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I am a little confused about the following concept from Real Analysis:

suppose $\{a_n\}$ is a sequence in $\mathbb{R}$ which is Cauchy. then it is convergent, and it is shown as follows:

first we show it is bounded (this I understand how to do). then, from the Bolzano-Weierstrass theorem(every bounded sequence of real numbers contains a convergent subsequence) we deduce that there exists a convergent subsequence $\{a_{n_k}\}$ with limit $a$ (say).

we now claim that $\lim_{n \to \infty} a_n = a$.

what we need to show then is that for any given $\varepsilon > 0$ we can find $N > 0$ such that $$\tag{1} n \geq N \Rightarrow |a_n - a| < \varepsilon\,. $$ by the triangle inequality we know that the following inequality is true, for all $n, n_k \in \mathbb{N}$: $$\tag{2} |a_n - a| \leq |a_n - a_{n_k}| + |a_{n_k} - a| \,. $$ i know that the usual argument is now to say we control the first term using the Cauchy property of $\{a_n\}$ and the second term using the convergence of $\{a_{n_k}\}$ to $a$.

and here is where I am struggeling: why are we allowed to constrain the $a_{n_k}$? this is not clear to me, given the statement ${(1)}$. it looks like we have to make a $second$ choice, namely for the right hand side of ${(2)}$ to be $< \varepsilon$, the $n_k$ must be large also.

in particular, this choice cannot (at least as far as i understand) be resolved in a straightforward way (such as simply taking the larger of two constants) because the subsequence is a sequecen indexed by the subscript $k$, not the subscript $n$ any more - so choosing $K$ so that $|a_{n_k} - a| < \varepsilon/2$ whenever $k \geq K$ cannot be combined with choosing $N$ so that $|a_n - a_{n_k}| < \varepsilon/2$ whenever $n \geq N$.

I added this last paragraph because I sense this is where my misunderstanding is coming from - what am I missing? Thanks a lot for your help!!

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The basic idea is that to pick an $N$ for which $|a_m - a| \leq \epsilon$ for all $m \geq N$, you

  1. Use the convergence of a subsequnce to find a $k$ such that $|a_{n_m}-a| \leq \epsilon$ for all $m \geq k$

  2. Use the cauchy sequence property to find an $l$ such that $|a_p - a_q| \leq \epsilon$ for all $p,q \geq l$

  3. Pick $K$ such that $K \geq k$ and $n_K \geq l$, and let $N = n_K$.

You then have for an arbitrary $m \geq N$ $$ |a - a_m| = |a + (a_N - a_m) - a_N| \leq |a - a_N| + |a_N - a_m| \leq \underbrace{|a - a_{n_K}|}_{\leq \epsilon\text{ by (1)}} + \underbrace{|a_N - a_m|}_{\leq \epsilon\text{ by (2)}} \leq 2\epsilon $$

Now, for cleanliness, you might want to go back and replace $\epsilon$ with $\frac{\epsilon}{2}$ everywhere ;-)

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You can make the two choices consistently.

First choose $N$ large enough so that $|a_n - a_m|< \epsilon/2$ for all $n,m \ge N$. Then choose $k$ such that $n_k \ge N$ and $|a_{n_k} - a| < \epsilon/2$. This can be done because any sufficiently large $k$ has both of these properties. Then letting $m = n_k$, the right hand side of your inequality (2) is $< \epsilon$.

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The idea, if you want $|a_n - a| < \epsilon$, is to choose an $N_1$ such that for any $n_k > N_1$, $|a_{n_k} - a| < \frac{\epsilon}{2}$. This can be done by convergence.

Separately, you want to find an $N_2$ such that for any two $n, m > N_2$, $|a_n - a_m| < \frac{\epsilon}{2}$. This can be done by Cauchy. If you now choose $n_k, n > \max(N_1, N_2)$, then by the triangle inequality (letting $a_{n_k}$ take the place of $a_m$), $$ |a_n - a| \leq |a_n - a_{n_k}| + |a_{n_k} - a| < |a_n - a_{n_k}| + \frac{\epsilon}{2} < 2\frac{\epsilon}{2} = \epsilon $$ where the first strict inequality is valid since $n_k > N_1$, and the second one is valid since $n, n_k > N_2$.

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