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First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, $\{a_{n_{k}}\}$ that converges to a limit, say $L$. Now we use this to prove that all Cauchy sequences are convergent.

So an $N_1$ exists such that $$\left|a_{n_{k}}-L\right|<\frac{\epsilon}{2}$$ for all $ k > N_1$, and an $N_2$ exists such that $$\left|a_m-a_n\right|<\frac{\epsilon}{2}$$ for all $n,m > N_2 $.

Pick any $k > N_1$ such that $n_k > N_2$. Then for every $n > N_2$, $$\left| a_n - L \right | \leq \left| a_n - a_{n_{k}} \right| + \left| a_{n_{k}} - L \right| < \epsilon/2 + \epsilon/2 = \epsilon$$.

So $$\lim_{n \rightarrow \infty}a_n = L$$

I'm fine with this proof until the last part - I'm confused as to why we can pick an abitrary $k$ like we do? Should the limit not depend only on $n$? Now it appears like it also depends on $k$ and if we pick a $k < N_1$, the inequality isn't true. Can someone clarify this for me please?

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    $\begingroup$ The basic part of the definition of subsequence $\{a_{n_{k}}\}$ is that $n_{k} \to \infty$ as $k \to \infty$ and hence it is possible to choose $k$ such that $k > N_{1}$ and at the same time $n_{k} > N_{2}$. Also note that $n$ is not dependent on $k$, we just need to have $n > N_{2}$. Only $n_{k}$ is dependent on $k$ and we only need to choose a suitable value of $n_{k}$ such that $n_{k} > N_{2}$. Don't mix $n$ with $n_{k}$. $\endgroup$ – Paramanand Singh Jul 7 '13 at 5:09
  • $\begingroup$ Yeah I get that, but why can we choose a value of k? $\endgroup$ – Matthew Hampsey Jul 7 '13 at 5:24
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At that point in the proof you’re trying to show that if $n>N_2$, then $|a_n-L|<\epsilon$. If you can find a real number $x$ such that $|a_n-x|<\frac{\epsilon}2$ and $|x-L|<\frac{\epsilon}2$, the triangle inequality will give you the desired result, so the proof boils down to finding such an $x$.

  1. What things do we know are close to $L$? Terms $a_{n_k}$ of the subsequence, provided that $k$ is sufficiently large.
  2. What things do we know are close to $a_n$? Terms $a_m$ of the original sequence, provided that $m$ and $n$ are sufficiently large.

We take care of (1) first: there is an $N_1$ such that $|a_{n_k}-L|<\frac{\epsilon}2$ whenever $k>N_1$. This means that we can take our $x$ to be any $a_{n_k}$ with $k>N_1$, and we’ll have $|x-L|<\frac{\epsilon}2$.

Then we take care of (2): there is an $N_2$ such that $|a_n-a_m|<\frac{\epsilon}2$ whenever $m,n>N_2$. This means that since we’ve already specified that $n>N_2$, we can take our $x$ to be any $a_m$ with $m>N_2$, and we’ll have $|a_n-x|<\frac{\epsilon}2$.

Can we combine the two requirements? Is there an $a_{n_k}$ with $k>N_1$ that is also an $a_m$ with $m>N_2$? Equivalently, is there an $a_{n_k}$ with $k>N_1$ such that $n_k>N_2$? Sure: the sequence $\langle n_k:k\in\Bbb N\rangle$ is unbounded, so its tail $\langle n_k:k>N_1\rangle$ is also unbounded and contains a term $n_k>N_2$. Thus, we can set $x=a_{n_k}$ and satisfy both requirements, so that we have

$$|a_n-L|\le|a_n-x|+|x-L|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

as desired. Note that this calculation doesn’t actually depend on a specific value of $k$: we could set $x=a_{n_\ell}$ for any $\ell\ge k$, the the calculation would be the same. As I said at the beginning, we’re really just trying to find one number $x$ that we can use to ‘tie’ $a_n$ to $L$ to within $\epsilon$; it turns out that there are lots of them, and it doesn’t matter which one we use.

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  • $\begingroup$ That was exceedingly clear, and I see what my misunderstanding was. Thank you so much! $\endgroup$ – Matthew Hampsey Jul 7 '13 at 6:40
  • $\begingroup$ @user46080: You’re welcome, and thank you. $\endgroup$ – Brian M. Scott Jul 7 '13 at 6:41
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We can choose a value of $k$ because we know that such an $N_1$ exists. This is an infinite sequence, so provided that we can choose the $N_1$ with the desired properties, we must simply choose $k \ge N_1 + 1$.


A good way of thinking about this is somewhat in reverse.

Suppose we couldn't pick any arbitrary $k$ greater than $N_1$? What would that mean?

It would mean that either a.) $N_1$ is the maximum of some set, or b.) $N_1$ doesn't exist.

Can you rule out both of these scenarios? (Hint: by definition of a Cauchy sequence, they are both ruled out).

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  • $\begingroup$ Cool, thanks!!! $\endgroup$ – Matthew Hampsey Jul 7 '13 at 5:46
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The real problem is to understand the logical construction of the real numbers from Dedekind cuts. This is where the LUBP(least upper bound property) comes from, i.e. $\mathbb{R}$ is order-complete.

The LUBP easily implies that every monotonic bounded sequence converges.

From this $\mathbb{R}$ has the finite-intersection property, i.e. If $[a_n,b_n]_{n\in {\mathbb N}$ is a nested sequence of closed bounded intervals then its intersection is a closed bounded interval or a single point.

From any Cauchy sequence $(x_n)$ you can systematically define a nested sequence of closed bounded intervals $(I_n)$ with $x_n \in I_n$ and such that the lengths of the intervals goes to zero. Thus, the intersection of all such intervals is $\{x\}$ for some $x$ by the finite intersection property and this $x$ must be the limit of the Cauchy sequence.

If you don't go through an axiomatic justification of our number systems in a Discrete math class and don't see it in real analysis, your math program is fundamentally flawed. When you work hard to understand these formal logical underpinnings, all of analysis becomes much easier.

The above approach works for any metric space, i.e. a metric space is complete if and only if it has the appropriately defined version of the finite-intersection property which again is just that every Cauchy sequence is convergent. It just happens that R has a natural linear order but this is not important for metric completeness but requires a little help from general topology to guarantee that arbitrary intersections of closed sets are closed because of the general definition of an open set in a topological space.

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