• Cauchy sequence: “we don't know where we are heading, but we agree with each other more and more”.
• Cauchy sequence is equivalent to convergent sequences…. in $\R$.
• Limit sets and subsequences.

First, the definition. Let $(a_n)$ be a sequence in $\R$, we say $(a_n)$ is Cauchy, if for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$.

Lemma : Convergent sequence is Cauchy.
Pf: suppose $a_n \to a$. We need to show that for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$. Let $\epsilon_1 = \epsilon /2 $, then by convergence of $a_n$, we have $N_1>0$, such that $|a_n - a| < \epsilon_1$ for all $n > N_1$. Thus for any $n,m > N_1$, we have $$ |a_n - a_m| \leq |a_n -a | + |a_m - a| \leq \epsilon_1 + \epsilon_1 = \epsilon $$ Thus, let $N=N_1$ would work.

Lemma : If $A_n \geq a_n \geq B_n$, and $\lim A_n = \lim B_n = a$, then $\lim a_n = a$.
Pf: for any $\epsilon > 0$, we have $N>0$ such that, for all $n > N$, $|A_n - a| < \epsilon, |B_n - a| < \epsilon$, then $$ a_n \leq A_n \leq a+\epsilon, \quad a_n \geq B_n \geq a-\epsilon $$ Thus ,$|a_n - a| \leq \epsilon$ for all $ n > N$, hence $a_n \to a$.

Lemma : Let $(a_n)$ be a bounded sequence. $(a_n)$ is convergent if and only if $\limsup a_n = \liminf a_n$.
Pf: Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$, we know $A_n, B_n, a_n$ all convergent, hence $$ \lim A_n \geq \lim a_n \geq \lim B_n. $$ Suppose $(a_n)$ is convergent to $a$, then