I need help understanding this proof that a Cauchy sequence is convergent.

Let $(a_n)_n$ be a Cauchy sequence. Let's prove that $(a_n)_n$ is bounded. In the definition of Cauchy sequence: $$(\forall \varepsilon>0) (\exists n_\varepsilon\in\Bbb N)(\forall n,m\in\Bbb N)((n,m>n_\varepsilon)\Rightarrow(|a_n-a_m|<\varepsilon))$$ let $\varepsilon=1$. Then we have $n_1\in\Bbb N$ such that $\forall n,m\in\Bbb N (n,m>n_1)\Rightarrow(|a_n-a_m|<1)$. From there for $n>n_1$ we have $|a_n|\leq |a_n-a_{n1+1}|+|a_{n1+1}|(*).$ Now $M=\max\{|a_1|,...|a_{n1}|,1+|a_{n1+1}|\}$ such that $|a_n|\leq M,\ \forall n\in\Bbb N.$

Bounded sequence $(a_n)_n$ has a convergent subsequence $(a_{p_n})_n$, i.e. there exists $a=\lim_n a_{p_n}$. Let's prove $a=\lim_n a_n$. Let $\varepsilon>0$ be arbitrary. From the convergence of subsequence $(a_{p_n})_n$ we have $n'_\varepsilon\in\Bbb N$ such that $$(n>n'_\varepsilon)\Rightarrow(|a_{p_n}-a|<\frac{\varepsilon}{2}).$$ Because $(a_n)_n$ is a Cauchy sequence, we have $n''_\varepsilon\in\Bbb N$ such that $$(n,m>n''_\varepsilon)\Rightarrow(|a_n-a_m|<\frac{\varepsilon}{2}).$$ Let $n_\varepsilon=\max\{n'_\varepsilon, n''_\varepsilon\}$ so for $n>n_\varepsilon$ because $p_n\geq n$ we have

$$|a_n-a|\leq|a_n-a_{p_n}|+|a_{p_n}-a|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon \ (**)$$ i.e. $a=\lim_n a_n$.

$(*)$Where did $|a_n|\leq |a_n-a_{n1+1}|+|a_{n1+1}|$ come from? I understand why that inequality is true, but I don't see the point in writing in like that.

$(**)$ Why is $|a_n-a_{p_n}|<\frac{\varepsilon}{2}?$


2 Answers 2


$(*)$ The fact that $|a_n|\leq |a_n - a_{n1+1}| + |a_{n1+1}|$ is used to prove that $|a_n|\leq M$, meaning that the sequence is bounded. You need to first show that the sequence is bounded so that way you know it has a convergent subsequence.

The fact itself follows from $a_n=a_n-a_{n1+1} + a_{n1+1}$ which means that $$|a_n|=|a_n-a_{n1+1}+a_{n1+1}|\leq |a_n-a_{n1+1}| + |a_{n1+1}|$$ by triangle inequality.

$(**)$ Because $n>n'_{\epsilon}$. $n_\epsilon'$ is defined as the integer such that $$(n>n'_\varepsilon)\Rightarrow(|a_{p_n}-a|<\frac{\varepsilon}{2}),$$

and since $n>n_\epsilon=\max\{n'_\epsilon, n''_\epsilon\}$, you know that $n>n'_\epsilon$.


Here is a quick proof using only the supremum property :

Let $(a_n)$ be a Cauchy sequence of reals. It is bounded [ There is an $N$ such that $ a_N, a_{N+1}, \ldots $ are in $ (a_N - 1, a_N + 1) $. Now $ \max \{ |a_1|, \ldots, |a_{N-1}|, |a_N|+1 \} $ is $ \geq $ each $ | a_n | $ ].

So $ \alpha_{j} := \sup\{a_j, a_{j+1}, \ldots \} $ are well-defined, bounded (and decreasing). Therefore they converge, to $ \alpha := \inf\{ \alpha_1, \alpha_2, \ldots \} $.

Let $\epsilon > 0 $. There is an $ N (=N_{\epsilon}) $ such that $ a_N, a_{N+1}, \ldots $ are in $ (a_N - \epsilon, a_N + \epsilon) $. So $ \alpha_N, \alpha_{N+1}, \ldots $ are in $ [ a_N - \epsilon, a_N + \epsilon ] $, and hence so is $ \alpha $.

Finally $ a_N, a_{N+1}, \ldots $ and $ \alpha $ are all in $ [ a_N - \epsilon, a_N + \epsilon ] $, ensuring each $ | a_N - \alpha |, |a_{N+1} - \alpha |, \ldots $ is $ \leq 2 \epsilon $.


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