• 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: Assume $\limsup a_n = \liminf a_n = a$. Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$. By preview lemma, we know $a_n \to a$.

Assume $(a_n)$ is convergent to $a$. Then, for any $\epsilon > 0$, there is an $N>0$, that for all $n > N$, $|a_n - a|<\epsilon$. In particular, we know $\limsup a_n \leq a+\epsilon$. Since this is true for all $\epsilon>0$, we have $\limsup a_n \leq a$. Similarly, $\liminf a_n \geq a$. On the other hand, since $A_n \geq B_n$, we have $\limsup a_n = \lim A_n \geq \lim B_n = \liminf a_n$. Hence, $\limsup a_n = \liminf a_n = a$.

Lemma: Cauchy sequence is bounded. (Exercise)

Lemma: If $\limsup a_n = A$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - A| < \epsilon$. Similarly, if $\limsup a_n = B$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - B| < \epsilon$.
Pf: Since $A_n = \sup_{m>n} a_m$ is a monotone decreasing sequence with limit $A$, then for any $\epsilon>0$ and $N>0$, we can find $M > N$ such that $A+\epsilon/2 > A_{M} \geq A$. By definition of $A_{M}$, there exists some $a_n$ with $n \geq M$, that $A_{M} \geq a_n > A_{M} - \epsilon/2$. Now, we have $$ |a_n - A| \leq |a_n - A_M| + |A_M - A| \leq \epsilon/2 + \epsilon/2 = \epsilon $$ The result is proven.

Theorem : Cauchy sequence in $\R$ is convergent.
Proof: Let $(a_n)$ be a Cauchy sequence. By previous lemma, it is bounded. Let $A = \limsup a_n$ and $B = \liminf a_n$, we only need to show that $A = B$ to show $\lim a_n$ exists. We know $A \geq B$ for all bounded sequence $a_n$, suppose $A > B$, and let $\epsilon = (A-B)/3$. Then, by Cauchyness of $(a_n)$, we there is an $N>0$, such that for all $n,m> N$, we have $|a_n - a_m| < \epsilon$. By previous lemma, there exists $n>N$, with $|a_n - A|<\epsilon$, and $m>N$ with $|a_m - B| < \epsilon$. Hence $$ |A - B| \leq |A-a_n|+|a_n - a_m| + |a_m - B| < 3 \epsilon = |A-B|, $$ notice the inequality is strict, hence we have a contradiction. Thus $A = B$.