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Lemma 2.6.3 $\implies (x_{n})$ is bounded. So use the Bolzano-Weierstrass Theorem to produce a convergent subsequence $(x_{n_{k}})$ . Set $x= \lim x_{n_{k}}.$ So $(x_{{n_{k}}}) \to x. \quad \Large{(♪)}$. The idea is to show that the original sequence $(x_{n})$ converges to this same limit. Once again, use a triangle inequality argument. $\color{brown}{\text{ We know the terms in the subsequence are getting close to the limit $x$ } }$. $\color{forestgreen}{\text{ $(x_{n})$ is assumed to be Cauchy, so the terms in the tail of $(x_{n})$ are near each other. } \qquad (♫)}$ Thus, make each of these distances less than half of the prescribed $\epsilon$.

1. Where are the 'ideas' of this proof on top from? I cannot predict things, and I don't want to memorize all these steps to remember how to prove this on the exam. Is there any intuition?

Let $\epsilon >0$. Because $(x_{n})$ is Cauchy, there exists $N$ such that $|x_{n}-x_{m}|< e/2$ whenever $m,\ n\geq N$.

By $(♪)$, choose a term in this subsequence, call it $x_{n_{m}}$, with $ n_{m}\geq \color{red}{N}$and $ |x_{n_{m}}-x|<\frac{\epsilon}{2}.$

2. Why $ n_{m}\geq \color{red}{N}$?
$(♪)$ implies that for all $\varepsilon/2 > 0$, there exists $N_1$ such that $n_m \ge N_1 \implies |x_{\Large{n_{m}}} -x|< \varepsilon/2$.
But $N_1$ doesn't relate to $\color{Red}{N}$? Why didn't proof write what I wrote in this box?

$n_{m}$ has the desired property for the original sequence $(x_{n})$, because if $n \ge n_{m}$. then $\begin{align} |x_{n}-x|\ & =\ |x_{n}-x_{\Large{n_{m}}} \; + x_{\Large{n_{m}}} - x| \\ & \leq\ \color{forestgreen}{|x_{n}-x_{\Large{n_{m}}}|}+ \color{brown}{|x_{\Large{n_{m}}}-x|} \\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2} \end{align}$

3. How does the green sentence $\color{forestgreen}{(♫)}$ signify $\color{forestgreen} { |x_{n}-x_{\Large{n_{m}}}| }$ ?

I think gowers.wordpress.com expatiates on this here.

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  • $\begingroup$ 2. They are different $N$s, yes. (3) What?? You add and subtract...(4) $N$ is large. $\endgroup$ – IAmNoOne May 3 '14 at 19:35
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    $\begingroup$ This is extraordinarily hard to understand, because of the the weird coloring, the use of symbols in place of words, and above all the nonstandard English. $\endgroup$ – Potato May 3 '14 at 19:41
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    $\begingroup$ The issues that user Potato points out were described at meta.math.stackexchange.com/questions/12885/… . There is nothing wrong with speaking English as a foreign language - but please use English as simple as possible to express the question, avoid complicated formatting, and avoid lengthy quotes. $\endgroup$ – Carl Mummert May 3 '14 at 19:57
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    $\begingroup$ @Potato what's hard? please write me. $\endgroup$ – Analysis May 3 '14 at 20:15
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    $\begingroup$ @CarlMummert thanks. anything else to better now? i don't like complicated formatting, and lengthy quotes $\endgroup$ – Analysis May 3 '14 at 20:16
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For 1, the intuition really comes from just being comfortable with what the definitions are really saying. The proof can be summed up fairly nicely as

  • if our sequence $(x_n)$ is Cauchy, then (by definiton) the terms in the sequence, the $x_n$, are getting closer and closer to each other,
  • Bolzano-Weierstrass gives a convergent subsequence $(x_{n_k})$, so the Cauchy property of $(x_n)$ (which includes the $x_{n_k}$) means the terms are all getting closer to one another, in particular, to the elements of $(x_{n_k})$ which are getting closer to the limit of the convergent subsequence.

So the terms of $(x_n)$ should be tending to the limit of the $x_{n_k}$ as well. As you've written, the definitions make this all precise, but it's really just a matter of reflecting on what Cauchy/convergent mean: points getting close to each other/a limit.

For 2, it's perhaps a little sloppy to do this if they haven't gone over this before, but you are correct that the $N$ and the $N_1$ do not have to be related. However, the condition is for all $n \geq N$ and for all $n_m\geq N_1$, so set $N' = \max(N,N_1)$ and you can use the same $N'$ for both the original sequence being Cauchy and the subsequence converging.

Why? Any $n\geq N'$ is certainly greater than $N$ by the choice of $N'$, and similarly any $n_m\geq N'$ must be greater than or equal to $N_1$, so the conditions we want hold.

They seem to have glossed this here and just set $N$ to be something large enough, like the $N'$ chosen above. You can always pick a larger $N$ because the condition is for all $n$ larger anyway, so this is a detail that will likely be glossed over for the rest of the book and is good to be comfortable with!

For 3, $(x_n)$ is Cauchy. So, you know that for $n,m\geq N$ we have $|x_n - x_m| < \epsilon/2$. So if $n,n_m\geq N$, the last line follows. But we chose them so that $n \geq n_m \geq N$, and everything works out!

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  • $\begingroup$ thanks. in your 2e bullet, what sequence do you refer to when you write 'the Cauchy property means the terms are all getting closer to one another' $\endgroup$ – Analysis May 14 '14 at 10:44
  • $\begingroup$ and in your 2e paragraph, which limit do you refer to, 'So the sequence should be tending to the limit as well' ? $\endgroup$ – Analysis May 14 '14 at 10:44
  • $\begingroup$ Please apprise me in your answer? Not in comments please. $\endgroup$ – Analysis May 14 '14 at 10:45

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