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12:00 AM
@AkivaWeinberger sorry. Just goofing around. :-)
 
Then, we define $\omega+1$ to be the order type of $\{0,1,2,\dots,\omega\}$, with $\omega$ greater than all of the other elements
And $\omega+2$ is the order type of $\{0,1,2,\dots,\omega,\omega+1\}$
(You can think of it as the orderings $\{0,1,2,\dots\}$ and $\{\omega,\omega+1\}$ stuck together; those have order types $\omega$ and $2$, hence, $\omega+2$)
And we can even continue to $\omega\cdot2=\{0,1,\dots,\omega,\omega+1,\dots\}$
and even to $\omega^2$, and $\omega^\omega$, and $\omega^{\omega^\omega}$, and …
$\omega^{\omega^{\omega^{\dots}}}$, with an infinite power tower, can even be defined. It's called $\epsilon_0$.
Because $\epsilon$ usually means a small number.
In any case, it keeps on going, each one representing another well-ordering.
$\{-1,-1/2,-1/3,\dots,0\}$ from before has order type $\omega+1$.
All of the ordinals I've mentioned so far are countable.
There's a natural ordering on the ordinals; the first uncountable ordinal is the order type of all of the countable ordinals.
The Continuum Hypothesis says that it's equinumerous with $\Bbb R$. This has been proven to be independant of ZFC.
 
@AkivaWeinberger questions regarding the mathematical continuum are best asked to users on sci-fi stack exchange
...or is that the space-time continuum I'm thinking of.
 
12:18 AM
@AkivaWeinberger Good lord...
 
Yeah I do not endorse this
 
You have a fanbase, lol
 
Infinitely many mathematicians walk into a bar. The first says, “I’ll have a beer.” The second says, “I’ll have half a beer.” The third says, “I’ll have a quarter of a beer.” Before anyone else can speak, the barman fills up exactly two glasses of beer and serves them. “Come on, now,” he says to the group, “You guys have got to learn your limits.”
 
Three logicians walk into a bar. The bartender says, "Do you all want a beer?" They respond, "I don't know," "I don't know," "Yes."
 
12:25 AM
@AkivaWeinberger What's equinumerous with $\Bbb R$?
 
The first uncountable ordinal (assuming CH)
Written $\omega_1$
 
An engineer, a physicist and a mathematician are staying in a hotel. The engineer wakes up and smells smoke. He goes out into the hallway and sees a fire, so he fills a trashcan from his room with water and douses the fire. He goes back to bed.

Later, the physicist wakes up and smells smoke. He opens his door and sees a fire in the hallway. He walks down the hall to a fire hose and after calculating the flame velocity, distance, water pressure, trajectory, etc. extinguishes the fire with the minimum amount of water and energy needed.
@AkivaWeinberger I didn't
 
@AkivaWeinberger Is there anything special about $\omega_2$?
 
@AkivaWeinberger nvm, I thought you meant $\omega_0$ for some reason
 
lol @TheGreatDuck
 
12:27 AM
It's consistent with ZFC that it's equinumerous with $\Bbb R$ instead, I think
 
or just $\omega$
 
@Brody I thought $\omega=\omega_0$?
 
I think it's been proven that ZFC proves that $|\Bbb R|\le\aleph_{\omega_4}$
That's the cardinality of $\omega_{\omega_4}$.
 
lol
That's, um... ok then
But is there anything interesting about $\omega_2$?
 
@SimplyBeautifulArt right
 
12:29 AM
Wait, no, I stated it wrong
 
1:03 AM
Does Well-Ordering Theorem mean that all sets can become well-ordered or only ordered sets?
 
1:18 AM
@Jasch1 It means all sets can be well-ordered. (It doesn't matter if we're talking about ordered sets, because we're giving it a new order anyway.)
 
ok
 
Also, you need the axiom of choice for it.
Fun fact: Without the axiom of choice, you can't prove that $P(\Bbb R)$ (the power set of the reals) can be ordered at all, much less be given a well-order.
 
Does the set of perfect squares have the same density as set of naturals?
Sorry for so many questions, I'm just looking back on a paper and I'm confused about some stuff I wrote
 
1:37 AM
@Jasch1 How many perfect squares in the first 100 positive integers? How many up to 200 and so on?
 
i get that the size of the sets are the same
 
Sorry, you mean natural density or something else?
 
natural/asymptotic density
 
The cardinality of the continuum can be pushed to basically any cardinal. The only obstruction is cofinality @AkivaWeinberger
 
Oh.
I remember there being some theorem involving $\aleph_{\omega_4}$, though
 
1:42 AM
yah, I think I have seen what you are talking about, but I think there are other things going on and I don't remember what it is
 
Do you know the density of $\Bbb N$? @Jasch1
You're not wrong.
 
Im not?
 
What proportion of the first $n$ positive integers are positive integers? And the limit is obvious. :)
 
Oh, OK: If $\kappa<\aleph_{\omega}$ implies that $P(\kappa)<\aleph_{\omega}$, then $2^{\aleph_{\omega}}<\aleph_{\omega_4}$
 
1.Let f : [0, 1] → R satisfy |f(x) − f(y)| ≤6 |x − y| for all x, y ∈ [0, 1]. Show that f is continuous and that for all ε > 0, there exists a piece-wise constant function g such that sup x∈[0,1] |f(x) − g(x)| <6 ε.
2.For all integers n > 1, let un = $$∫^1_0$$f(t) cos(nt)dt. Show that the sequence (un) converges to 0.
 
1:45 AM
That first bit is independent of ZFC, though.
 
I have problem with part 2
any hints?
 
$u_n=\int_0^1f(t)\cos(nt)dt$?
And we want to show $u_n$ goes to zero?
 
yep
 
I assume $f$ has to satisfy all of the conditions of the first one?
 
yep
 
1:47 AM
Ah yah. It is sort of funny you can have the continuum be $\aleph_1$, $\aleph_{\omega_1}$ but you cant have it $\aleph_\omega$
 
Hm. What's $\int_0^1\cos(nt)dt$? It might be relevant
Like, without the $f$
 
I tried this @AkivaWeinberger
$$Let \(\epsilon>0,t\in[0,1],n>1\)
\[|\cos(nt)|\le 1\\\Rightarrow|f(t)\cos(nt)|\le |f(t)|\\\Rightarrow \int_0^1|f(t)\cos(nt)dt\le\int_0^1|f(t)|dt\\ \Rightarrow \left|\int_0^1 f(t)\cos(nt)dt\right|\le\int_0^1|f(t)|dt\]$$
 
apply Cheb. polynomial and integrate?
 
$\sin(n)$ @AkivaWeinberger
 
So $\int_0^1\cos(nt)dt=\sin(n)/n$, which goes to $0$ as $n$ goes to infinity.
@aquire Off by a factor of $n$
 
1:50 AM
i see
 
Hm. What if $f$ is positive everywhere?
 
Thanks @AkivaWeinberger
 
I mean, it probably isn't, but what if
Call its maximum $M$. Then $0\le\int_0^1f(t)\cos(nt)dt\le\int_0^1M\cos(nt)dt=M\frac{\sin(n)}n$?
It has a maximum because it's continuous and defined on $[0,1]$
Oh, wait, that doesn't work
Sorry
The middle inequality doesn't have to hold.
 
Is how to generate a satisfying tuple from an index (1 giving the lexicogrraphically least tuple, 2 the next, etc) a math.se question? Or would it be more cs?
 
$\cos(nt)$ is sometimes negative, so $f(t)\le M$ doesn't imply $f(t)\cos(nt)\le M\cos(nt)$
 
1:54 AM
I try to teach my stuff to a yellow duck on my desk made of plush. Maybe, just maybe, you should do the same?
 
So, never mind @aquire
 
Ah finally found the theorem, Easton's theorem @AkivaWeinberger
 
we can deal with absolte values @AkivaWeinberger
right?
 
Ooh, probably
But we might need to know what $\int_0^1|\cos(nt)|dt$ is.
I doubt that that goes to $0$.
 
you are right
 
1:58 AM
It's definitely less than $1$, at least
 
yep
 
Just throwing ideas out there — maybe see what happens at $f(t)=t$?
Like, what's $\int_0^1t\cos(nt)dt$
Oh, never mind, it's complicated
 
What's the simplest example of $f:A\to B,\; A^* \subset A$ such that $f(A)\setminus f(A^*)\ne f(A\setminus A^*)$?
 
$\frac{\sin n}n-\frac{1-\cos n}n^2$
@Brody $A^*$ is any subset?
 
@AkivaWeinberger Yeah.
 
2:03 AM
may be taylor expansion of cos might help
 
hmm, I have one that's simple but it doesn't involve $A^*$ empty
 
$f:\Bbb R\to\Bbb R,~x\mapsto x^2$, with $A=\Bbb R$ and $B=\Bbb R_{\le 0}$
$f(A)\setminus f(A^*)=\emptyset$, but $f(A\setminus A^*)=\Bbb R$.
I think we need to use the conclusion to part 1 @aquire
About the piecewise function
 
makes sense
 
Oouu, $\Bbb R_{\le 0}$ is nice notation
 
It's the same as $(-\infty,0]$
just like $\Bbb R$ is the same as $(-\infty,\infty)$
 
2:06 AM
Yeah, it looks better than some alternatives
imo
 
we use $\mathbb{R}^-_0$
 
e.g. apparently $\Bbb R^+, \Bbb R_0^+$ can respectively mean $(0,\infty), [0,\infty)$ or vice-versa depending on the text/region/etc.
 
Where I wrote $B=\Bbb R_{\le0}$ I meant $A^*=\Bbb R_{\le0}$.
 
that terribly confused me for a sec but I had noted :P
and the image of the set difference is a typo too
 
It should be $\Bbb R_{>0}$, shouldn't it
 
2:13 AM
that
Are you going for the greatest disparity, so to speak?
 
The what?
 
I am very intimidated by the math speak here.
I like it.
 
By simplest I was thinking simple (co)domain or something like that, e.g. $f:{0,1}\to{0};x\mapsto 0$ and consider the subset ${0}$. The disparity is between $\emptyset$ and ${0}$.
 
Oh, the constant function, derp
 
as opposed to say between $\emptyset$ and $\Bbb R$, which though (no more or less) unequal than the above are indeed the most unequal in some sense @Akiva
 
2:18 AM
That's a good example
 
domain $\{0,1\}$, latex derp
 
How about the function from robbers to the banks they rob?
The function on everyone but Robert isn't necessarily the same as all the banks but the one Robert robs
say, if someone else also robs the same bank.
 
barring that a robber may rob multiple places
 
Stopping Rob won't necessarily save his bank, essentially.
@Brody Assume spherical cows
 
I don't know how common that kind of serial robbery is irl. the cows perhaps needn't be that round
oh, and "Robert". lol
 
2:24 AM
Little Bobby Tables, we call him
 
quite an unfortunate path he's gone down D:
 
$$
We can say ${-1\over n}\le\int_0^1 \cos nt dt\le {1\over n}\\\Rightarrow \left|\int_0^1 \cos nt dt\right|\le{1\over n}$
That will do Thanks a lot @AkivaWeinberger
 
3:22 AM
TIL my nose is a wave.
Latin is weird.
 
Is it possible to have an endofunction on $\Bbb R$ such that a subset and its complement both have images $\Bbb R$? Don't think so but could be wrong...
 
What's an endofunction @Brody
 
I just mean $\Bbb R\to \Bbb R$
 
Is it required to be continuous?
 
no restrictions
 
3:31 AM
Oh, $x\sin x$. Positives and negatives.
Doesn't matter which one you put $0$ into.
Actually, any cut into $(-\infty,a]$ and $(a,\infty)$ would work.
(And same for switching open and closed.)
 
nice, thanks
this is an example where $f[A] \setminus f[A^*]=\emptyset$ and $f[A\setminus A^*]=\Bbb R$
 
4:10 AM
hi chat
 
@Akiva So you have what's called an "endomorphism"
Which is a morphism from an object to itself
If you're in the category of sets with your morphisms being functions, then you get an endofunction
But you could also make that case for a homomorphism from a group to itself, continuous function, and so forth, depending on the type of object
 
What are the sine curves that pass through $(2,\sin2)$, $(8,\sin 8)$, and $(14,\sin14)$ other than $y=\sin x$
 

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