-r} \leq \epsilon. $$
For integer $n \geq 1$, we have $$ a_{n+1} - a_n = (3 + \frac{1}{a_n}) - (3 + \frac{1}{a_{n-1}}) = \frac{a_{n-1} - a_n}{a_n a_{n-1}} $$ Note that, for any $a_0 > 0$, we have $a_1 > 3$ and all subsequence $a_n > 3$, hence for $n \geq 2$, we have $$ |a_{n+1} - a_n| \leq \frac{|a_{n-1} - a_n|}{9}. $$ Throw away the first term, we have $(a_n)_{n=1}^\infty$ satisfies the condition of problem 1, with $r=1/9$, hence it is a Cauchy sequence.
We need to show that, given any $\epsilon > 0$, there is an $N>0$, such that for all $n > N$, $|a_n - b_n| \leq \epsilon$. Since $(a_n)$ is Cauchy, for the given $\epsilon >0$, there exists an $N_0>0$, such that for any $n, m> N_0$, $ |a_n - a_m| \leq \epsilon$. In particular, if we take $m=2n$ and set $N=N_0$, then we get the desired result.
Since Cauchy sequences are bounded, there exists $M>0$, such that $|a_n| < M$ and $|b_n| < M$ for all $n$. Suppose $\epsilon > 0$ is given, we need to show that there exists $N>0$, such that for all $n,m>N$, $|a_n b_n - a_m b_m| \leq \epsilon $. We let $\epsilon_1 = \epsilon_2 = \epsilon / (2M)$, then since $(a_n)$ is Cauchy, there is a $N_1 > 0$, such that for all $n,m > N_1$, $|a_n - a_m| < \epsilon_1$. Similarly, using $(b_n)$ is Cauchy, we get $N_2$. Let $N = \max(N_1, N_2)$. Then for any $n,m> N$, we have $$ |a_n b_n - a_m b_m| = |a_n b_n - a_n b_m + a_n b_m - a_m b_m| \leq |a_n| |b_n - b_m| + |a_n - a_m| |b_m| \leq M |b_n - b_m| + M |a_n - a_m| \leq M \epsilon_1 + M \epsilon_2 = \epsilon. $$
We need to prove that, for any $\epsilon>0$, exists $N>0$, such that for all $n>N$, we have $$ | a_n b_n - c_n b_n | \leq \epsilon. $$
Since Cauchy sequences are bounded, there exists $M>0$, such that $|b_n| < M$ for all $n$. Let $\epsilon' = \epsilon / M$, then since $(a_n) \sim (c_n)$, there exists $N>0$, such that for all $n>N$, we have $ |a_n - c_n| < \epsilon'$, hence $$ |a_n b_n - c_n b_n | \leq |a_n - c_n| |b_n| \leq |a_n - c_n| M \leq \epsilon' M = \epsilon. $$