# Problem in understanding a proof of Cauchy's general principle of convergence I am studying Real Analysis on my own. I came across this proof of Cauchy's general principle of convergence. I cannot understand the logic behind the line indicated in blue: "Therefore there exists a natural number $$q>m$$ such that $$|u_q-l|<\frac{\epsilon}{3}$$. Why is $$q$$ greater than $$m$$? Any help would be appreciated.

• The inequality holds for infinitely many values of $q$. How many integers $q$ are less than or equal to $m$? – Kavi Rama Murthy Sep 16 at 11:13

Note that a sequence is a mapping $$f:\Bbb N\to \Bbb R$$, and we write $$x_n:=f(n)$$. Now, for any strictly increasing function $$i:\Bbb N\to \Bbb N$$, the composition map $$f\circ i:\Bbb N\to \Bbb R$$ is called a subsequence and we denote $$x_{i_k}:=f\big(i(k)\big)$$.
Now, $$l$$ is a sub sequential limit of a sequence $$\{x_n\}_{n\in \Bbb N}$$ if there is a subsequence $$\{x_{i_k}\}_{k\in \Bbb N}$$ converging to $$l$$.
Hence, for any fixed positive integer $$m$$ we have $$i(m)+1>m$$ as $$i$$ is strictly increasing.
Also for given any $$\epsilon'>0$$ as $$\lim_{k\to \infty}x_{i_k}=l$$ we have $$k_0\in \Bbb N$$ such that $$k\geq k_0$$ implies $$\big|x_{i_k}-l\big|<\epsilon'$$. Hence, considering $$p:=\max\{k_0,m\}+1$$ and $$q:=i(p)$$ we are done.