Prove that $\{x_n\}$ is a Cauchy sequence iff $\forall \epsilon > 0\ \exists N: \forall n > N \implies |x_n - x_N| < \epsilon$

Let $$\{x_n\}$$ denote a sequence. Prove that: $$\{x_n\}\ \text{is fundamental} \iff \forall \epsilon > 0\ \exists N: \forall n > N \implies |x_n - x_N| < \epsilon$$

Let $$P$$ be a statement that: $$P = \forall \epsilon > 0\ \exists N: \forall n > N \implies |x_n - x_N| < \epsilon\tag1$$

Let $$Q$$ be a statement that $$x_n$$ is a Cauchy sequence: $$Q = \forall \epsilon > 0\ \exists N \in \Bbb N:\forall n,m > N \implies |x_n - x_m| < \epsilon \tag2$$

We want to show two things: $$P \implies Q$$ and $$Q \implies P$$.

$$\Box$$ Start with $$P \implies Q$$. Then by $$(1)$$: $$\forall \epsilon>0\ \exists N_1 \in\Bbb N : \forall n > N_1 \implies |x_n - x_{N_1}| < \epsilon \tag3$$

At the same time: $$\forall \epsilon>0\ \exists N_2 \in\Bbb N : \forall m > N_2 \implies |x_m - x_{N_2}| < \epsilon \tag4$$

If we now choose $$N = \max\{N_1, N_2\}$$ both statements are true and we obtain: $$\forall\epsilon > 0\ \exists N =\max\{N_1, N_2\}: \forall n, m> N \implies \begin{cases} |x_n - x_N| < \epsilon \\ |x_m - x_N| < \epsilon \\ \end{cases}$$

Consider the sum of the inequalities: $$|x_n - x_N| + |x_m - x_N| < 2\epsilon \\ |(x_n - x_N) - (x_m - x_N)| \le |x_n - x_N| + |x_m - x_N| \\ |x_n - x_m| < 2\epsilon$$ Now if we choose $$\epsilon = {\epsilon_0 \over 2}$$ in $$(3)$$ and $$(4)$$ we get: $$\forall \epsilon_0>0\ \exists N =\max\{N_1, N_2\} : \forall n, m > N \implies |x_n - x_m| < \epsilon_0\ \Box$$

$$\Box$$ Proceed to $$Q\implies P$$. This is where I got stuck. So far I've got that if a sequence is Cauchy then it must be bounded.

What should be my steps to show the second implication? Also I'm not quite sure about the first part is correct, could you please confirm/reject it as well?

$$Q\implies P$$ is evident from definition of Cauchy sequence:

$$\forall \epsilon > 0\ \exists N \in \Bbb N:\forall n,m \ge N \implies |x_n - x_m| < \epsilon$$

if we let $$m=N$$, we get $$P$$.

while for $$P\implies Q$$, by carefully picking $$N$$ so that for given $$\epsilon>0$$, $$n>N$$ implies $$|x_n-X_N|<\cfrac {\epsilon}2$$, we have:

$$|x_n-x_m|\le|x_n-x_N|+|x_m-x_N|<\epsilon$$, when both $$n, m>N$$

hence $$(x_n)$$ is Cauchy or Q is true.

• thank you for the answer, looks i've overcomplicated things a lot – roman Dec 20 '18 at 9:20