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17:00
@TheGreatDuck it looks like his question was answered adequately in his opinion.
So who cares?
Don't harass the poor guy.
one possibility is that the two intervals don't overlap at all. that'll occur when $x<0$ or $x-1>2$.
@AlessandroCodenotti must every cofinality be $\omega_\alpha$ or $1$ or $0$?
and in that case the integral is just plain zero. so the integral is only interesting for $x\in(0,3)$.
Another case is that the interval $[x-1,x]$ lies entirely within $[0,2]$. That'll occur if $0<x-1$ and $x<2$ i.e. $x\in (1,2)$.
@Dodsy I'm not. It is common practice to share posts one spots as still needing deletion and close votes...
What about $x\in(0,1)$ or $x\in (2,3)$ ?
17:02
well
@AkivaWeinberger So ok this would be possible that the whole thing is negative.
@AkivaWeinberger actually every infinite subset of $\omega$ is cofinal
it will be partly in and partly out :P
@TheGreatDuck "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level."

"If you do not add context within the next several hours, we will be forced to close and delete your question."

"Are you going to respond?"
Right.
Let's move this to another room, actually, so that we don't interrupt other conversation.
17:03
yeah... I'm warning them that the post needs to be fixed.
not sure how to do that but okay
@LeakyNun no. A poset with $n$ maximal elements and no infinite ascending chain will have cofinality $n$
actually, I'm trying to drop it, but you guys were still bringing it up...
It should be true for cardinals though
@AlessandroCodenotti ordinals?
@AkivaWeinberger The real question behind this is that with this relation i should be able to say something about infimum, supremum, minimum, maximum of U
You'll never believe how easy it is to find questions needing closure. Just go through the list of users with only 2-6 reputation.
That's called lhf
I don't get why that would be a particular past-time for you, Duck.
it is not
17:08
Low Hanging Fruit
i just got curious to see how many questions are needing closure
@TheGreatDuck "Are you going to respond?" It could be that he hasn't seen that post or the post before. Many people are not on the computer 24/7.
@Dodsy why? Because all my posts need to be deleted?
Let him see it when he sees the automatic notification.
Relax guys
17:09
@TheGreatDuck Sorry, not sure what you're talking about?
oh nothing.
@Dodsy (pastime)
it's just that 90% of my posts were written knowing that they were duplicates and asking them anyways.
:p
@AkivaWeinberger >:E
(From the word "pass" rather than "past")
oh
@AlessandroCodenotti why?
I guess that makes sense.
I always thought it was past-time, silly me.
I don't know why there's only one 's'
I thought it was 'pass-time' until I looked it up
17:12
Because they're linearly ordered, but it requires some justification
@AlessandroCodenotti also, how to prove that every countable limit ordinal has cofinality $\omega$?
But it does come from the word 'pass'
English does weird things.
@LeakyNun they are cofinal in themselves
So you need to prove that they don't have a finite cofinal subset
@AlessandroCodenotti Yeah but it could conceivably be a larger ordinal of the same cardinality
I mean, that never happens, but why
(We'd require the function to be increasing, I guess)
17:14
Because cofinality is defined as "the cardinality of the smallest..."
@AlessandroCodenotti what if we define it as the order-type of the smallest...?
I don't think it makes a difference, countable limit ordinals are limits of sequences of length $\omega$ and I don't see why it wouldn't work for higher ones
@AlessandroCodenotti how to prove the former statement?
I know it's intuitively obvious, but I don't know how to prove it.
17:30
Hm, I feel there should be an obvious proof but I can't come up with it right now
I'll blame the fact I'm cooking dinner in the meantime :P
Huh, the smallest ordinals of each cardinalities are equal to $\omega_\alpha$ for some $\alpha$, except for the finite ones.
The cardinalities go $0$, $1$, $2$, $\dots$, $\aleph_0$, $\aleph_1$, $\dots$, $\aleph_\omega$, $\dots$
@AkivaWeinberger you're never gonna list them all out this way; they are unlistable :p
I'm trying to remember a property/method of proving something in the measure theory subject. I think there something along the lines of 'if I want to prove something for the borel sigma algebra, I only need to prove for those sets(open intervals, or close intervals) which will create the sigma algebra through countable intersection, union and complementar op
erations
17:44
Like a sort of method of induction
Uh huh.
That posit you just removed.
hmm, posit?
17:47
@Anoldmaninthesea. those generate the $\sigma$ algebra
If your property is preserved by complements and intersection you're done
perhaps you were thinking of financial math.
@AlessandroCodenotti yes that's it
@theDoctor You posted "Posit: negative numbers ... [insert pre-17th century view of negative numbers here]"
where can I find a formal statement of that fact
?
Peano arithmetic doesn't include negative number, yes?
17:49
@theDoctor yes
my minds a little unsettled about the negation function vs. an implicit subtraction symbol.
@theDoctor Well, no. Peano is all about $\mathbb{N}$ (not including $0$ for any frenchies).
is there a binding order (or preference) to the use of a negation symbol "-" with an exponent? -x^2 -- which is grouped: the square to the symbol "x" or the negation?
@theDoctor -x^2 = -(x^2)
@theDoctor subtraction is not defined for all pairs of elements in Peano arithmetic
-3^2 = -9
17:52
@theDoctor PEMDAS. Exponent before Minus.
@AlessandroCodenotti where can I find a formal statement of that fact
?
-1^2 = -1 is an interesting one.
not really interesting
-3 = -1*3
17:54
It's perhaps interesting because $1^2 = 1$
I think this is an arena for (WHAT I CALL) domain theory.
The minus doesn't do anything interesting, though.
something is transcending domains implicitly. The exponent operator is NOT to be part of $bb \
$bb \Z$.
@SteamyRoot cries
@theDoctor why not?
17:55
for the very f'n reason why those equations don't look right.
@Astyx I know, I know. Here in Belgium we also learn that $0 \in \mathbb{N}$.
Then we share this pain
@theDoctor I'm not getting what you mean.
What do you mean by "the exponent operator is not to be part of $\Bbb Z$"?
17:56
It's a foreign symbol that didn't evolve with basic math.
It only gets worse once you have to do research where $0 \notin \mathbb{N}$, and teaching, where $0 \in \mathbb{N}$.
Physics started inserting it.
@theDoctor it may not have originated from math itself, but I'm very certain that the operator is studied extensively throughout the centuries
$x \mapsto e^x$ is invariant under differentiation, as an example.
@theDoctor wut
Sanity check : If $X$ and $Y$ are independant random variables, then $E(XY) = E(X)E(Y)$ right ?
17:57
The notation of putting an exponent at the top-right of another thing was started by Descartes.
@SteamyRoot wow, how do you even know that
yes @Astyx
I think zero can be considered a natural number, like the inferred subject in some sentences.
newton was a physicist
.
@theDoctor it is widely considered a natural number
@theDoctor descartes isn't newton
Let's just make it so.
17:59
-3^2 = -1(3^2)
@SteamyRoot Really?
where did Descartes talk about exp?
I was going to ask for source; thanks!
@Liad Thanks. Do you know how I can prove that if $X = \sum a_n X_n$ where the $X_n$ are Rademacher variables, then $E(e^{tX}) \le e^{t^2/2}$ for all $t\in \Bbb R$ ?

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