« first day (5260 days earlier)      last day (56 days later) » 

leslie townes
leslie townes
00:50
@psie continuity is necessary for existence. maybe as a (hopefully straightforward) exercise, prove that any linear functional with a rule given by the form f mapsto <f, g>, with g in L^2, will be continuous (when the domain is given the norm topology of H, which is the implicit notion of 'continuous' here), so it would not be possible to represent other linear functionals in this way
for reasons why the assumption of continuity is automatic in the finite dimensional case, and why it is not vacuous in the infinite dimensional case, see literally any functional analysis textbook (neither of these phenomena is what i would regard as an elementary consequence of any relevant set of definitions)
 
2 hours later…
pie
pie
03:23
I was going to write a question about "Given side lengths of n sided polygon $ l_1, \ l_2, \dots, \ l_n$, What is an upper bound for its area?" but found this 11 year old question without answers, should I ask a new question? if I did would it get answers? or maybe take this question to MO? what do you guys think
5
Q: Upper bound for area of polygons

Taladrisis there a formula for an upper bound for the area of a polygon, knowing the length of its edges? In the ideal situation, the answer would be a function $f_n(l_1,l_2,\dots,l_n)$ of the edges lengths (the number $n$ of edges is fixed). In a less ideal situation, the answer would be a function $g...

geometry estimation area
Maybe it is an open problem? Since the question didn't get any answers, so the question is very hard, if so how can I search for it?
 
1 hour later…
Jakobian
Jakobian
04:37
How can anyone act with accordance to their feelings when the world simply doesn't want it that way
@XanderHenderson Xander mad they didn't let him operate in a hospital based on his credentials
@pie you can be the first
leslie townes
leslie townes
05:15
pie: i wouldn't assume anything about the difficulty of the question from the fact that it doesn't have answers. a lot of non-open questions posted to MSE don't have answers, and a lot of unsolved problems posted on MSE nevertheless have answers providing partial solutions and commentary. in this case someone left a comment saying it "seems" to be hard for cyclic polygons, but they don't cite anything.
i don't know if there is any procedure that the network intends to be used simply to 'bump' an old, unanswered question that you would like people to reconsider (i can think of ways of doing this, but they may not be ways that are supposed to be used for this purpose). if you did post a new question i would, at a minimum, include a cite to the previous question and include any other context you can think of, e.g. any special cases you have solved or at least have observations about
or any links you have to more information that are not included in that post.
pie
pie
@Jakobian That is the kind of motivation I need, I will prove the RH but not being able to reed my proof lol
@leslietownes Do you think this question would be good for MO?
I asked similar question on MO a long time ago
8
Q: What are the $\inf$ and $\sup$ of the area of quadrilateral given its sides length?

pieI asked this question on MSE here. Given a quadrilateral with side lengths $a,b,c$ and $d$ (listed in order around the perimeter), t's known that the area, is always less than or equal to $\frac{(a+b+c+d)^2}{16}$ i.e This establishes an upper bound for any quadrilateral with these side lengths....

mg.metric-geometry euclidean-geometry
leslie townes
leslie townes
05:30
pie: i dunno about MO vs. MSE [that MO post is something else i would cite as context if you were to ask it in either forum]. there is a kind of tradeoff, the average MO reader might be more knowledgeable of what is known or easy to prove than the MSE readership, but the average MO reader also might be less likely to respond with partial information
Jakobian
Jakobian
05:42
and also the average MO user will scold you for posting known facts on their site
so unless you feel educated in a topic, I'd advise against using that site
its for researchers first and foremost
@pie I'm surprised such elementary looking question survived there without any scolding
you should also let @XanderHenderson explain you crossposting
as far as I can see, you've asked your question on math.se, and then just two days later you asked on mathoverflow
nickbros123
nickbros123
06:29
guess ill be doing probability theory in the summer
Jakobian
Jakobian
@nickbros123 that's nice. Advanced probability is something that I have a secret desire to explore
nickbros123
nickbros123
06:49
Nice. Probability I think is a very neat application of analysis theorens
Jakobian
Jakobian
I don't know why you would make such statements
please learn what probability theory is first
nickbros123
nickbros123
07:06
@Jakobian it's an opinion? Based on my experience with basic probability?
I'm not limiting probability to just applied analysis, of course
Jakobian
Jakobian
07:24
@nickbros123 In case you didn't understand, I am saying that you should first learn what probability is before making such opinions. They are borderline offensive
usually reducing one field to being just an application of another field is offensive, and in this case this is especially true since probability theory is incredibly rich
nickbros123
nickbros123
07:38
I wasn't, but ok
psie
psie
08:37
@leslietownes thanks leslie :)
psie
psie
09:07
Folland calls a Borel measure $\nu$ on $\mathbb R^n$ regular if
(i) $\nu(K)<\infty$ for every compact $K$;
(ii) $\nu(E)=\inf\{\nu(U):U\text{ open},E\subset U\}$ for every $E\in\mathcal{B}_{\mathbb{R}^n}$
He mentions that condition (ii) is implied by condition (i). I'm wondering; could we simply replace $\mathbb R^n$ with an arbitrary metric space $(A,d)$? Would the implication still hold?
Jakobian
Jakobian
09:31
(i) is called locally finite, and (ii) is called outer regular
@psie here is a counter-example
sorry no this example is outer regular but not inner regular
Jakobian
Jakobian
09:57
user image
oddly enough, this is a very similar example but its locally finite but not outer regular
but for some positive results, it should be true that any locally finite measure on a Polish space that is a complete separable metric space, is regular
this example doesn't have that - the space is complete, but not separable
mo-_-
mo-_-
10:21
hello
Jakobian
Jakobian
10:43
@nickbros123 well, I took it that way, and I'm not sure how to interpret how you said differently. So forgive me if I was mistaken, since I'm not sure what you mean otherwise
either way I hope my intentions were understood
@mo-_- hi
Pizza
Pizza
@SineoftheTime anyway I asked in the end it was how do I define θ = 0
Sine of the Time
Sine of the Time
my explanation was ok?
Pizza
Pizza
yes I understand
one potato two potato
one potato two potato
11:26
@ModularMindset where does this come from?
 
1 hour later…
nickbros123
nickbros123
12:55
@Jakobian I can understand how it could come off as offensive, but my intention was not to disrespect. I meant to address the parts of probability that derive from analysis, and wanted to say that I find these things neat.
ILikeMathematics
ILikeMathematics
user image
user image
But there is a problem.
We need to assume $A^{\operatorname{tr}} \in O_n(G, K)$ for that to work
nickbros123
nickbros123
what is G here? some arbitrary matrix?
ILikeMathematics
ILikeMathematics
@nickbros123 It's the Gram matrix of a bilinear, non-degenerate $\beta$
non-degenerate $\implies$ $\det(G) \neq 0$ $\implies$ invertible, that's why we're allowed to do that
Jakobian
Jakobian
@Thorgott I've tweaked results that I've obtained to be a bit more general. First, say that $X$ has property $D$ if every closed discrete subset can be separated from any disjoint closed set. If $X$ has property $D$ and $x\in \beta X\setminus X$, then $x$ cannot be approach by sequences in $X$
this generalizes the assumption of normality, pseudonormality and countable compactness, under one simple definition
property $D$ does exist in literature, although not in this form and more rarely (in a form equivalent for first countable spaces)
other than that I left everything unchanged
Note that there exist spaces $X$ for which there exists $x_n\in X$ and $x\in \beta X$ such that $x_n\to x$
(deleted) Tychonoff plank is one such example
nickbros123
nickbros123
13:25
@ILikeMathematics K is an arbitrary field?
Jakobian
Jakobian
I think I have an idea of an example of a path-connected, locally compact space such that there is $x\in \beta X\setminus X$ and $x_n\in X$ with $x_n\to x$, but $\beta X$ is not path-connected.
psie
psie
Let $[c,d]$ be a compact interval and $F$ increasing and right-continuous. Let $((x_i,y_i))_{i\in\{1,\ldots,N\}}$ be a finite cover of $[c,d]$. Is it true that $$F(d)-F(c)\leq\sum_1^N(F(y_i)-F(x_i))?$$Appreciate any help on how to prove this. I'm not really sure where to start.
Jakobian
Jakobian
user image
basically something like this, its a bit like Warsaw circle, long circle, long line and Tychonoff plank, all in one
Vladimir Lysikov
Vladimir Lysikov
@ILikeMathematics If you do $A^* A = G^{-1} A^{\operatorname{tr}} G A = G^{-1} G = I$, you don't need to assume anything
nickbros123
nickbros123
@VladimirLysikov to show A^-1 falls back into O_n(G,K) we may need something, unless im missing something
Jakobian
Jakobian
13:38
@psie If $[z, w]\subseteq [x, y]$, what can you say about $F(w)-F(z)$ and $F(y)-F(x)$?
ILikeMathematics
ILikeMathematics
@nickbros123 yes
psie
psie
@Jakobian certainly $F(w)-F(z)\leq F(y)-F(x)$ because $w$ is closer to $z$ than $x$ is closer to $y$, and $F$ is increasing. Right?
Jakobian
Jakobian
yeah
Vladimir Lysikov
Vladimir Lysikov
$A \in O_n(G, K)$
Jakobian
Jakobian
now if you have any partition of $[c, d]$ say $z_1 = c, z_2, ..., z_m = d$, then $F(d)-F(c) = \sum_{k=1}^{m-1} F(z_{k+1})-F(z_k)$
ILikeMathematics
ILikeMathematics
13:42
@VladimirLysikov Ok yes this is it, thanks a lot!
psie
psie
that makes sense
Vladimir Lysikov
Vladimir Lysikov
Alternatively, you don't even need the expression for $A^{-1}$, just use
$A \in O_n(G, K) \Rightarrow G = A^\top G A \text{ invertible} \Rightarrow A \text{ invertible}$
and
$A \in O_n(G, K) \Rightarrow A^\top G A = G \Rightarrow G = (A^{-1})^\top G A^{-1} \Rightarrow A^{-1 }\in O_n(G, K)$
Jakobian
Jakobian
so the task is to, given some cover of $[c, d]$ by open intervals, to find such points so that each $[z_i, z_{i+1}]$ is a subset of unique $(x_l, y_l)$
Vladimir Lysikov
Vladimir Lysikov
(In the last line I multiply by $A^{-1}$ on the right and $(A^{-1})^{\top}$ on the left)
Jakobian
Jakobian
or emm, you know, that there is a bijective correspondence between intervals $[z_i, z_{i+1}]$ and those intervals $(x_l, y_l)$ so that it assigns each interval so that its a subset
psie
psie
13:48
sounds technical :)
ILikeMathematics
ILikeMathematics
@VladimirLysikov Thanks!
Jakobian
Jakobian
@psie I think that it's good to assume that there cannot be a further subcover by those intervals, that is we can't take out any of them
since it doesn't make the sum any bigger
psie
psie
@Jakobian hmm, well, if $(x,y)$ is the left-most interval in the finite cover of $[c,d]$, then $x<c$. So doesn't the sum get smaller if we decide to replace $x$ with some $z$ such that $x<z<c$?
Jakobian
Jakobian
@psie so what if it gets smaller
psie
psie
@Jakobian well, by the same argument, we could make the sum bigger by replacing $x$ with $z<x$
Jakobian
Jakobian
14:03
so what if we make it bigger
psie
psie
13 mins ago, by Jakobian
since it doesn't make the sum any bigger
Jakobian
Jakobian
yes, this is my comment
psie
psie
ok, well, let's just assume we can't take out any intervals of the finite cover
nickbros123
nickbros123
I think one can arrange the open covers in the order of thier lower bound, and ignore those intervals that are subsumed into bigger intervals and only pay attention to the remaining ones.
Jakobian
Jakobian
@nickbros123 this is basically what we're doing here, yes
Now any interval $(x, y)$ from our collection has the property that either $y$ is contained in some unique interval of our collection, or $d\in (x, y)$
so the notion of "next interval" makes sense
so now $F(y_{k+1})-F(y_k)\leq F(y_{k+1})-F(x_{k+1})$
of course if we reorder $x_k$ and $y_k$
nickbros123
nickbros123
14:12
im working with a vague picture here, forgive me for any imprecisions: So i guess we can arrange these intervals like $I_1 I_2 I_3 ...$, and choose such an $n_1$ so that $I_1 \cap I_2 \cdots \cap I_{n_1}$ is non empty, but $I_1 \cap I_2 \cap \cdots I_{n_1 +1}$ is empty (if such doesnt exist also its fine i guess?). From this intersection, we can take $z_1$. For $ z_2$, we can do the same thing but for $I_2, I_3$ and so on. for $z_3$, we take the intersection fo $I_3, I_4, I_5$ and so on.
am i makin sense?
the above is post deletion of intervals that are subsumed by bigger intervals
psie
psie
need to use bathroom real quick, be right back
Jakobian
Jakobian
@nickbros123 maybe, but it looks awfully complex
but I guess I'll let you guys handle this one
psie
psie
@Jakobian I get this inequality, but I don't see how it helps us?
Jakobian
Jakobian
because $F(y_1)-F(c)\leq F(y_1)-F(x_1)$ and $F(d)-F(y_{N-1})\leq F(y_N)-F(x_N)$ as well
so you add both sides together and puff, $F(d)-F(c)$ is smaller than the sum you wanted
psie
psie
14:30
ok 👍
 
1 hour later…
nickbros123
nickbros123
15:44
@Jakobian I only have a rough idea of the argument, havent formalized, maybe there r some edge cases
there may be some isues with well defined-ness of this procedure that im employing
hm yeah i think its fine.
(although this algorithm may be computationally expensive xD)
 
1 hour later…
pie
pie
16:58
Do guys think this question deserves a bounty ?
2
Q: Why are powers of one thousand the foundation for naming large numbers, and how did this system originate?

pieIn our modern number system, large numbers are named based on powers of one thousand. For example, "thousand", "million", "billion", and so forth. Numbers in between these powers are named using combinations of smaller units, such as "hundred thousand" or "ten million." This structure also shapes...

mathematics terminology notation numbers
nickbros123
nickbros123
@nickbros123 I think Ill write a C++ code for this one, looks like this algorithm must be 1) union the intervals that have the same lower-bound 2) A kind of search algorithm to compare all the intervals and delete the ones that are contained in another 3) Sort the final list by lower bound. Then we can look for the $\{z_i\}$ that make the partition
ModularMindset
ModularMindset
@onepotatotwopotato It's a definition I constructed that lays the foundation for my paper. Would you be interested in reading it? I could email the pdf to you
psie
psie
17:40
My memory is a little rusty on this, and I'm not asking for a proof, just a reminder of how one goes about it; how does one show that the Lebesgue sigma algebra of $\mathbb R$ is strictly larger than the Borel sigma algebra?
Is it about cardinality? If so, I think I know where to look.
Jakobian
Jakobian
@psie I assume that's what you mean by "strictly larger"
psie
psie
@Jakobian yeah, I'm confused myself :D basically, the Borel sigma algebra is a subset of the Lebesgue sigma algebra. How does one show this?
Jakobian
Jakobian
do you mean, proper subset?
psie
psie
yes
Jakobian
Jakobian
there is continuum Borel subsets of $\mathbb{R}$
57
A: Cardinality of Borel sigma algebra

user83827Let's say the $\sigma$-algebra on $X$ is generated by the sets $A_i \subseteq X$. For each subset $I$ of the natural numbers, consider the set $B_I = \bigcap_{i \in I} A_i \cap \bigcap_{i \notin I} (X \setminus A_i)$. For distinct sets $I$ and $J$, the corresponding sets $B_I$ and $B_J$ are dis...

But there is $2^\mathfrak{c}$ Lebesgue measurable subsets of $\mathbb{R}$
since any subset of ternary Cantor set is Lebesgue measurable
psie
psie
17:55
@Jakobian ok, let me ask a basic question then. If the cardinality of the Lebesgue sigma algebra has cardinality > than the cardinality of the Borel sigma algebra, does this imply that the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra?
pie
pie
@Jakobian I thought 2 days were enough it turns out I need to wait at least a week.
psie
psie
Since the power set of any set has cardinality greater than the cardinality of the set, I'd say $\operatorname{card}(B)>\operatorname{card}(A)$ does not imply $A\subset B$, or?
Ben Steffan
Ben Steffan
...do you not know that every borel set is a lebesgue set???
nickbros123
nickbros123
@Jakobian damn!!!!!!!!!!
Jakobian
Jakobian
@psie yes
psie
psie
18:00
@BenSteffan I do know this is true :) sounds true at least, but I've forgotten (and forgotten where to look) how to show this
Jakobian
Jakobian
whats your definition of Lebesgue measurable set
Ben Steffan
Ben Steffan
@psie I'm sure you'll have no trouble finding a proof somewhere. But then why do you ask questions to which you know the answer already?
Xander Henderson
Xander Henderson
@Jakobian A set which can be measured using Lebesgue, duh!
Jakobian
Jakobian
well-either way I assume that your definition is that those are sets which satisfy this
nickbros123
nickbros123
@BenSteffan somehow I know this, not cuz I know the details or anything, just that, in a probability book I was reading, it was an en passant statement. probably the book was motivating why I shud care for borel sets
psie
psie
18:05
@Jakobian let $\mathcal{M}(\mu^\ast)$ be the set of all $\mu^\ast$ measurable sets, i.e. the one's satisfying the Caratheodory condition. Set $$\lambda^\ast(A)=\inf\left\{\sum_{i\in\mathbb N}(b_i-a_i):A\subset\bigcup_{i\in\mathbb N}(a_i,b_i)\right\}.$$Then the Lebesgue sigma algebra is $\mathcal{M}(\lambda^\ast)$.
Ben Steffan
Ben Steffan
@nickbros123 good, good
if it's a statement about measure theory and I know it then it must be fundamental and/or easy
Jakobian
Jakobian
the definition of Lebesgue measure is that using this theorem
and part of statement of that theorem is that it extends your pre-measure to a sigma-algebra generated by your ring of sets
and that includes all open intervals
which generate all Borel sets
so that's why its kind of obvious that the inclusion holds
nickbros123
nickbros123
borel sets was my first (and last) encounter with transfinite induction. when I saw the wikipedia page i immediately closed the webpage.
Ben Steffan
Ben Steffan
transfinite induction is good :)
Xander Henderson
Xander Henderson
@BenSteffan LIES!
Ben Steffan
Ben Steffan
18:10
weird way to admit you're scared of unions :^)
nickbros123
nickbros123
capitalist moment
@BenSteffan I faced it early in my undergrad without any math maturity and got scared lol
Ben Steffan
Ben Steffan
I literally learned about it for the first time like 4 weeks ago lol
Jakobian
Jakobian
transfinite induction appears a lot in topology
for example in the proof that every monotonically normal space is countably paracompact, or that every point-finite open cover of a normal space admits a shrinking
Xander Henderson
Xander Henderson
@BenSteffan Not at all. My great-grandmother was a labor organizer, as were my grandparents.
nickbros123
nickbros123
@BenSteffan write a tutorial on it, prefarably for sub room temperature IQ students xD
or something like transfinite induction for dummies
Jakobian
Jakobian
18:14
that I think about it, all those transfinite induction proofs are used in proofs involving covering properties
Ben Steffan
Ben Steffan
@nickbros123 there's really not that much to it tho
it's ordinary induction + limit cases
nickbros123
nickbros123
I just cannot bring myself to learn it
Ben Steffan
Ben Steffan
you'll get around to it eventually
Xander Henderson
Xander Henderson
Transfinite induction isn't really anything special. It is normal induction, plus an extra step. Nothing to be afraid of. I just don't work in an area of mathematics where it is a very useful tool all that often.
Ben Steffan
Ben Steffan
but you can successfully avoid for a long time, as evidenced :)
Jakobian
Jakobian
18:16
there is something to point out in transfinite recursion that people don't tell you though
which is slightly different but adjacent topic
namely, in transfinite recursion you need uniqueness, but there is a way to get around that - well-order the set under consideration, that way you can always choose points in unique way
so even if its not unique, you can still use it
Ben Steffan
Ben Steffan
can't you just invoke choice
Jakobian
Jakobian
infinitely many times?
that's not formal in any way
Xander Henderson
Xander Henderson
@BenSteffan Indeed. I have "learned" transfinite induction, but I have never needed transfinite induction.
And I've published paper!
(And really, really need to get around to writing a couple more.)
Ben Steffan
Ben Steffan
@Jakobian I guess I don't understand your use case
Is applying choice once not enough?
Jakobian
Jakobian
@BenSteffan what do you mean
Ben Steffan
Ben Steffan
18:21
In fact that's what you're doing by choosing a well-ordering, no?
Jakobian
Jakobian
@BenSteffan and how do you do that
Ben Steffan
Ben Steffan
you produce some set of candidates at each stage, and then take a pick at the end?
Jakobian
Jakobian
10 mins ago, by Jakobian
that's not formal in any way
you can't just invoke axiom of choice infinitely many times like that
and the choice of every next term depends on the previous one
you want to do things uniformly
the ordering in this approach is wrong
instead, you first pick a well-ordering, so that you only invoke axiom of choice once
and then you choose your points based on that well-ordering
and isn't this precisely the issue? To make things formal? So yes, its not just pedanticism
 
5 hours later…
Thorgott
Thorgott
23:13
@Jakobian yeah nice, that property is clean

« first day (5260 days earlier)      last day (56 days later) »