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3:09 PM
Can I get some good refs/resources of point-set topology... or something broader...
what I wanted to know -->

we define
**Boundary point**: Let $u \subset \mathbb{R}^n$, $x\in \mathbb{R}^n$ and $\forall r>0$, $B_r(x)\cap u \neq \phi$ and $B_n(x) \cap u^c \neq \phi$, then we say $x$ is a boundary point.

Here, is the set $u$ a region?...like continuous and connected etc....(words not to be taken with mathematical rigour)...
say I define some scattered points in $u$...then how do I define a boundary and an interior
 
@Snapdragon-X If you are looking for references, Munkres Topology is a pretty standard text. There is also a book by Armstrong (Basic Topology) which I think is reasonable.
 
Thanks!
 
In the definition you give, $u$ is just a subset of $\mathbb{R}^n$. No other properties appear to be assumed a priori.
Also, \empyset or \varnothing should be used for the emptyset, not \phi.
 
so I can take...say 4 points?
 
Yes, $u$ could be a set containing four points (in which case, $u = \partial u$).
 
3:22 PM
Okayy, so given finite random points $\subset \mathbb{R} ^n$ all the points will be $\partial u$
True?
 
What do you mean by "random"?
 
@Koro Dragon ball z >:)
 
If you take any finite set in $\mathbb{R}^n$ with respect to the usual topology, that set will be equal to its boundary.
So if $u$ is a finite collection of points, and $x \in u$, then $x \in \partial u$.
 
random...just omit it ...it's not needed
Yupp! Got it thanks
 
often random is used in place of arbitrary, I've noticed.
 
3:27 PM
@robjohn And "arbitrary" is a word which seems to confuse students. Better not to use any adjective, and just stick with universal quantification. :D
Whelp... time to teach. We're going to see what my calc students think of induction today...
 
isn't arbitrary $\equiv$ random
 
I once had a native French speaking TA who pronounced it ar BIT ra RY with the BIT the main accent.
@XanderHenderson Tesla coils today?
 
@Snapdragon-X if you say "choose a random variable", "random variable" means something different in probability contexts
 
True...lol, nice example
 
3:43 PM
Consider a bounded real $3-$manifold $M=(0,1)^3.$ The slices of $M$, with $z=c$ for some positive constant $c\in(0,1)$ are $\Bbb R^2_{\gt0}$ equipped with flat metric $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2},$ for $x,y\in (0,1).$ This is diffeomorphic to a component of $\Bbb R^{2}$ (the third quadrant). $M$ must have cusp singularities at $(0,1,1)$ and $(1,0,0).$ What's a construction of $M?$ What sort of geometry might it have?
 
3:54 PM
XanderHenderson the horror. i still think it's black magic
@SAJW sdp win, congrats
 
4:23 PM
@geocalc33 hey
@leslietownes I'm putting the PyQt5 (desktop) app on hold and getting back into this web solution I coded a while back: youtu.be/kKJySXo9BDA
What's nice about this approach is you'll be able to browse CD's (commutative diagrams) from your phone, immediately out-of-the-box when I release the production site.
Hosting is not free and I want the site to be fast, so I should probably charge 1$ / month per mathematician. That's not asking too much. If it is, then I'll charge like 6 doll hairs / year lol
But think about it: You take a grid diagram with one node reading "rows exact" in the q.uiver.app editor. Then there exists a rule in the graph database that will convert that to include a row of kernels and their maps that are compatible and add to the commutative squares in the grid. All diagrams are assumed to be commutative unless "non-commutative" is seen in the diagram
But multiply that by the totality of all logical diagram-chasing rules. The video shows that I've already got the search and apply working (with variable substitution awareness!)
@leslietownes do you think you will use or recommend this tool once it's ready?
 
i won't use it (my job involves no math whatsoever), dunno about others. i do wonder if people already use/think in diagrams may be less of an audience for this than people learning that
 
4:39 PM
@leslietownes I'm not sure what your last sentence means :)
 
i.e. i wonder if "mathematicians" may be less likely to want/need that than students in their first big algebra class
 
So it's not a general math tool albeit. It's for the CD world. Both for students and for pros.
 
web solutions are cool, i think of stuff like desmos, geocalc's 3d grapher and the various complex graphers on the first page of google's complex grapher search
 
They're already using and liking the Quiver editor which can output to TikzCD for their LaTeX papers. It will be like a calculator for diagram chasing proofs. You don't have to use it, paper is always best, however, if you do want help on a proof, it won't hold back - it will give you the full proof, lol. So grad students could cheat with it, but imho understanding the proof is still the hard part.
 
the matrix calculators engineers students always use also
 
4:45 PM
@shintuku Imagine seeing all steps in a diagram-chasing proof, no exercises left - unless someone of course uses the database + an API for creating a gamification of learning homological algebra. So it's open-ended what you can do once you have an editor and a database. It's like any other content. The web is full of text content that is stored in a DB, so why stop there and exclude CD's.
They should be stored as well. The perfect storage method would be a graph database, so Abstract Spacecraft (what I'm calling the new version of Quiver Database) uses Neo4j, the premiere graph db technology.
@shintuku have you ever worked on a diagram chase? They're fun, so I think users will have fun entering in content to the database. It's kind of like a useful thing. Like an OEIS but for CD's.
If you're an editor, you get a cut of the site's profit. That's the only way this thing will be fair, futuristic, and mathematically green.
Of course users can then use the database and sell their apps, it's under MIT licensing so open or closed source is available to developers
 
nop, all i'm familiar with is those commutative diagrams from linear algebra
 
So think of MSE except your rep isn't just a number, it represents how smart you were getting into the CD editing game while there were still low-hanging enough fruit. But once everyone has the db, and we're on the frontier of CD & HA knowledge, people can work together and collaborate and see where we can take this kind of math.
@shintuku what are those?
So if you were the one who entered in the first isomorphism theorem, and it happens to be the version that everyone chooses when they go to search, you could make some dough.
I don't like the idea of money, so I'm not going to make this about money, so it's not going to be a company like SO that makes money off of us viewing ads, to self-sustain and give everyone reps and moderator priveleges for further sweat non-equity. Not to badmouth MSE. It's a useful tool, but money bounties should have been in place a long time ago. A math entusiast has go to eat, lol.
 
this stuff
 
Yes, that's precisely what a CD is, except more formalized. E.g. q.uiver.app/…
Okay, so you can see mathematicians really like the mneumonic
It's a lot easier to memorize a 3x3 grid and "rows exact" than to memorize a text list of all equalities, in fact doing that is not really feasible.
@shintuku if you have a sequence $A \xrightarrow{f} B \xrightarrow{g} C$, then we say that the sequence is exact at $B$ if $\text{im} f = \ker g$. Where kernel is from lin. alg.
But you can also have $\text{im} f \subset \ker g$ strictly and now it was recently discovered you can have $\text{im} f \supset \ker g$ and still compute a functorial homology. See here:
0
A: Is "reverse homology" $\ker g \subset \text{im} f$ possible?

SmokenSieEinBitteChebaHitBitsWe define a reverse chain complex $X_{\bullet}$ of $A$-modules and $A$-module homomorphisms $d_{i} : X_i \to X_{i-1}$ to be a usual chain complex of $A$-modules except the condition $\operatorname{im} d_i \subset \ker d_{i-1}$ gets reversed, that is: $\ker d_{i-1} \subset \operatorname{im} d_i$. ...

I haven't yet found an application of the idea that anyone will say ooh ahh about yet. But I'm pretty sure there is one, since the idea is so fundamental. If you have an inequality in one direction, by $\subset$, then what happens when you reverse it, and it turns out an analogous thing happens and it's quite miraculous.
@leslietownes have you seen that reverse-homology post I made? Isn't it strange that no one in math history has made use of it?!
 
5:11 PM
If p, q, r, s are distinct non-zero real numbers such that
(p^2+ q^2 + r^2)x^2 - 2 (pq + qr + rs)x + (q^2+ r^2+ s^2) ≤ 0, then p, q,r, s are in ap, gp or hp?? Please share its solutions
 
5:21 PM
@SmokenSieEinBitteChebaHitBits
hey
 
@geocalc33 do you know what homology is? There's a variant called de Rham cohomology used in manifold theory
 
I'm reading about that in my manifold book actually @SmokenSieEinBitteChebaHitBits
 
Take a sequence $\dots \xrightarrow{d_0} M_1 \xrightarrow{d_1} M_2 \xrightarrow{d_2} M_3 \xrightarrow{d_3} \dots$
When $d_i(M_i) \subset \ker d_{i+1}$ you get what's defined as a differential complex
 
smoke, not my field, i dunno. not clear to me that it hasn't been done before or that it isn't equivalent (in some goofy abstract way) to something done before. if new, probably hard to interest people in it without new results (or new proofs of old results)
but again, completely not my field, i have no idea
 
That's just another way of saying $d^2 = 0$
 
5:26 PM
@leslietownes how many years have you spent on a research problem? (curious)
 
It hasn't been done before. It can handle a sequence of surjections or a sequence of injections, i.e. places that aren't usually forming differential chain complexes in the normal sense
 
I am asked to determine whether $A : L^2[0,1] \to L^1[0,1]$ defined by $$A(f)(x) = (f(x))^2 + f(x) $$ is a linear operator. It's pretty clear that it isn't; just take $f$ to be the constant $1$ function and consider the scalar $a=-1$. However, I'm having trouble seeing why the function $A$ maps into $L^1[0,1]$. That is, if $f \in L^2[0,1]$, why is $f^2 + f \in L^1[0,1]$?
 
So for $\Bbb{Z}_{p^{n+1}} \twoheadrightarrow \Bbb{Z}_{p^n}$ you can compute its reverse homology but not its usual homology.
 
geo, hard to know when to start counting. i.e. when is one no longer preparing for formulating a problem vs. actually working on it. and how specific is a 'problem.' 2-3 years, maybe.
 
@geocalc33 what is your manifold book titled?
 
5:29 PM
An introduction to manifolds
second edition
by loring w. tu
 
Loring W. Tu? I own a hard copy of that (softcover)
I haven't ever finished it
I should though :)
Well, you see, those cohomology CD's is where HA takes off from the ground.
 
user: int |f^2 + f| <= int |f|^2 + int |f|. int |f|^2 is finite because f is in L^2 (this is the definition of L^2). int |f| is finite e.g. because L^2[0,1] is contained in L^1[0,1] (more generally L^b(X) is contained in L^a(X) for any 1 <= a <= b <= infty, for any finite measure space X, e.g. via holders inequality)
 
@leslietownes Oh, yeah...Of course, thanks!
 
5:46 PM
@smoke I want to extend a certain univariate probability distribution into a multivariate one - so I can use it to assess multidimensional risk, and asset inequality among populations
 
Nice, sounds like random vectors / matrices
 
Arnold, Taguchi and many others are working on it and it's extremely difficult for them
@Smoke yeah it's statistics, probability distributions, economic theory etc.
 
Well, you have billionaires and you have zeroillionaires. And for some reason the billionaires keep telling us to work hard like them even though we're doing that
Billionaires wield too much power, but what do I know...
 
billionaires control the structure of space-time
 
Ikr
I wonder if Elon Musk does DMT and talks to the machine elves lol I never have, but that could be why I'm poor.
 
5:55 PM
@geocalc33 you an econ student?
 
@geocalc do you partake in ganja?
I will send you a pic of this year's harvest over email, if you want. Have a few more days of trimming to do.
@geocalc33 is that topic related to the 3D model you showed me?
 
@shintuku I am not an Econ student but on weekends I try to study the Lorenz distribution and possible extensions (from the viewpoint of differential geometry). I am relating it to Lorentz surfaces (Riemann surfaces but for Lorentz-Minkowski spaces) and type changing metrics etc.
@SmokenSieEinBitteChebaHitBits yes that sculpture I developed and made about 5 years ago
 
@Smoke the sculpture obviously intersects itself in dimension 3. but that's because the higher dimensional structure doesn't embed into dimension 3
 
6:17 PM
@geocalc33 what medium do you use to sculpt in dimensions higher than 3?
 
@robjohn what do you mean?
 
If it has to be explained, it's not worth it.
 
not a complete waste. it made me giggle.
 
6:32 PM
@robjohn okay
 
@leslietownes I'm glad someone got it.
 
I got it - just not in a joking mood
 
Sep 3 at 16:48, by Ted Shifrin
We do not joke in this room.
I forgot
 
@robjohn Transparent aluminum?
Oh, or time crystals?
In condensed matter physics, a time crystal is a quantum system of particles whose lowest-energy state is one in which the particles are in repetitive motion. The system cannot lose energy to the environment and come to rest because it is already in its quantum ground state. Because of this the motion of the particles does not really represent kinetic energy like other motion, it has "motion without energy". Time crystals were first proposed theoretically by Frank Wilczek in 2012 as a time-based analogue to common crystals, whose atoms are arranged periodically in space. Several different groups...
 
6:56 PM
I've been trying to get rid of some space-time crystals on the black market, but I will have never been able to have done so.
@XanderHenderson how do I know he wasn't the one to discover it?
 
@robjohn You don't. :D
"Computer? Hello, computer?!"
 
Use the keyboard
 
What I love about that scene is that he has no idea what a mouse is, but can type, like, a thousand words per minute.
 
@XanderHenderson and can navigate the early HFS on the Mac
 
Indeed.
 
7:11 PM
suspension of disbelief is required for many movies.
Especially those starring Nicolas Cage
 
the buses in the bay area were nowhere near as clean as the one in the movie, although i think the graffiti and punk guy were intended to make it look menacing
 
@robjohn I mean, Nic Cage is the embodiment of suspended disbelief.
 
@leslietownes the punk guy was an associate producer.
 
@leslietownes Movies are made for the midwestern audience. For a midwesterner, who doesn't even really know what a "public bus" is, it was pretty menacing. :D
 
hah, perfect.
 
7:14 PM
Also, you can't really capture the smell of urine on screen.
 
there's an episode of 'streets of san francisco' that has a chase through a perfectly clean (and newly constructed) BART station. it looks like something from the future. why can't we can't have nice things.
 
@XanderHenderson there was a skit in Kentucky Fried Movie with Feel-a-round, however.
 
@leslietownes There is some show from the 60s or 70s (TJ Hooker, maybe?) in which scenes were filmed on the 210 east of LA before it opened. So there is never any traffic.
Which contrasts severely with my experience of the 210.
 
haha, that's great. i like old movies set in LA where there is ample street parking around business areas and dense residential neighborhoods.
or the silents where glendale is like a dirt road with four houses on either side of it
 
@leslietownes Certain things used to be better. :D
 
7:24 PM
some people watch for the scripts, the costumes, the set design, the performances of classic hollywood. i watch to indulge fantasies about traffic and parking.
 
My mother went to school at Pomona College. We drove by there a few years ago, and she was shocked that all of the orange groves were gone.
 
hah. my father-in-law has memories of riding horses through a riverbed in compton.
 
@leslietownes On the other hand, my little brother was watching Wonder Woman from the 70s recently. There was a scene shot at Ontario International. There was so much smog that you couldn't see Baldy (maybe 10 miles, as the crow flies).
Now, Baldy is visible from Riverside most days (20-30 miles, as the crow flies).
 
oh yeah, the air quality is something you also notice. there was a lot of that on the rockford files, even near the ocean.
 
@leslietownes Wow!
@leslietownes Oh my gawd! That show! James Garner is the best Garner.
 
7:29 PM
there are a few thin man movies that have some great locations. they drive across the bay bridge in one of them very shortly after it opened (i think the top level was SF to Oakland then and not the other way around). they visit the race track and you can see albany hill in the background.
in another one they fly out of the long beach airport. the front building of which looks the same today.
i wish i could live in the rockford files.
 
@TedShifrin hey Ted!
 
@StanShunpike Umm...
or is he asleep on the Knight Bus?
 
Is there a geometric interpretation of the geometric mean of a probability distribution? Like, if I take the arithmetic mean of a bunch of samples I will get the center of mass of the underlying PMF/PDF; does taking the geometric mean of a bunch of samples correspond to any nice feature of the PMF/PDF?
 
The arithmetic mean of the log of the data...
 
@leslietownes For all of its silliness, the show Nash Bridges (from the 90s?) very much lived in the Bay area.
 
7:40 PM
So you have to transform the PMF/PDF logarithmically and look at the center of mass of that?
 
@user10478 I don't know if there is a better way to look at the geometric mean of a probability distribution. If there is, let me know.
 
okay, ty
 
If $0$ has a non-vanishing density, the whole thing collapses, or perhaps not...
$\int_0^\infty\log(x)\,e^{-x^2}\,\mathrm{d}x$ is not $-\infty$, so I guess it is more of a concern in discrete distributions.
 
@Xander I used to watch that from time to time. one of the last shows i can think of that actually filmed on location in the bay area
 
@leslietownes But Vancouver is almost the same as SF, right? Only cheaper! NO ONE can tell the difference, right?
 
7:50 PM
hahaha. visiting toronto for the first time was a real trip for me. so many locations eerily familiar from movies where it stands in for every large american city.
 
@leslietownes There was a pretty good cops'n'robbers show set in Toronto a few years back...
 
I am back, with more pointset topology...
We define boundary set as
**Boundary of a set**: $u\subset \mathbb{R}^n$, we define the boundary of $u$, denoted as $\partial u$, to be the set of all boundary points of $u$.

here $\partial u$ is set of all boundary points of $u$.
I am wondering if my understanding is correct
[True]
And closed set as
**Closed set**: $u\subseteq\mathbb{R}^n$ is called a closed set if $\partial u\subseteq u$

Also I read https://math.stackexchange.com/q/392212/811225
 
Flashpoint, known as Critical Incident and Sniper for its former working titles, is an action drama television series that premiered on July 11, 2008, on CTV in Canada and on CBS in the United States, before it was moved to Ion Television. It was created by Mark Ellis and Stephanie Morgenstern. It was an international joint between Canada and the United States. The series starred Hugh Dillon, Amy Jo Johnson, David Paetkau, Sergio Di Zio, and Enrico Colantoni. The show focuses on a fictional elite tactical unit, the Strategic Response Unit (SRU), within a Canadian metropolitan police force (styled...
 
my wife watched that. i never saw it but because of its setting i would joke about its plot lines. "what's this week's crime? someone driving 10 mph too fast? parking in a reserved space? not returning library books? being impolite?"
dramatic courtroom scenes where the defendant says he's sore-y
etc
 
8:15 PM
@Snapdragon-X I am not sure that I understand your question.
 
8:25 PM
I am just asking if the red region is a correct closed set example..

because i read somewhere (linked above) that a closed set --> iff the complement is an open set

(also here https://people.maths.bris.ac.uk/~maxmr/opt/closedandconvex.pdf_
 
@Snapdragon-X Depending on which book you are reading, that is taken to be the definition of a closed set. That is, a closed set is defined to be a set whose complement is open.
 
Okay so the definition i stated in the first message is still valid
 
In your picture, it is not clear what the red set is supposed to be.
It can be shown that the two definitions are equivalent, yes.
 
red set is the region between the two red circles
i tried to shade with lines
 
What are the two red circles? and what do they have to do with the unit ball?
 
8:28 PM
so might as well ignore the red lines
wait
if the closed set overlapped everything
 
pictures are tough vehicles for understanding issues like this. it's hard to graphically represent whether edges of regions are, or aren't, intended to be part of whatever sets without adding the thousand words that the picture is intended to replace.
 
@XanderHenderson i dont see how they are equivalent...the complement of the red region spills to the white place...which is not holding true for open set (i.e. not interior of open ball)
 
this is why many books use toy examples in R^2 defined using algebraic conditions or other formulas, which may then be depicted, instead of just pictures
 
$B_2^2(0,0)$ be an open ball, $1<x^2+y^2<9$ be the region i want to call closed set

Am i correct in calling that
..yeah, this was simple
lol
 
@Snapdragon-X that region is not closed.
 
8:32 PM
{(x,y): 1 < x^2 + y^2 < 9} is not a closed set (using the usual notion of 'closed' for subsets of R^2).
for example, to use one of your definitions, (1,0) is a boundary point for that set, but is not an element of that set.
i don't see a definition of 'boundary point' above; you might want to review that if it was presented previously. if it's a new concept, that formulation of closedness might not be much help.
 
why why...am i not following all required conditions...the boundary is the subset
 
Every neighborhood of the point $(3,0)$ intersects that region, yet that point is not in that region.
$(3,0)$ is a limit point of that set, yet not in that set.
 
but i dont need that
defined:
43 mins ago, by Snapdragon-X
I am back, with more pointset topology...
We define boundary set as
**Boundary of a set**: $u\subset \mathbb{R}^n$, we define the boundary of $u$, denoted as $\partial u$, to be the set of all boundary points of $u$.

here $\partial u$ is set of all boundary points of $u$.
I am wondering if my understanding is correct
[True]
And closed set as
**Closed set**: $u\subseteq\mathbb{R}^n$ is called a closed set if $\partial u\subseteq u$

Also I read https://math.stackexchange.com/q/392212/811225
 
$(3,0)$ is a boundary point of the set, yet is not in the set.
 
that defines "boundary" of a set in terms of a concept of "boundary point". that concept also has a definition.
 
8:35 PM
so that set is not closed
 
a definition that does not appear in the quoted material above, but might be crucial for understanding the definition of 'boundary' of a set.
 
but isnt the set...just the ball? of radius 2
 
@Snapdragon-X Compare $\{ x : \|x\| < 1\}$ and $\{x : \|x\|\le 1\}$. These are both balls. One is open. The other is closed.
 
then why are you mentioning $\left\{(x,y):1\lt x^2+y^2\lt9\right\}$?
 
that is the set i want to call closed....can I not do that?
 
8:37 PM
which is the set you are looking at?
what is the space you are looking at? is it just the ball of radius $2$?
 
@XanderHenderson yes exactly...by my definition, it has all the boundary...but similarly...take the open balls complement and restrict the radius to 5...wouldnt it still be closed?
 
so that all the points outside the ball of radius $2$ don't matter (or exist for all intents and purposes)?
 
edited: scratch that, did not see it is an ordered pair
 
what about the point $(1,0)$ that is a boundary point that is not in your set
 
Yes, let set of consideration be $u \in B_2^2(0,0)$
call closed set as $\{(x,y): 1<x^2+y^2<9\}$
 
8:40 PM
@Snapdragon-X is $B_2^2(0,0)$ open or closed ball?
 
open ball
but does that matter?
 
then it doesn't intersect your other set
 
I am absolutely sorry, i meant radius 1 to radius 3
instead of 2 to 3
now the region intersects with the boundary of the ball,...be it closed or open
 
So, your entire space is $B_2^2(0,0)$?
$\left\{(x,y):1\lt x^2+y^2\lt9\right\}=\left\{(x,y):1\lt x^2+y^2\lt4\right\}$?
 
My space is $\mathbb R^2$
in which i am considering that open ball
 
8:43 PM
then the annulus you mention is open and not closed.
 
OF which i am finding the closed set
 
that ball is open.
wait, either you are in $\mathbb{R}^2$ or you are in $B_2^2(0,0)$
which is your space?
You can look at the intersection of the ball and the annulus...
 
Are you trying to find the closure of $\{ x : \|x\| < 2 \}$? If so, that is $\{x : \|x\| \le 2\}$.
 
I am not very clear...I am working in $\mathbb R^2$

I created a set....say $A$ that being the ball
then i made the red region I wanted to call the closed set

now...here...what wrong assumptions did i make?
 
you don't get to 'call' a set open or closed. you get to define what the set is, and then the definitions will determine whether it's open or closed.
2
 
8:47 PM
you can't just call a set closed, unless you are defining a non-standard topology.
 
This is supposed to be point-set topology
 
yes, but with the standard metric topology
 
is there a goal in mind? like a specific problem? or are you just thinking through various examples to explore the definitions?
 
$\left\{(x,y):1\lt x^2+y^2\lt9\right\}$ is an open set. It's complement is closed.
 
The goal is to understand and make examples of stuff I learned in multivariate calc. course today...we started off with some notions of point-set topology ...and yeah...these are the examples I found contradicting online to some defintions
 
8:50 PM
around any point in that annulus, you can put a ball, centered at that point, that is completely inside that annulus.
 
but the complement doesnt enclose the boundary at $r=2$?
 
the complement includes the boundary
 
how..I think the open set includes the boundary... i.e. $x^2+y^2 = 4$ which lies between 1 and 9
 
Oh, the boundary of $B_2^2(0,0)$ is included in the annulus
but what does that matter?
but that does not make the annulus closed
the annulus needs to contain its boundary to be closed.
it doesn't matter that the annulus contains the boundary of some other set
 
Hmm I see, I learnt some stuff, maybe there is a bit of conflict between definitions... I can probably resolve them now! Thanks
 
8:56 PM
if taking a multivariable calculus class, it's likely that you will only be dealing with simple surfaces and other regions that are closed, and the 'boundary' coincides roughly with an intuition that can be developed by looking at a number of examples. delving further into the general concepts might not be that helpful. a lot of web resources will be discussing far more complicated sets than the ones you meet in multivar.
 
@Snapdragon-X There is no conflict between the definitions. They are equivalent.
 
A set is called closed if (1) it contains all of its boundary points or (2) if its complement is open. These two definitions are equivalent.
 
OHHH I got my problem lol
I was defining closed set OF a closed set
 
not sure that is a real thing.
 
8:58 PM
that's why the open ball was confusing everyone
 
you can look at an intersection and see if that is open or closed.
 
yeah...now i see... the main point was that i was getting a set of a set...like a function lol
but it does seem a bit interesting..
consider the boundary of the boundary of the boundary... recursively,
 
the boundary of the boundary of $X$ is the boundary of $X$ (unless the boundary has an interior)
 
I meant infinitesimal regions i was assuming...it'd look like a wave
Also, see ya, it's 2:38am
 
9:52 PM
What applications do we have you think from detecting palindrome numbers PLEASE?
123321 is palindrome number b/c we can read it both ways and still read it the same way
 
10:23 PM
@robjohn. Hello, do you think please there is any prerequisite to measure theory class?
Do you think once can proceed with the course without strong background in any math course that should predate it?
 
It depends on the direction of the class. It might need topology, or real analysis, or other prereqs. The best thing to do is to see what prereqs are listed for the course.
 
10:46 PM
I know that if vector spaces (over field F) V and W are finite dimensional then dimension of the space of all linear maps from V to W i.e, L(V,W) is dim V dim W. If U is finite dimensional vector space then dual of U is $U^*$= L(U,F). It follows from the aforementioned formula for dimension that dim U=dim U$^*$. I don’t understand how it follows.
 
dim F = 1
e.g. {1} is a basis for F as a vector space over F
 
Ahh, I was thinking what if F were Q, which is not finite dimensional.
But we’re looking at F(F) - a vector space over field F.
which indeed has dimension 1 as you say.
Thanks a lot @Leslie :)
 
11:10 PM
Hey, guys. Long time no see.
I found a slowly converging series for computing reciprocals. desmos.com/calculator/xsuf4jy3yh
Any ideas on how I can make it converge faster?
 
@Avra Based on 40 years of advising, I say both some real analysis and some topology (working with unions and intersections of sets, some abstraction) is necessary.
 
AMDG: positive encouragement? stock options? foosball in the break room?
 
@leslietownes I think I'm too inexperienced to get your joke XD
Clearly I gotta slap it into submission and then it will have absolute convergence with finitely many terms
 
that's the other approach. the carrot or the stick.
i didn't realize that my daughter's cast made her something of a celebrity at day care. on dropoff this morning she was instantly surrounded by kids who wanted to talk to her about it.
 
I imagine that I might be able to squeeze out more precision by integrating the series.
Then I can just use $\exp(\ln(a) - \ln(b))$.
 
11:22 PM
@TedShifrin. Thank you. I will keep that in mind. I did not take topology though. I dropped before taking it.
 
Or, I can just add an arbitrary number of recursive steps to compute natural logarithm and then the reciprocal, then use that as the basis for the terms, rinse and repeat.
 
How do you thinkg please about sum of $ 1+2+4+⋯+n/4+n/2+n \in O(n),$
I see clearly that this is a geometric series of sum $2^{n+1}$
I got answer saying that summing 1+2+4+16+32... is different from summing $1+2^i...$!
 
I fail to see how the substance is different from $2^n$.
Unless of course we're talking about a different $n$ or beginning, in which case it is a unique and entirely different sum.
 
@TedShifrin. Have you studied or taught measure theory before please?
Some students say it's very hard and similar to AA
(abstract algebra)
 
11:40 PM
All math PhDs have studied it and taken exams on it. Abstract algebra is far simpler. The complexity of quantifiers in measure theory is very challenging. Algebra is the simplest in that regard.
 
I will testify that I find algebra easier than anything to do with analysis.
 
I always told grad students to take point set before measure theory/integration theory. Everyone thanked me.
There are lots of joke analysis courses in the US, but I do not know of joke measure theory.
 
I am taking first year analysis as part of my grad school courses. It's sort of slow but following Rudin's real and complex analysis, so we started off with abstract measures and integration, no examples.
 
so much of measure theory just doesn't make sense or feel motivated unless you have some idea of the kinds of pathologies that it's set up to avoid or deal with. e.g. why the riemann integral has bad formal properties. which gets you right into mild counterexamples in analysis, and sets that look weird and aren't just unions of intervals.
 
I should read some serious measure theory at some point.
@leslietownes That or if you have some prior exposure in probability.
 
11:49 PM
yes, although some stat books kind of sidestep the technicalities entirely. there are a whole lot of joke stat courses in the US.
which can be fine for some applications, but doesn't help to motivate measure theory
 
Ah yes
Whatever working knowledge of measure theory I have I attribute to Durrett's classic text on probability theory
Which is not really what commonfolk understand by probability and statistics.
 
Interesting. Someone is downvoting some of my old answers… incomplete answers marked hints, for example. These no longer conform to community standards, I suspect. Screw that.
 
if someone is analyzing stochastic processes in continuous time they probably have a good grasp of the motivation. someone who is collecting data and modeling it with a normal distribution, whether or not that is in any way appropriate, not so much.
 
Right
 
11:52 PM
if there is batch downvoting of your stuff, particularly across a wide time period, that sounds like targeted voting, not mere enforcement of standards.
 
i think enforcement people would leave a comment
 
Not enough for “batch” yet.
 
I upvoted to reverse the effect. Actually, one of them is not even a hint, it's a complete answer.
You have left far shorter hints
 
You can see my votes ?
 
I can see on your activities tab, yes
 
11:56 PM
Interesting. The Frenet-Serret one was complete, esp. with discussion in comments. The topology one was pretty complete, too. Some hater.
 
Must be yes
 
A few times people have asked pertinent questions to point out why answers were not sufficient. Sometimes I agree and try to amend.
Sometimes, eight years later, I no longer remember what I had in mind. But that’s a valid complaint.
 
I get those as well, but I seem to have completely stopped being active in MSE.
 
Seem?
 
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