Mathematics - 2023-06-14 (page 1 of 3)

Ted Shifrin

00:05

@D.C.theIII There’s an extra subtlety. What norm are you using to measure $\delta$?

D.C. the III

Well I set it up through $\|f(A+H) - f(A) \| = \|(A+H)^{-1} - A^{-1}\|$. I wanted to ask about the $A+H$ notion as well. But for the current moment I used that to represent a general invertible map in the space. So that whole set is less than $\|A^{-1}\| \frac{\epsilon}{1- \epsilon}$ which I have from part (b) of the question.

THE_CRANIUM

@shintuku Do you think an explicit function $f(n) = \sum_{i=0}^{n-1} 10^i$ would work?

D.C. the III

There is also thr $\|A^{-1}H\| \leq \|A^{-1}\|\|H\| < \epsilon < 1$ relationship I have as well

shintuku

@THE_CRANIUM yeah there's a bijection between natural numbers made out of only 1 and the strings made out of only 1

take that as an axiom and you can prove your desired bijection @THE_CRANIUM

PM 2Ring

@Jakobian It's very unclear. But from their comments, I guess they want some way to evaluate the divergent Taylor expansion of $$\lim_{v\to c}\,1/\sqrt{1-v^2/c^2}$$ similar to 1+2+3+4+...=-1/12.

D.C. the III

00:18

So another idea from the same part (b) which can also be applied is that if $\|H\| < \epsilon / \|A^{-1}\|$ then $\|(A+H)^{-1} - A^{-1}\|< \frac{\epsilon}{1- \epsilon}$ but this time $\epsilon = \epsilon / \|A^{-1}\|$

Ted Shifrin

OK, so you’re using the norm, rather than usual Euclidean length. They’re related by an exercise in section 5.1. I don’t get your last thing.

Are we trying to choose $\delta$ so that the norm of the difference of the inverses is less than $\epsilon$?

D.C. the III

I used the relationship you allude to from section 5.1 when I did 6 (c): $\|A\| \leq (\sum_{i,j}a_{i,j}^2)^{1/2} \leq \sqrt{n} \|A\|$. And to your second question that's what I'm attempting. THe last part was just me playing around with what I know

well actually I didn't but I know that it was the hint...

Ted Shifrin

What hint?

How do you get $\epsilon$ instead of over $1-\epsilon$?

D.C. the III

the hint was if $(\sum h_{ij}^2)^{1/2} < \delta$, then $\|H\| < \delta$, but that was from 6(c)

Ted Shifrin

Oh.

D.C. the III

00:34

Originally I had said $\delta = \frac{1- \epsilon}{\|A^{-1}\|}$

Ted Shifrin

That has to be wrong. Why?

D.C. the III

but that doesn't exactly work with my inequality relationship that I have because what I want is $ \|(A+H)^{-1} - A^{-1}\| < \delta$, but what I have is: $ \|(A+H)^{-1} - A^{-1}\| < \|A^{-1}\| \frac{\epsilon}{1- \epsilon}$ as a relationship. But going back to the original inequalities if I can bound $\|H\|$ properly then I'll be in business

Ted Shifrin

No, that is certainly not what you want!

D.C. the III

yikes

Ted Shifrin

What’s the $\delta$-$\epsilon$ definition, again?

D.C. the III

00:44

for all $\epsilon > 0 $, and for all $A \in X$ there exists a $\delta > 0$ such that for all $H \in X$ if $\| H - A \| < \delta$, the $\|f(H) - f(A)\| < \epsilon$. Here $X$ is just the set of invertible matrices

I do wonder if I should have used $A+H$ instead of just $H$ in the defintion.

Ted Shifrin

I’d rather stick with my notation. We’re comparing $A$ and $A+H$. So where did your what I want is come from?

@D.C.theIII Indeed.

D.C. the III

So if I use the $A+H$ instead I will have $\|H\| < \delta$

Ted Shifrin

Which makes sense.

D.C. the III

yea, this part of it was me not being 100% clear on how to treat the objects.

Ted Shifrin

So how do you end up with $\|f(A+H)-f(A)\|<\epsilon$?

D.C. the III

00:51

just fiddling around to get it...I've not diappeared

Ted Shifrin

No prob

D.C. the III

01:14

sigh...I can't seem to isolate an $H$ to find a $\delta$....

Ted Shifrin

This is just basic algebra. You want to make $bx/(1-x)=\epsilon$.

D.C. the III

amateur mistake....not remembering that the two $\epsilon$'s are not the same. So what I get is $\epsilon^{'} = \epsilon/(\|A^{-1}\| + \epsilon)$.

so if $\delta = \frac{\epsilon}{(\|A^{-1}\| + \epsilon)\|A^{-1}\|}$ then $\|(A+H)^{-1} - A^{-1}\| < \frac{\epsilon^{'}}{1 - \epsilon^{'}} = \epsilon$

Ted Shifrin

01:30

One should never have been $\epsilon$ :)

Right. Well done.

D.C. the III

I was just trying to see how the $\delta$ creates the implication in the whole idea. it is because we have $\|A^{-1}\|\|H\| < \epsilon$, but in this case $\epsilon$ is my $\epsilon^{'}$ so the relationship that I established in part (b) can be used.

Ted Shifrin

Part of the trick of real analysis is parsing the logic of your epsilons.

D.C. the III

So back to this $A+H$ idea........so are you implying from what I proved, that whatever fixed invertible matrix $A$ that I take, if I take a square matrix $H$, as long as the condition $1/ \|A^{-1}\|$ is satisfied then $A+H$ will be invertible?...

more so what I'm trying to get at is if this characterizes the set $GL(n)$?

Ted Shifrin

01:49

Yes, but this goes back to earlier things. Not the continuity we’ve just been through.

D.C. the III

Yes yes...I wanted to ask about it before, but just wanted to get the continuity idea completed.

Ted Shifrin

That’s an empirical argument that $GL$ is open. There are abstract ways just with continuity of det (discussed in chap 7).

D.C. the III

but it is necessary that the norm of the square matrix $H$ be less than $1$. So these square matrices can expressed as infintie series?

Ted Shifrin

No, look carefully.

D.C. the III

So from 6(a) I got that $(I-H)^{-1} = \sum_{k = 0}^\infty H^k$ and from 7(b) I had gotten $(I- A^{-1}H)^{-1} = \sum_{k=0}^\infty (A^{-1}H)^k$.

copper.hat

02:26

Any nilpotent $H$ will also converge, regardless of norm.

Ted Shifrin

We’re trying to get an open ball, @copper.

So it’s $A^{-1}H$ whose norm you need to control. Whence ….

copper.hat

An open ball around $I$ so that $I+H$ is invertible?

Ted Shifrin

Or, more generally, around an invertible $A$.

copper.hat

Useful result.

D.C. the III

02:51

That was the second part of my question too.....everything was going to be dependent on the invertible matrix $A$.

Ted Shifrin

As you expect, of course.

D.C. the III

hmmm......this idea of open is something I have to get comfortable with.....interesting idea though.

Ted Shifrin

It’s no different from chapter 2.

D.C. the III

feels like it. Mainly because of the objects and how they are defined.

Ted Shifrin

They’re still elements of a vector space.

D.C. the III

02:56

if I think of the matrices as just "points" then I understand what you mean

as you just said above

Ted Shifrin

Yup. In this context, that’s how you should think.

D.C. the III

03:12

I'm at a stage where I have to start blending the various subjects more fluidly: Analysis, Linear Alg, Topology, etc......

Ajay

@TedShifrin Could you check my working in this answer: math.stackexchange.com/questions/1581136/…

I don't understand the downvote.

Ted Shifrin

Who knows. But your solution is just an expanded version of achille hui’s from 8 years ago.

shintuku

03:36

munkres has nothing on finite topological spaces

where are these

copper.hat

03:51

@D.C.theIII Matrices with the Frobenius norm make the connection to the 'usual' vectors obvious.

Ted Shifrin

04:20

I didn't teach DC that term. But he did the exercise comparing that norm to the operator norm (which term I also did not use) :D

Thomas Finley

Guys, need some help with this, please :

0

Q: Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c , d$ are real numbers and $a <b, c<d.$

Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c,d$ are real numbers and $a <b, c<d.$ This is a well-known classical question in real analysis, if I am not wrong. The thing is, these problems are new to me and I am interested in learning some general approaches to establi...

real-analysis

leslie townes

k, jiang has given maybe the most explicit solution in the comments. there is a bijection between these intervals whose graph is a line segment and the formula for it is in that comment.

breaking it down as a series of bijections into other spaces is fine too. i haven't checked your formulas (or k. jiang's)

Thomas Finley

@leslietownes Then, you might consider checking my work ?(If that interests you, coz I need to know what I did was at all legit or not. A concise and apt clarification, is what I am lookin for)

leslie townes

something missing from your outline is that all of the families of 'transformations' you describe are, in fact, bijections. this is very close to the surface but it is definitely part of the problem. writing one function as a composite of "simpler" maps is a useful general idea, but the only reason it produces a bijection in this case is because each of the simpelr maps are bijections.

you say "all of these transformations are bijections," and that is certainly true, but a lot of people might also regard that as of roughly the same level of 'difficulty' as the original problem.

so what you describe is a 'legit way' in the sense of how it is someone might approach the problem and come up with the formula in that comment. but arguably, it is missing a moment where one proves (rather than just says) that one or all of those basic transformations are bijections.

Thomas Finley

04:36

@leslietownes So you are saying to prove them to be bijections ? It seems doable, but won't it make the answer huge? I mean, I never verified it myself, whether they all are bijections or not, but it seems to be true intuitively. I think, a better idea, is to prove the final transformation, I wrote at last, to be a bijection. Do you agree?

leslie townes

coming up with formulas for those functions is arguably most of the "work", because you can use the formulas to prove bijectivity in a straightforward way.

one potato two potato

@XanderHenderson May I ask about what your research field is?

Thomas Finley

@leslietownes Ok, let me see, if I can attach a proof of bijectiveness of the final formula, at the bottom.

leslie townes

thomas: i agree with you there. once you have the formula for that map that you want, proving that it gives a bijection is about the same amount of work as it would be to prove that any one of the ingredient 'transformations' is a bijection.

Thomas Finley

@leslietownes yes, and the later would be simpler in my opinion.

leslie townes

04:38

i think the logic of thinking through the map as a chain of simpler transformations is maybe helpful in coming up with that formula, but yeah, once you've got it, you can forget how you got it and just prove that it does what you want.

Thomas Finley

@leslietownes That's a good idea.

leslie townes

note that sometimes the difficulty can go the other way around. e.g. if someone gives you some random formula from a calculus book and asks you to prove that it's continuous at all real numbers, it might actually be shorter to generally prove "a sum of continuous functions is continuous," "a composition of continuous functions is continuous," etc. and then verify continuity of basic ingredients (such as the identity function) by hand, and note how the function is built up from those ingredients.

if you just say "given epsilon" and try to choose delta that works for that formula, it's likely going to be a mess if you want to 'customize' the delta for that given epsilon.

but here, having an explicit formula for something makes it easier to do a "one-off" proof that doesn't detour through intermediate results.

Ted Shifrin

@Thomas This is still the easy one. I asked you to do $(a,b)$ and $[c,d)$ or $[c,d]$.

Good luck trying to find a linear function!

@leslie Phbhhhhbt!

one potato two potato

Author says 'introductory' I say 'introductory for people who already know'

Ted Shifrin

More whining in here. What subject/book?

one potato two potato

04:54

Just a joke it's not whining.

Ted Shifrin

OK … You haven’t traditionally been a whiner!

copper.hat

basic does not necessarily mean easy

good night folks!

Ted Shifrin

Night, copper!

shintuku

consider the set $\{a,b,c\}$ with the topology $\{\emptyset, \{c\}, \{b,c\}, \{a,b,c\}\}$. am i correct in saying that $\{a,c\}$ is neither closed nor open?

one potato two potato

a detail description of the book Xander linked says 'Background needed for a potential reader of the book consists of a working knowledge of real and complex analysis on the level of first- and second-year graduate courses.' How much should I know exactly? Full knowledge of RCA for example?

shintuku

05:03

@shintuku yes turns out this is a thing that can happen

Thomas Finley

@leslietownes I made the modufucations and added a section in it, where I formally prove it. If you're still interested, your remarks regarding it, will be much helpful.

@TedShifrin As @leslietownes mentioned, I should've done the thing, more formally. It was just now, I have done it and attached it to the OP. So, you might say, I am occupied with this stuff, rather than proceed towards the general case. You might check it out, too ( If you're interested). Your remarks will also be helpful!

Koro

Thanks @copper.hat

leslie townes

ted: bphtbtpt

Thomas Finley

@leslietownes Do think the proof's formal "enough"?

Koro

@shintuku yes, you are.

You can even say that the set with that topology is not metrizable.

Because metric spaces are Hausdorff but this space is not: it is not possible to separate b and c by disjoint open sets.

Thomas Finley

05:17

@Koro I did a proof, of the fact that $(a,b]$ and $(c,d]$ does have the same cardinality, which we're talking bout a few hrs ago. You might consider giving it a look.

0

Q: Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c , d$ are real numbers and $a <b, c<d.$

Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c,d$ are real numbers and $a <b, c<d.$ This is a well-known classical question in real analysis, if I am not wrong. The thing is, these problems are new to me and I am interested in learning some general approaches to establi...

real-analysis

shintuku

at koro: ty

am studying nonhausdorff spaces 😎

Thomas Finley

Still no answer to this:

0

Q: Absurdity while calculating the $p$ discriminant.

Find the $p-discriminant $ of the differential equation $F(x,y,p)\equiv 4xp^2-(3x-1)^2=0.$(Note: $p=\frac{dy}{dx}$) We proceed to calculate the p-discriminant of $F(x,y,p)=0,$ where $F$ is the function representating the differential equation in the question. So, $F(x,y,p)=4xp^2-(3x-1)^2=0\implie...

calculus ordinary-differential-equations

(Strange!)

leslie townes

is that another question from your 19th century ODE book? it's possible that nobody uses the 'p discriminant' or 'eliminant' language anymore, even if the concepts are still alive.

the terms are unrecognizable to me.

one potato two potato

lol

Any recommendation on functional analysis textbook leslie?

leslie townes

the chapters in folland's real analysis books are a good intro to the essentials. i forget if they have everything but i think they cover most of the major things. certainly all the L^p theory is in there.

peter lax's functional analysis book is really good, but also i think pretty expensive.

riesz and sz.-nagy have a book available in super cheap dover reprint that is really good, but also pretty 'old fashioned.' there might be stuff about the p discriminant in there for all i know.

one potato two potato

05:30

aha I was looking for some functional analysis textbook after folland

leslie townes

lax is the best book i have seen.

i should say, i didn't learn directly from that book, i got it later, but every time i needed to look something up, it was usually in there, and with better proofs than i had seen before.

conway's book is a fine overview but i didn't learn out of that one either and the exposition is not as good as lax.

if you have specific stuff you are interested in maybe there are more specific books. lax has a few nods to ODE/PDE but i don't think he does a ton of it. if you want to learn weird shit about abstract banach spaces you are not going to find that in a survey book.

Thomas Finley

@leslietownes Nope, that's a thing in my coursework.

But, what about my proof-writing? Did it match ur expectations?

@leslietownes I am aware of this.

leslie townes

thomas: the proof looks fine to me. i did not check every expression but it looked like the way i would expect a proof to go.

Thomas Finley

05:55

@leslietownes Ok. Thnx.

leslie townes

i appreciated the detail about, does this formula do the right things at endpoints and really map one interval to the other. that's the main place where the choices of coefficients in your formula "really matter."

if you like to mix and match ideas, note that you're giving an example of a bijection between those things that is "continuous" in the usual sense. it isn't significantly easier in this case to 'write down' formulas for bijections that aren't continuous, but sometimes it is, if all you want is a bijection. and sometimes it will not be possible to find a continuous bijection even if a bijection exists.

Ted Shifrin

@ThomasFinley Not strange. The language/approach is not in the least consistent with modern-day mathematics.

Thomas Finley

@leslietownes Yes, I took an extra care to show that the formula does not have any element in (a,b] , outside $(c,d]$ i.e it maps every element of (a,b] to a point in (c,d] only. And also, I did not even bother to show that the function is well-defined as, I proved it to be a bijection, which should end any possible debate, about it's well-defineness whatsoever.

Ted Shifrin

@leslietownes Like the “interesting case.”

Thomas Finley

@TedShifrin Do you mean, terms like "p-discriminant" have gone extinct ? That' s strange as well. I felt it's something crucial to find singular solutions of a 1st order ODE. Without that, conecept, I dont think it's really feasible to find one ( sing soln) or at most not possible to do it using elementary ways.

3 mins ago, by Thomas Finley

@leslietownes Yes, I took an extra care to show that the formula does not have any element in (a,b] , outside $(c,d]$ i.e it maps every element of (a,b] to a point in (c,d] only. And also, I did not even bother to show that the function is well-defined as, I proved it to be a bijection, which should end any possible debate, about it's well-defineness whatsoever.

Also, I think this might be a good strategy to skip proving whether a function is well-defined or not

i.e, by showing that it is a bijection/injection/onto. Isn't it, @leslietownes ?

@leslietownes Yes, but tbh, I think I haven't encountered such cases as of yet, or if I have, then it wasn't too significant either.

leslie townes

06:09

i'm not sure what you mean by 'well defined.' i guess it's clear that the formula makes sense as a map from [whatever the domain is] to some subset of R, and in proving that the function is one-to-one, the algebra basically doesn't notice or care what the range of the function is. in proving that it's onto, you do need to say, okay, for y in [this specific set] we can produce x in [this specific set].

so you do need to deal with that somewhere.

where you deal with that is up to you.

Thomas Finley

@leslietownes By well-defineness I meant, that for a particular "input" of a function, it doesn't give two different outputs, i.e produces only one value for each input (in domain).

leslie townes

yeah, i don't see this problem as really presenting that issue.

Thomas Finley

So, I was just saying, we don't really have to bother about this, since we have already proved the function to be a bijection, and so, it closes all the debates.

@leslietownes More because of this ↓↓↓. Isn't it?

2 mins ago, by Thomas Finley

So, I was just saying, we don't really have to bother about this, since we have already proved the function to be a bijection, and so, it closes all the debates.

leslie townes

i don't see it as being in the function "being a bijection." it's more in, you have a background understanding of the formulas used to define the function. you have to know what addition and multiplication of real numbers are to make sense of that. and if you do know that, you know it's producing only one output for each input.

the only conceivable static around that might be if your formula had denominators that might be zero or something. but it doesn't.

with more complicated formulas, or functions that didn't have 'algebraic formulas' at all, i could see their being issues.

e.g. if you purport to define "f(x)" by using the decimal expansion of x, there would be some detail checking around whether your purported definition assigns the same value to 0.9999... as it does to 1.0000..., things like that.

or if the formula for f is a complicated composition of things that maybe aren't defined everywhere. you see this in calculus books, e.g. ln(sin(bleh)) + sqrt(blah)), you have to wonder, well wait, for x in my domain, does this formula even have meaning, e.g. is it asking me to ln something that isn't a positive number or sqrt something that isn't nonnegative.

Thomas Finley

@leslietownes When I give it a thought, what you said it seems true. I was making this complicated here. The only thing gurantees that two distinct outputs doesn't occur, just follows immediately from the fact that addition and multiplication, division and subtraction all are well-defined binary operations in real numbers. My bad, for over -complicating and over- analysing it!

leslie townes

06:19

but just with f(x) = mx + b, i don't see there being issues in, okay, that defines only one thing for any given x. the issue maybe being, given x, does it define not just some real number (which it clearly does and maybe this isn't worth comment) versus does it return an element of [c,d] (or whatever, which absolutely is worth comment, but is maybe covered in your proof of f being surjective, or maybe separately covered in an argument specific to mapping properties of f)

Thomas Finley

@leslietownes yes, I agree.

leslie townes

its pretty common in calculus/analysis/whatever to easily write down functions and intervals where you can't easily say what the range is. e.g. to find what f([a,b]) is even if f is continuous you need to actually identify the max and min values of f and that could be difficult even if f has a 'nice formula.'

thats not directly related to your problem, just a random observation. its not that 'common' to be able to identify the mapping properties of a randomly given real valued function on R, even if it has a nice formula.

Thomas Finley

@leslietownes Hmm, I know what you mean.

Is the above question at all valid ?

It basically implies for any two arbitrary sets, $A,B$ in real nos, $cl(A)\subset cl(B).$

This is absurd.

By cl(A) - I mean closure of A, which is also sometimes expressed as $\overline{A}$

Am I inferring the scenario correctly ?

leslie townes

06:39

there is a missing hypothesis, presumably that A is a subset of B.

they like nesting binary relations more than i do but this might even be a typo, "A, B subset R" instead of "A subset B subset R"

Thomas Finley

07:13

@leslietownes Yes, that might be the only possible explanation. If A is a subset of B, then it's very trivial to prove cl(A) is a subset of cl(B). This is easy as, A' is a subset of B'. So, cl(A) which is the union of A and A' and cl(B) which is a union of B' and B, implies, cl(A) is a subset of cl(B).

Koro

what's an example of a space X which is sequentially compact but not limit point compact?

@Jakobian

Are sequentially compactness and limit point compactness equivalent?

leslie townes

07:28

koro: topology.pi-base.org/…

Thomas Finley

@leslietownes What do you think about the example I cited here?

1

Q: Let $A,B\subset \Bbb R.$ Then prove that $\overline{A\cap B}\subset \overline{A}, \overline{B}.$

Let $A,B\subset \Bbb R.$ Then prove that $\overline{A\cap B}\subset \overline{A}\subset \overline{B}.$ Give example to show, $\overline{A\cap B}\neq \overline{A}\subset \overline{B}.$ I think, $\overline A$ denotes the closure of the set $A.$ Assuming this, notation, being used here, I feel the q...

real-analysis

Balarka Sen

I think probably the smallest uncountable ordinal with the order topology works, @Koro

A conceptually easier version is the one-point compactification of the long ray. The point at infinity is clearly a limit point, but not limit of any sequence

Any sequence, arranged in increasing order, stops at a countable ordinal

I have not thought very hard about this though

Koro

@BalarkaSen I think that's true by ordinal definition.

leslie townes

thomas: in your example, A and B are not disjoint, in fact B is a subset of A (e.g. because 1/k = 1/(2k) + 1/(2k))

Koro

I'll think about this.

Koro

07:52

31 mins ago, by Koro

what's an example of a space X which is sequentially compact but not limit point compact?

no such example exists.

because seq. compactness \implies l.p.c

I see why the converse may not be true at least in case the space is not 1st countable.

@BalarkaSen here, if one takes increasing unbounded sequence?

2

A: Comparison of sequential compactness and limit point compactness.

Sequential compactness implies limit point compactness in general. Let S be an infinite set. Since S is infinite you can have a sequence such that $x_n \neq x_m $ for all n,m integer. Sequential compactness implies this guy has a convergent subsequence and limit x of this subsequence is a limit ...

what is 'right order topology'?

I don't think (a, \infty) form a basis for some topology on R.

if they did, then (a,b] is open (being \cap of (a,\infty) and (b,\infty) assuming b>a) but for any $x\in (a,b]$, there is no c such that $x\in (c,\infty)\subset (a,b]$.

Balarka Sen

i havent thought about these things in a couple of years

id have to work to engage with your questions more meaningfully

Jakobian is probably the person to ask

Koro

I tagged him already. Let's wait for his inputs.

leslie townes

koro OP may mean subbasis instead of basis. have you tried parsing the example if that is the case?

Koro

I was just thinking about that

leslie townes

also, isn't the intersection of (a, infty) and (b, infty) just (b, infty) when b > a

Koro

08:07

oh my bad, I don't know what I was thinking

leslie townes

indeed isn't bigcup (a_i, infty) = (inf a_i, infty) and bigcap (a_i, infty) = (sup a_i, infty) so the thing is even closed under arbitrary (not just finite) intersections

Koro

oh, then I think we are done.

leslie townes

koro if you look at the site i linked above, it would not surprise me if it knows at least most of the examples from the 'counterexamples in topology' text, so you can often order it to offer up examples within that universe

Koro

That gives the example: limit pt. compact but not sequentially compact.

thank you so much

@leslietownes I looked at it but found the navigation there confusing :(.

(like they nowhere seem to be using the term 'limit point compactness')

leslie townes

OK. like some sources, they say "weakly countably compact" instead of "limit point compact."

you can click on any terms it uses to see what definitions it is using, and when it can't find an example of something with a combination of properties, it can often spit out a proof of why no such example exists.

if you do care about point set topology (and i'm not suggesting that you ought to, and might even recommend against it), it might be a good resource

Koro

08:19

@leslietownes thanks, I think you had shared that link earlier also here.

leslie townes

there used to be some other site which was less 'interactive' but maybe also easier to read. i've definitely linked that other one to somebody.

it was a huge chart, instead of this thing where you type in properties.

Thomas Finley

08:36

@leslietownes That 's admittedly a nice observation. Do you have any suitable example in your mind ?

leslie townes

someone has given one there. if you don't like "weird" topologies, you can find examples with subsets A, B of R with the usual topology where A intersect B is empty while cl(A) = cl(B) = R. you can even find such examples where B = R \ A, where the desired property is just "A is dense and its complement is also dense"

冥王 Hades

^My honest reaction when our professor cracks a “joke”

Thomas Finley

@leslietownes tbh, I only know about Basic Topology in R and the general notion of topology, topologic space, metric space are unknown to me at this stage.

But yes, I came up with an idea, looking at the answer you're talking about. If we consider A={a} and B={b} such that, a\neq b, then, A'( the set of limit pts of A) is an empty set, and so is B'(the set of limit points of B). So, cl(A)={a} and cl(B)={b}. Now, $A\cap B=\emptyset$ and $(A\cap B)'=\emptyset$ and so, $cl(A\cap B)$ is an empty set, but cl(A)={a} and cl(B)={b} and so, cl(A\cap B)\neq cl(A)\neq cl(B). Is this example simple enough ?

@leslietownes Yes, your example is nice as well.

But what do you think bout the one I gave now?

Shaun

09:10

Please would someone suggest improvements for the following (as it was downvoted and I don't know why)?

1

Q: What is the connection between algebraic groups and topoi?

I have a longstanding interest in topos theory. (See this MSE search of my questions about topos theory.) I am studying for a postgraduate research degree in linear algebraic group theory. Naturally, I wonder what connections there are between the two. A quick Google search produces this page, in...

group-theory category-theory algebraic-groups topos-theory

Jakobian

@Koro no, all types of compactness are not equivalent, but for $T_1$ spaces limit point compactness is just countable compactness

Koro

yes, figured.

Jakobian

You can find examples on pi-base like Leslie or counter-examples in topology where those examples are from

Most of the spaces on pi-base are from that book

Koro

Leslie: have you watched tv-show 'from'?

@冥王Hades

I thought there won't ever be any show like Haunting of hill house or Haunting of Bly manor, but from is so amazing.

it has that 'post-apocalyptic' kind of setting that I like.

Jakobian

Also for $T_4$ spaces countable compactness is just pseudocompactness

Koro

09:23

why is local connectedness formulated differently than local compactness?

Jakobian

What?

Do you mean connected in-kleinen

There are two inequivalent types of connectedness at a point

They coincide when they hold for all points

Oh wait I read connected twice

Koro

X is l. ctd at p if for every nbd. U of p, there is a connected open set $V\subset U$ containing p. X is l. cpt. at if p if there is a compact subspace C that contains a nbd. of p.

I am tempted to say: X is locally connected (resp. locally compact) at p if p has a connected nbd. (resp. compact nbd.).

But I see the problem with this: if X itself is connected, then X would become locally connected.

I know this is not true.

local compactness definition that I wrote above (the one involving C) is along the lines of the one given in Munkres.

Jakobian

Well there are different definitions of locally compact, equivalent for Hausdorff spaces

Koro

(note how it is different from connectedness)

Then I see the following definition for locally compact at p: if p has a compact nbd.

Jakobian

X is locally compact at p iff p has a neighbourhood basis of compact sets for example

Koro

09:31

further to my last message: this would imply every compact space is locally compact.

which I think is not true. (I'm sure that connectedness may not imply local connectedness)

and it feels weird to call an open set compact.

I'm convinced now that every compact space is locally compact.

Jakobian

Locally compact has like 4 definitions

X is connected im Kleinen at p iff p has a neighbourhood basis of connected sets

Koro

X is locally connected at p if p has a connected nbd. (why is this not correct?)

I know this is not correct, but what went wrong.

john doe

Does anyone have a hint on how to prove this?

Show that the set of real numbers that are solutions of
quadratic equations $ax^2 + bx + c = 0$, where a, b, and c
are integers, is countable.

Koro

@johndoe given a,b,c integrs, what is the maximum no. of solutions of the equation?

(a,b,c)\in Z^3 and Z^3 is countable.

john doe

well in any case if its a quadratic equation the number of solutions is always two right?

Koro

09:39

atmost two

(but that doesn't matter here. only finiteness is enough.)

Jakobian

I mean, why expect an equivalence for connected sets when they don't have all the properties that compact sets do

Koro

because I was expecting some uniformity in the definitions involving 'local'.

Jakobian

Anyway the proper way to say X is locally connected at p is to say it has a neighbourhood basis of open connected sets

Koro

something like: X satisfies property P locally at p if there is a nbd. U of p on which X has property P.

@Jakobian .

Jakobian

The definition of locally compact you cite is just the easiest to check

john doe

09:42

@Koro @Koro indeed $Z^3$ is countable but I need to prove that the set of real solutions is countable. How can I do that>

Jakobian

The more "canonical" one is using neighbourhood basis of compact sets

Koro

@johndoe suppose S(a,b,c)= the set of solutions of ax^2+bx+c=0. You agree that S(a,b,c) is finite. Consider the countable union $\cup_{(a,b,c)\in Z^3} S(a,b,c)$. So you have countable union of countable sets. What you are looking for is a subset of this.

Jakobian

So it'd be more like: X satisfies property P locally at p if there is a neighbourhood (or open neighbourhood) basis U of p with property P

Koro

3 mins ago, by Koro

something like: X satisfies property P locally at p if there is a nbd. U of p on which X has property P.

Jakobian

I typod I meant basis

Koro

09:45

yeah, makes sense.

Jakobian

Anyway open in this definition varies like with connected im Kleinen and locally connected

Koro

I don't yet know im Kleinen

Jakobian

For something like compactness we can't really expect for U to be open

@Koro well I explained it like 5 times above there

Koro

@Jakobian ..

Jakobian

12 mins ago, by Jakobian

X is connected im Kleinen at p iff p has a neighbourhood basis of connected sets

Koro

09:48

so connected im Kelinen = locally connected.

Jakobian

No

At a point they differ

Koro

where?

Jakobian

There are spaces with points connected im Kleinen but not locally connected at that point

Koro

locally connected for me is: $X$ is locally connected at p if for every nbd. U of p, there is a connected open subset V (subset of U) containing p.

Jakobian

Yes

Koro

09:51

collection of all such V's give me the nbd. in im Kleinen. no?

Jakobian

A neighbourhood need not be open

Koro

of course, I should say $V_U$'s.

ohh, nbds. are open for me.

Jakobian

Then compact neighbourhoods would be open too

Koro

although, I understand what you are referring to. S is a nbd. of p if S contains an open set containing p.

but I don't use that definition. I think it is Nicholas Bourbaki's definition.

Munkres also doesn't use it.

Stephen Willard does.

@Jakobian allowed in the definition in the book I'm studying currently.

But I see that will create problem if X is Hausdorff.

john doe

@Koro @Koro Noted. I have one more notation question though. Is $S$ a function?

Sorry for the double mention I'm still figuring out how the chat works

Koro

09:55

X is Hausdorff + locally compact would often mean X is not connected as the compact nbds. would be clopen, I think.

Jakobian

Anyway those are all the differences between locally property P, either neighbourhood basis of open sets, or just neighbourhood basis, all satisfying property P

Koro

@johndoe does it matter? But if you like: S can be considered from Z^3 to a set of sets.

so that S(a,b,c) is a set (you know it is finite as we discussed above).

john doe

do I need to prove that the set of the polynomials of this form is countable?

Koro

but that's set-isomorphic to Z^3, right?

22 mins ago, by Koro

@johndoe suppose S(a,b,c)= the set of solutions of ax^2+bx+c=0. You agree that S(a,b,c) is finite. Consider the countable union $\cup_{(a,b,c)\in Z^3} S(a,b,c)$. So you have countable union of countable sets. What you are looking for is a subset of this.

countability of Z^3 is enough here.

@Jakobian I should use this definition.

(S is a nbd. of p if S contains an open set containing p.)

Jakobian

10:12

It's usually better imo. There are some instances when you want to assume open, but then you can just say open neighbourhood

Jakobian

10:22

I think I had a moment like this once "here it's important to assume this neighbourhood is open"

When writing a proof

Jam

Suppose k a complex root of an irreducible polynomial over Q. Does Q(k) contains also the complex conjugate of k?

i mean if a+bi in a field then $(a+bi)^{-1}=\frac{1}{a^2+b^2}(a-ib)$ is in the field

but to get that i multiplied $(a+bi)^{-1}=\frac{a-bi}{(a+bi)(a-ib)}$ which i dont know if it is an element of the field tho it is equal to $1=\frac{a-bi}{a-bi}$

Dannyu NDos

10:48

Is there a dedicated name for prime numbers of the form $n!+1$?

They gotta be good for performing Schönhage-Strassen algorithm.

27! + 1 is prime, for example.

Jam

search factorial primes

Thomas Finley

11:23

What does does it mean when they say"Let $ \sum u_n$ and $\sum v_n$ be two series of positive real numbers". Do they mean the terms of the sequences, $(u_n)$ and $(v_n)$ are all positive ? Or are they trying to say, all the partial sums of each series i.e $\sum u_n$ and $ \sum v_n$ are positive ?

Jakobian

11:52

the former

a series is just a sequence that we treat differently

Thomas Finley

@Jokobian You mean, that the phrase in general (say) "Let $\sum u_n$ be a series of positive real numbers", implies that the sequence generating the series, i.e , $(u_n)$ is such that all the terms of the sequence $(u_n)$ is positive i.e u_k\geq 0, \forall k\in \Bbb N$ ?

@Jakobian Yes, exactly

@Jakobian But it seems like the later one i.e the phrase means that the terms of the sequence of partial sums of (u_n) i e $(s_n)$ (defined by $s_1=u_1,s_2=u_1+u_2,s_3=u_1+u_2+u_3$ and so on) has all it's terms positive, i.e $s_n\gt 0$, $\forall n\in \Bbb N$.

Jakobian

why did you mention me 3 times in a row

Thomas Finley

12:18

@Jakobian becoz you mentioned something which seemed to be lacking context and details. That was confusing (to me)- A simple answer(?)

Jakobian

do you know what "the former" means

Thomas Finley

@Jakobian There is an option, called "ignore this user" when you click on someone's chat icon in here. You might consider enabling it, for me. ;)

Jakobian

you're jumping to conclusions

Thomas Finley

@Jakobian huh? Wdym?

@Jakobian I now understand that :

26 mins ago, by Thomas Finley

@Jokobian You mean, that the phrase in general (say) "Let $\sum u_n$ be a series of positive real numbers", implies that the sequence generating the series, i.e , $(u_n)$ is such that all the terms of the sequence $(u_n)$ is positive i.e u_k\geq 0, \forall k\in \Bbb N$ ?

is the right inference.

But needed only a clarification from you, in terms of a yes or no.

I felt that would make the communication much simpler and would prevent it from going haywire.

Jakobian

well that depends if you're using the French convention or not

$x$ is positive can mean either $x\geq 0$ or $x>0$

Thomas Finley

12:27

@Jakobian Ah, now that you mention, in my comment I made a latex error, i.e it should be $x\gt 0$ and not, $x\geq 0$

Jakobian

12:39

@ThomasFinley you're trying to pick fights with me for no reason at all. I don't like this attitude

Jakobian

12:54

Jun 3 at 17:02, by Jakobian

$X$ is Lindelof implies $e(X) = \aleph_0$ where $e(X)$ is the extent of $X$

@SouravGhosh turns out spaces with $e(X) = \aleph_1$ are called $\aleph_1$-compact

ZaWarudo

In one of the solutions given by my lecturer, there is written: "$\int_1^{+\infty} \frac{1-\cos x}{\left(\sqrt{1+x^2}-1\right)\arctan \sqrt{x}}dx$ diverges because $\frac{1-\cos x}{\left(\sqrt{1+x^2}-1\right)\arctan \sqrt{x}}=o\left(\frac{1}{\sqrt{x}}\right)$ as $x \to +\infty$". However, I don't think this is correct; because

$\int_1^{+\infty} \frac{1}{\sqrt{x}}dx$ diverges.

Is the solution incorrect or am I wrong?

Jakobian

See this paper

ZaWarudo

I mean, $\frac{1-\cos x}{\left(\sqrt{1+x^2}-1\right)\arctan \sqrt{x}}=o\left(\frac{1}{\sqrt{x}}\right)$ as $x \to +\infty$ is true but this means that for "big positive $x$" we have $\frac{1-\cos x}{\left(\sqrt{1+x^2}-1\right)\arctan \sqrt{x}}<\frac{1}{\sqrt{x}}$, but this does not imply the convergence of $\int_1^\infty \frac{1-\cos x}{\left(\sqrt{1+x^2}-1\right)\arctan \sqrt{x}}dx$.

sunny

I am stuck on this one definition of neighborhoods.

> Let $X$ be a metric space . A set $U\subset X$ is a neighborhood of $x\in X$ if $B_r(x)\subset U$ for some $r>0$.

Now, is a neighborhood necessarily an open set? It seems not.

Jakobian

It doesn't have to be

sunny

13:01

Right, thanks @Jakobian.

Rudin for example defines neighborhoods differently, i.e. his definition of neighborhood is equivalent to an open ball.

john doe

I need to prove that the function $ f: Z^+ \times Z^+ \rightarrow Z^+$ is injective. It's defined as $f(m,n) = \frac{(m+n-2)(m+n-1)}{2} + m$. I tried using $(a,b) \ne (c,d) \implies f(a,b) \ne f(c,d)$ and got

$a^2 + 2ab - a + b^2 - 3b = c^2 + 2cd - c + d^2 - 3d$.

Does anyone have an idea how to proceed?

Thomas Finley

13:17

@Jakobian Don't get me wrong. I never tried picking up fights with you. Moreover I appreciate how you helped me get over the dilemma. This is the problem of online media as they say, "Tone is hard to decipher."( Sheesh... I was at one point, simply joking with you, haha.)

Jakobian

actually this is ambiguous because a lot of people use $\aleph_1$-compact to just refer to Lindelof spaces

Thomas Finley

My impression to you is not hostile in any way and forgive me, if you felt it that way, but never did I intend it. Oh, and here's to it, you were correct, I was getting confused by convolutions! Thanks a lot! Cheers!!!

Shaun

14:32

5 hours ago, by Shaun

Please would someone suggest improvements for the following (as it was downvoted and I don't know why)?

5 hours ago, by Shaun

1

Q: What is the connection between algebraic groups and topoi?

I have a longstanding interest in topos theory. (See this MSE search of my questions about topos theory.) I am studying for a postgraduate research degree in linear algebraic group theory. Naturally, I wonder what connections there are between the two. A quick Google search produces this page, in...

group-theory category-theory algebraic-groups topos-theory

shintuku

15:31

sounds like mathoverflow material btw

PNDas

I want to find $\frac{\partial}{\partial t}\int_0^t f(x+(t-s)w,s)\,ds$. Can I apply Leibniz rule? $f$ is a function of two variables?

Here consider $x,w$ to be constants.

Ted Shifrin

@PNDas That’s precisely what the rule is about.

PNDas

So the derivative is $f(x,t)+w\int_0^t f_t(x+(t-s)w,s)ds$?

Bill Dubuque

15:49

Today Scientific American published an article about the definite integration "oracle" Cleo and an interview with Ron Gordon on the matter. Nothing at all new for those familiar with the matter - simply a popularized account.

3

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$Mathematics$ Mathematics