Corollary of Cauchy Criterion for Convergence

Question

I encountered a corollary of the Cauchy Criterion for convergence that reads:

A sequence $(x_n)_{n=1}^{\infty}$ in $R$ diverges if and only if $\exists$ $\epsilon_0 > 0$ $\exists$ $(n_{k})_{k=1}^{\infty}, (m_{k})_{k=1}^{\infty}$ sequences in $N$ such that: $(i)$ $n_1 < n_2 < n_3 ...$ $(ii)$ $m_1 < m_2 < m_3 ...$ $(iii)$ $|x_{m_k}-x_{n_k}| \ge \epsilon_0$ $ \forall$ $ k\in$ $N$

I'm having trouble understanding how this follows from Cauchy's criterion for convergence which states:

A sequence in $R$ converges if and only if it is a Cauchy Sequence

Answer 1

Cauchy says: Convergence $\iff \forall \epsilon > 0 \; \exists N\; \forall i,j > N \; |x_i - x_j| < \epsilon$

So since "$\sim P \iff \sim Q$" is logically equivalent to "P $\iff$ Q", you get

Divergence $\iff \sim$Convergence $\iff \sim (\forall \epsilon > 0 \; \exists N\; \forall i,j > N \; |x_i - x_j| < \epsilon)$

$\iff \exists \epsilon > 0 \; \forall N \; \exists i,j > N \; |x_i - x_j| \ge \epsilon$

So now just show that $\forall N \; \exists i,j > N \; |x_i - x_j| \ge \epsilon$ is the same as the existence of increasing sequences as in the original statement.