For larger and larger $a$, $x_n \to \pm \infty$, as $n \to \pm \infty$ respectively. (being a strictly monotone sequence). [in case of $x_{-1}, x_{-2},... ,$ we modify the recursive formula a little bit by defining $x_n=x_{n-1}-(\eta-\epsilon)$]

What are $x_i,x_j$? How do you define the sequence?

So, the sequence becomes $x_n = x_{n-1} + \eta -\epsilon$ for $n \in \mathbb{Z}^+$, $x_0=0$, $x_{n-1}= x_{n}-(\eta -\epsilon)$ for $n \in \mathbb{Z}^- \cup \{0\}$

@Thorgott $x_i , x_j \in I$

$\epsilon$ is a very very small positive quantity (suitably adjusted according to the value of $\eta$ and $\delta$)

I see, then I'd say $x_i,x_j$ is an unfortunate notation. Either way, that $\varepsilon$ looks superfluous to me. Also, I don't follow your limiting process; the sequence $x_n$ is different for each $a$, so which is the final sequence supposed to be?

okay, let me try some more. What if I keep everything same, and let $\eta$ depend on both $a$ and $\delta$

Letting $a \to \infty$ would do, right?

we can get to the $\epsilon$ later

$\eta$ already depends on both $a$ and $\delta$. I think working with increasing intervals is gonna be a lot harder than the suggested approach. A diagonal argument might work, but I'm not sure

@Thorgott why should it be any harder?

Notice that $|x_n-x_{n-1}|<\eta$

for any $n$

Because you construct different sequences for each $a$ and it is not clear what the final sequence should be. If you go by writing $\mathbb{R}$ as union of non-overlapping, compact interval, you can obtain a finite number of terms for each of them and just stitch them together to get the final sequence; this is essentially a one-linear once the idea has been had, so I'd certainly say it's easier to the other approach, which I cannot even complete.

so, $|x_{n-1}-x| \leq |x_n-x_{n-1}| < \eta$, where $x_{n-1} \leq x \leq x_{n}$

@Thorgott that's the only little problem here. The $a$

Yes, except this isn't a little problem, but the crux of the argument.

Oof

That a continuous function on a compact interval can be approximated by usual step functions was already clear.

Is stone-Weierstrass is of any use here?

@Thorgott Now, no two consecutive compact intervals can be completely disjoint [I believe]

Because, we are going to miss out some $x$ between $x_n$ and $x_{n+1}$

That's why I said non-overlapping (meaning having disjoint interior). Covering $\mathbb{R}$ with non-overlapping, compact intervals is possible (see my earlier post). Which value you choose at the end points doesn't matter, so this is of no concern.

Now, I understand why $a$ plays a key role here. Letting $a \to \infty$ would destroy the guarantee that we can find any such $\eta$ at all

[Otherwise, my line of argument will imply that $x^2$ in uniformly continuous on $\mathbb{R}$, which is false]

desmos.com/calculator/iqr5nq5ktf

why do all these desmos graphs have almost the same plot?

You can find an $\eta$ for each $[-a,a]$, but there is not necessarily one for $\lim_{a\rightarrow\infty}[-a,a]=\mathbb{R}$, yes (because $g$ is not necessarily uniformly continuous on $\mathbb{R}$). A limiting argument might still work, but the issue is more subtle. It might work if you construct a sequence for each $a$ and then show that they have a pointwise limit as $a\rightarrow\infty$ (not sure if that's true, but it would suffice).

@Mathphile Because $x^y$ is much larger than $\pm x$ or $\pm y$ when $x$ and $y$ grow large

@Thorgott sort of like sequence of sequences?

all i see

That's what you would get (granted you focus only on numbers of the form $1/n$ for $a$)

@Thorgott ????? wouldn't that be $n$?

True, my bad, was mixing it up with something else I'm working on right now

@Thorgott the tru multitasker

teach me master

:p

Well, I think, for some functions $g(x)$, the final sequence would become ridiculous (according to my intuition). $|x_n -x_{n+1}|$ might become extremely large

or even extremely small

well, I am getting noober day by day

good night. Gotta hit da bed

bb

good night ^^

hi chat

Salut, M Astyx

Tu vas bien ?

Oui, merci, et toi?

Ça va ça va

OK :)

J'ai un examen sur la théorie de Galois mercredi prochain

Ah, bien intéressant ... pas toujours des maths appliquées.

Oui ça change

ça te plaît?

Oui ça va

Mais bon, les études commencent à me fatiguer, j'ai plus beaucoup de motivation

Ça arrive le mois de mai :)

Professor @TedShifrin is your AoPS course over

almost ... one more class and the final (although it remains to be seen if the kids show up for that).

What makes you think they will not show up?

What's in it for them? Not like they've been doing the weekly homeworks, either.

Sounds like none of them are really interested...

They were interested enough in class, but too busy otherwise.

And they all just took the AP test.

Anyhow, I'm done after this.

Done with teaching kids?

I guess so.

:-(