Don't understand proof that $x_n \rightarrow A$ $\iff$ every subsequence of $\{x_n\}$ converges to $A$ So we are given that for all $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, we have that $|x_n - A| < \epsilon$ and we want to show that for all $\epsilon' > 0$, there exists an $K$ such that for all $k \geq K$, we have that $|y_k - A| < \epsilon'$ (where $y_k$ is a subsequence of $x_n$).
All the proofs I see write the subsequence in terms of the original sequence, so it would look something like $x_{n_k}$, but to avoid ambiguity, I will write it as $(x_n)_k$ (since $x_{(n_k)}$ tells us nothing about the order the subsequence is in).
So the proofs I read say that we want to find such a $K$ as stated above, but make the (correct) guess of choosing $K = N$. So for all $\epsilon' > 0$, there exists an $N$ such that for all $k \geq N$, we have that $|(x_n)_k - A| < \epsilon'$, but we still need to show that the statement $|(x_n)_k - A| < \epsilon'$ is a true statement. So the proofs say that because $n_k \geq k$, we have that $n_k \geq N$, which I follow. Then they make the logical leap of saying that as a consequence of $n_k \geq N$, this makes the statement $|(x_n)_k - A| < \epsilon'$ true, which I don't follow why. Can someone explain why this is true?
 A: We know that if $n_k$ is a sub sequence of $s_n$, then $k \ge n$. By convergence, there exists $N$ so that $|s_n - A| < \epsilon$ whenever $n \ge N$. But, we have $n_k \ge n$ so that $$n_k \ge n \ge N.$$ Hence, $$|s_{n_k} - A| < \epsilon$$ whenever $n_k \ge N$. Hence, $s_{n_k} \longrightarrow A$.
In response to the comments below:
Consider the sequences $s_n$, and $t_k$ defined by:
$$
s_1 = 1 \\
s_2 = 2 \\
s_3 = 3 \\
s_4 = 4 \\
s_5 = 4 \\
\vdots \\
s_n = 4, \\
$$
and let $t_k = k+1$.
Clearly, when $n \ge 4$, $|s_n - 4 | < \epsilon$. So $N=4$ is an appropriate choice to show that $s_n \longrightarrow 4$. 
Now, when $N=4$, $t_N = 5$. It follows that given $\epsilon > 0$, $$ |s_{t_n} - 4| < \epsilon$$ because $n \ge 4 \Rightarrow t_n \ge 5$.
Furthermore, note that:
$$ (s_n)_{n \ge 4} = (4, 4, 4, \dots),$$
and
$$(s_{n_k})_{n_k \ge 4} = (s_n)_{n \ge 5} = (4, 4, 4, \dots).$$
Does that make more sense?
Intuitively, we can think of a sub-sequence as shifting, and/or deleting some of the terms of $s_n$. The definition of convergence says that there exists a "point of no return" where EVERY term after that point can be made arbitrarily small to $A$. But, by generating a sub-sequence of $s_n$, we have only shifted or removed some terms, so while the "point of no return" may be moved, we are still guaranteed that the point DOES exist, and that every term of $s_{n_k}$ after that point can still be made arbitrarily close to $A$. Thus, $s_{n_k}$ MUST converge to $A$.
A: Assume $x_n \rightarrow A$.
To take a sub-sequence , by definition, you have a strictly increasing function
$\sigma : \mathbb{N} \to \mathbb{N}$ and the subsequence is defined as
$x_{\sigma(n)}$
Note that for every natural number $n$, $\;\sigma(n) \ge n$. So, intuitively, the subsequence is 'converging faster' than $x_n$.
Suppose $\epsilon>0$ is given and $N$ is chosen so that for $n \ge N$, $x_n$ is no farther away from $A$ than $\epsilon$.
Then a fortiori, the same $N$ works for the subsequence $x_{\sigma(n)}$.
