Mathematics - 2023-02-19

Sine of the Time

00:24

chat sleeping today :(

ペガサス Seiya

02:00

Koro

03:59

@Thorgott Ohh I don't understand how.

Almost day 2$\frac12$ on the same question

Koro

04:18

I don't know why the link [,] (,) didn't render there.

Koro

04:31

It seems that my question is such that everyone but me knows the answer of but it's difficult/impossible to explain.

That explains why no book I saw goes into details of this. I looked into Munkres', Marco Manneti's, Rotman's, Breden's, Hatcher's, Armstrong 's, Lee' s manifolds etc. but none of these go into the details of my question.

Prithu Biswas

@Koro Hello!

Does the nested interval theorem hold true if we replace "nested intervals" with "nested sets of real numbers"?

Koro

@PrithuBiswas Hi Prithu

@PrithuBiswas It need not. Take $I_n=(0, 1/n)$. Then look at the intersection $\cap_n I_n$.

Prithu Biswas

04:52

@Koro Oh that is a nice counterexample =)

Cotton Headed Ninnymuggins

06:02

If I'm going to buy 20 donuts from 6 different flavors, I believe there is ${20+5 \choose 5}$ ways to do this (stars and bars). What happens when I require $x$ of each flavor? Lets

... say I need $1$ of each flavor, is the problem just a question of grabbing $20-6\cdot 1=14$ donuts of $6$ flavors?

SAJW

06:25

Can a Infinite Powertower equal a negative Numbers? I assume yes with a complex x, but I can Not find anything on the topic.

Also have a A1 day

robjohn

@Koro isn't this a way?:

$$
\begin{align}
F(0,t)&=(0,3t)&0\le t\le\tfrac13&&&F(0,t):\left[0,\tfrac13\right]\leftrightarrow\gamma\\
F(0,t)&=(3t-1,1)&\tfrac13\le t\le\tfrac23&&&F(0,t):\left[\tfrac13,\tfrac23\right]\leftrightarrow\beta\\
F(0,t)&=(1,3-3t)&\tfrac23\le t\le1&&&F(0,t):\left[\tfrac23,1\right]\leftrightarrow\delta^{-1}\\
F(1,t)&=(t,0)&0\le t\le1&&&F(1,t):[0,1]\leftrightarrow\alpha\\
F(s,t)&=(1-s)F(0,t)+sF(1,t)
\end{align}
$$

Koro

06:44

@robjohn I think that you mean H instead of F.

A. P.

Hi

robjohn

@Koro whatever letter... I'm just trying to find the map between the curves.

A. P.

I have a question about terminology. Does it make any sense when referring to a recurrence that grows very fast and is eventually bounded to say something like "the succession eventually stabilises"?

Koro

@robjohn Actually, I am not sure. I looked at your maps. I didn't understand them :(.

A. P.

I meant "sequence eventually stabilises"

Something similar like: "sequence eventually stabilises"= "the sequence a_n is constant for some large n"

?

robjohn

06:57

@Koro what don't you understand?

Koro

It may be that I have mixed up symbols. But F(0,t)=the curve of left edge of the square, which is $\gamma$.

And you seem to be defining this already defined thing again.

robjohn

@Koro No. F(0,t) is the left edge of the square for $0\le t\le\frac13$

Koro

And then the first line says: F(0,t):[0,1/3] <-> $\beta$. I don't understand what this means. Because the left edge is $\gamma$ so I don't understand how it is related to $\beta$.

robjohn

So $\beta$ and $\gamma$ are backwards

There. fixed

Koro

Happy to inform that my other two questions have been answered (one was answered above earlier, and the second one in the comments under my post on mse).

robjohn

07:07

The right hand column is a comment

Koro

@robjohn Now, this is the only remaining one that I want to understand.

Using the maps suggested, I define H(0,t)= F(0,3t) for t<=1/3. But since we require this to be a constant quantity for all t, it is not giving us that.

Koro

07:29

Viro, Ivankov topology book also doesn't go into the details of my question.

it looks like an open problem to me 😅.

Koro

08:00

I see a closely related lemma as lemma 7.12 in Lee's book, of course left as an exercise.

PNDas

08:38

Any idea what is the set in math.stackexchange.com/questions/1254176/… ?

Okay got it.

robjohn

@Koro why does it have to be a constant for all $t$?

Koro

08:54

because we are proving two maps say f and g to be path homotopic.

so they have to be fixed at s=0 and at s=1 for all t.

the question is eligible for bounty in 15 hours. Can the bounty be placed before 15 hrs?

robjohn

No, it is the other way around. $t$ is the variable that traces out the curves. $s$ is the variable that moves from one curve to the other.

Koro

or one must wait for 15 hrs is the only way. There aren't any exceptions?

Here is the definition that I am using: Suppose that f,g:I-->X are two paths in X. We say that f and g are path homotopic if there exists $F:I\times I\to X$, continuous such that $F(0,t)=f(0), F(1,t)=g(1)$ for all t in I, and that $F(s,0)=f(s), F(s,1)=g(s)$ for all s in I.

In my question: f= $\alpha, g= \gamma. \beta. \delta^{-1}$.

PNDas

I'm looking for an example of complete metric space X which has a countable $G_\delta$ subset which is dense.

Any ideas?

D Left Adjoint to U

@Koro found a twin, or even general $2k$-separated prime average pattern:
https://math.stackexchange.com/questions/4642118/conjecture-if-n-in-bbbn-setminus-1-is-not-a-twin-prime-average-then-n2

It turns out $n^2 - 1$ is square-free if and only if $n$ is a twin prime average!

Conjecture inside for the general case.

One direction is immediate since $\gcd(p_n, p_{n+1}) = 1$. The other direction is unproven or at least I don't know of the proof yet

Koro

@DLeftAdjointtoU Hi! I've been stuck at a homotopy problem for about 2 days now. Unfortunately, still no progress on that.

I wanted to do number theory this weekend (which has almost passed) but got occupied by this homotopy exercise.

D Left Adjoint to U

09:07

$n^2 - 1 = pq \iff n \pm 1 = p,q$ is a twin prime average (we know this much)

robjohn

09:44

@Koro bounty system

Koro

09:55

thanks :-).

robjohn

10:09

@DLeftAdjointtoU if $n=16$, then $n^2-1=3\cdot5\cdot17$ is square-free, but $16$ is not a twin prime average.

$n=34$ is another example where neither $33$ nor $35$ is prime.

ILikeMathematics

10:26

Is this proof valid?
Proof that $0 \leq a < b$ and $0 \leq c < d \implies ac < bd$.

We denote $\mathbb{R}^+$ as $P$.

If $a < b$, then $ca < cb$ for $c > 0$, since $0 < b - a$, $c > 0$, thus $c \in P$ and $(b - a) \in P$.
The product of two elements in $P$ is in $P$, thus $c(b - a) > 0 \implies cb - ca > 0 \implies cb > ca.$

Since $a < b$, we have $ac < bc$ and $bc < bd$ since $c < d$, both from $a < b \implies ca < cb$ for $c < 0$.
Thus $ac < bc < bd$.

Would you change anything in it?

ILikeMathematics

11:03

It seems cases are needed, since $c$ is not always in $P$.

Proof that $0 \leq a < b$ and $0 \leq c < d \implies ac < bd$.

Case 1. $a = 0$, $c > 0$

$ac = 0$, $b \in P$ and $d \in P$, so $ac = 0 < bd$.

Case 2. $a > 0$, $c = 0$

Interchange $a$ and $c$ in Case 1.
$ac = 0 < bd$.

Case 3. $a = 0$, $c = 0$

$ac = 0 < bd$.

Case 4. $a > 0$, $c > 0$
We denote $\mathbb{R}^+$ as $P$.

If $a < b$, then $ca < cb$ for $c > 0$, since $0 < b - a$, $c > 0$, thus $c \in P$ and $(b - a) \in P$.
The product of two elements in $P$ is in $P$, thus $c(b - a) > 0 \implies cb - ca > 0 \implies cb > ca.$

Thorgott

13:12

@Koro you will have to actually ask specific follow-ups if you want to get anywhere

Franklin

14:52

Can anyone please help me with this : math.stackexchange.com/questions/4642260/… ?

Jai Sri Krishna

15:42

Can someone help me with factorising $(1-x^n)$ with (1-x) as a factor?

I want to find the coefficient of $x^11$ in $(1+x^2+x^3+x^4+x^5)^5$

How can I do it?

Please help me I have my exam tomorrow :-(

Franklin

15:58

Can anyone tell me how to post pictures here?

@JaiSriKrishna well, $(x^n-1)=(x-1)(x^{n-1}+...+1)$, so multiplying -1 on both sides...

For the second part, try to factorize $(1+x^2+x^3+x^4+x^5)^5$...

Jai Sri Krishna

Oh So it should be $1-x^6$/1-x ?

Right?

Franklin

16:13

I didn't calculate it, though....

Koro

0

Q: Constructing a homotopy in a square

Let $F:I\times I\to X$ be continuous, $\alpha, \beta,\gamma, \delta$ be paths in $X$ as below Thus, $\alpha(t)=F(t,0), \beta(t)=F(t,1),\gamma(t)=F(0,t), \delta(t)=F(1,t)$. Then it is to be proven that $\alpha\simeq \gamma\circ\beta\circ \delta^{-1}$ rel {0,1}. Here $\circ$ represents concatenatio...

Can anyone please explain to me how to get the homotopy here?

@Jakobian you may be interested in this. Please advice how to get the homotopy when you have time. Thanks.

Franklin

@Koro Can you please help me wigh a query if you don't mind: how to post images here?

Koro

near the typing space in desktop version of chat, you'll see an upload button.

Franklin

Can anyone help me understand this? 😔

@Koro thanks!

Koro

np

Franklin

16:21

I actually posted this in the main site, but I think I will delete it now, since it's garnering downvotes

Actually, there in that image I dont get whether the increment $\Delta y$ as mentioned in the beginning, be any increment ? Like $\Delta y\in \Bbb R$ , or precisely it can be any real number, right? This is the thing i wanna know?!?

Thorgott

16:48

Koro, you've received multiple comments from both others and me already. I understand the situation may be frustrating, but just asking again without asking more precisely is not going to help.

Koro

16:59

Ohh I wanted Jakobian's view on this. And I have explained my confusion already. I have received an answer to my post, I discussed with the answerer in detail but I'm not understanding how they got the map using the picture.

leslie townes

franklin: it's basically, and with not a lot of precision, trying to "define" (that might not be the right word) the derivative. it might be getting downvotes because at many schools, understanding that is the subject of at least half of a year-long calculus sequence.

Koro

If you're interested in the discussion, you might just see the 'moved to chat' section below the answer. I don't want to put the whole discussion here again.

If not, then I hope Jakobian gives his view on the post.

I think my question is precise and is to the point.

I'm not in any hurry though. Let it take years to get the answer.

leslie townes

koro, one reason for the low interest may be that this is the kind of thing where the exact formulas are not the point, and two different people who approach the exercise may come up with different answers. so "explain how this one person chose to do it in this one way" is maybe not that fun to answer, at least when it is not addressed to that one person.

questions where there is more 'uniqueness' (not quite the precise word) in the solution are more likely to be of broader interest.

Koro

Leslie, I don't want a unique answer. I understand that there are several homotopies which will do the trick (two of them listed in my post). But I want to create atleast one of those many possible homotopies using the picture. That's all. 'Write H like this, and verify that this homotopy works' is not what I am looking for. Had it been the case, I wouldn't have posted my question as I already have the answer then.

leslie townes

OK, and maybe that's almost too much non-uniqueness in the answer for people to want to help. the problem does seem to be illustrating that a picture is worth a thousand words.

i don't care one way or the other, and wish you good luck :D

Thorgott

17:12

well, Thomas Andrews' comment and my follow-up thereto are left unaddressed and they yield one way of constructing such a homotopy (a way that you already proposed yourself two days ago in this chat when we first discussed the question, which is why it irks me that it's being ignored now)

Koro

I did address Thomas' comment and I said that it doesn't work because X is not a subset of R^n so I didn't understand how convexity makes sense there.

and I also replied to your comment suggesting to assume F= identity, to which I received no feedback if it was right or wrong.

So I thought it would have been wrong.

Thorgott

I left a comment on the question yesterday explaining that

Koro

@leslietownes it's in between. There are two maps there already. I hope I get to understand atleast 1 of those. :D

with respect, I did not understand the comment at all. I still don't understand what you mean by that.

I hope there is some other way.

Franklin

@leslietownes Exactly! But I have just one question: Can $\Delta y$ be any real number, (since this is the implied sense in the book), such that for each and every such $\Delta y$, if the function $y=f(x)$ can be expressed as $\Delta y=f (x+ \Delta x)-f( x)= A(x)\Delta x+\alpha(x,\Delta x)\Delta x,$ where $\lim_{\Delta x\to 0}\alpha(x, \Delta x)=0,$ then such a function is called differentiable at the point $x.$ Is this what is meant?

leslie townes

17:27

uh, there's a little too much going on in that question for me to answer. i'd expect that (as is pretty common in these discussions), x and Delta x are to be understood as 'independent variables' there, and Delta y just is whatever it is, as determined by that equation.

what it can be would, in general, be determined by what f is, as well as the choices of x and Delta x.

for fixed x, the structure of A and alpha will really only be a focus of attention for small Delta x. they're not interesting functions for large Delta x or interesting multivariable functions in general.

Thorgott

@Koro that's fine, but then ask a follow-up

the point is to think of the problem in two steps

you have a path in $I\times I$ that traverses the bottom edge, then apply $F$ to that to get the path $\alpha$ in $X$

you have a path in $I\times I$ that traverses the left edge, then the top edge, then the right edge. applying $F$ to that path yields $\beta.\gamma.\delta^{-1}$ (or whatever the order of the variables was).

Thomas Andrews comment points out those two paths in $I\times I$ are already homotopic

and that of course means that if you apply $F$ to those homotopic paths, the results are homotopic

that was also what I meant by my "pretend F=id" comment a while ago that I never elaborated on (sorry)

if you can do it in the interval, it also follows everywhere else. the two paths in the interval are a "universal" example, so to say.

Franklin

@leslietownes Ok, so for $\Delta x$ being any real number, we must have for each and every such $\Delta x$, if the function $y=f(x)$ can be expressed as $\Delta y=f (x+ \Delta x)-f( x)= A(x)\Delta x+\alpha(x,\Delta x)\Delta x,$ where $\lim_{\Delta x\to 0}\alpha(x, \Delta x)=0,$ then such a function is called differentiable at the point $x.$ is this what is meant?

leslie townes

17:44

sure. the person who wrote it may not be expecting to, uh, use that definition for very much. the key thing you might get out of it now is that (rearranging that slightly), [f(x + Delta x) - f(x)]/[Delta x] = A(x) + alpha(x, Delta x), where A(x) is something that depends only on x, and alpha(x, Delta x) goes to 0 as Delta x goes to 0.

that's a form in which it might more directly relate to a definition that you have seen before in that text, or elsewhere in some other text.

we can all ponder whether it's a good idea pedagogically to introduce both the derivative and differentials in the same paragraph.

Thorgott

(the answer is no)

Franklin

@leslietownes but then $A(x)$ is nothing but $f'(x)$ ?

So basically, they are asserting that if one can write $f(x+ Delta x)-f(x)=A(x)Delta x+Alpha(x,Delta x)Delta x$, such that Alpha(x,Delta x) goes to 0 as Delta x approaches 0, for all x in the domain of f then, we can say f is differentiable at $x$ , right?

Koro

hahaha, I saw one very funny username on mse.

after so much time, finally I understood the answer to my question. I have accepted the answer now :-).

leslie townes

18:00

franklin: yes, A(x) is f'(x). this definition doesn't seem to be caring too much about domains, and may only be focused on those equalities holding (x, Delta x) where x is in the domain of f and Delta x is small enough that x + Delta x is also in the domain of f, or maybe even Delta x smaller than that, but i don't know, and my guess is at this stage in the exposition, the author doesn't care.

Koro

@Thorgott yes, I know that result- re-parametrization does not affect the homotopy (path).

leslie townes

maybe they could have only one x in mind. i don't see any of the usual hand-wringing about domains that you might see in a more modern treatment, so it probably doesn't matter.

Koro

@Thorgott I'll think about this one. Thanks :-).

I was expecting that it would take me 3 years to understand the homotopy from the picture. Don't ask how 3 years was calculated. I was ready to memorize the maps across the way.

But fortunately, I understood it today itself.

Thorgott

there's no reason to ever memorize dumb formulae like that

Franklin

@leslietownes But is $A(x)$ the same for all x?

leslie townes

18:05

franklin: it's important that A(x) depends only on x, yes.

if that's what you mean by "the same for all x"

Koro

why are there downvotes on this?

the OP has provided context, I think. There are close votes too on the question.

Franklin

@leslietownes Actually,I meant to say, $A(x)$ is a single function , say $A(x)=x^2$ for every $ x$ considered.

Thorgott

the comments seem to indicate that pretty clearly

leslie townes

"why are there downvotes on this" is not really something the chat can answer without speculating

Koro

let's ignore the downvote part. It has no answer.

leslie townes

18:09

franklin: yes, A(x) is a function of x alone (and it does not for example depend on Delta x).

Koro

But the close votes may have valid answer.

leslie townes

i don't know. it seems like they didn't put a lot of effort into making the question readable. i forget if that's a bucket you can throw a 'vote to close' into.

the OP doesn't define "BV" or "T_F", and doesn't say what parts of the argument that they do understand. only what they don't. and even that isn't organized very well.

if you were the OP, koro, me saying this on chat might have a point, but as you're not, it's just me filling time. :D

Koro

hahaha

I can have dinner in piece now.

leslie townes

given that the only hypothesis of the lemma is that f is in BV, and that all of the conclusions of the lemma involve T_F, the definitions of BV and T_F seem, uh, somewhat conspicuous by their absence.

ok, he's put the definition of T_F in there, sort of. you have to wade through some confusion to get there. that should have come first.

"lacking details and clarity," if that is one of the close buckets, seems OK to me. i didn't vote to close, but i can see how someone would.

Thorgott

yes, and that was only added after people complained in the comments

and it's still image-based, which people don't like

questions should optimally be so self-contained that someone with experience in the subject can read them and immediately figure the answer, this is very much not close to that

leslie townes

18:20

i don't care so much myself about the image thing, although i understand why people do. for example, it is an accessibility nightmare. it does contribute to a general feeling that if you somehow don't have the time to pose the question, maybe it's not yet in a state where people can reasonably spend time on answering it.

has anyone ever posted a problem in video form? is that even possible? like you click in and its someone in selfie mode saying some stuff about their problem, and then holding their phone to a book ro screen so you can see some equations.

that would amuse me more than it would annoy me. i might take the time to answer that question.

Thorgott

it's also the easiest low-effort signifier

video form would be hilarious

leslie townes

maybe they're driving somewhere while they make the video

D.C. the III

19:09

Is there a difference between the notion of a complex number and an imaginary number?

leslie townes

the former is a little less ambiguous, and more inclusive. sometimes people use 'purely imaginary number' to refer to those nonzero complex numbers that have zero real part.

"imaginary number" is not really a term of art, though. i don't think you would see it defined in most complex analysis books.

D.C. the III

I ask this because I have been given that $T$ is a normal operator and I have to use the spectral theorem to prove that $T = -T^* iff every \lambda_i$ is an imaginary number

leslie townes

ok, they mean "a real multiple of i" there.

what i was calling "purely imaginary" above, except they are also including 0 in that.

D.C. the III

I had written out a proof and was stuck, but apparently where I was stuck was the conclusion. So assuming $T = -T^*$ and applying the spectral theorem I will end up at $\sum \lambda_i T_i = \sum -\bar{\lambda_i} T_i$. And apparently the fact that $\lambda = -\bar{\lambda_i}$ is enough to conclude it is imaginary

leslie townes

sure, yes. with a, b real, the conjugate of a + bi is a - bi. the negative of a + bi is -a - bi. these are equal iff a = -a and -b = -b, i.e., if a is zero.

i should say that i don't quite know what's going on with your T_i and T_j abnove, but i assume at some point you're diagonalizing this thing and getting to something that is the equivalent of comparing diagonal entries.

D.C. the III

19:18

Ok that decomposition makes more sense to me. I was getting lost in what it meant to be an imaginary number

leslie townes

too many darn ways to phrase the spectral theorem.

replace with "zero real part" or equivalently "lambda-bar = -lambda" and you're good to go.

D.C. the III

The $T_i$ are orthogonol projections, so yea. those are the eignevalues

of T

leslie townes

the less familiar analogue of a complex number being 'real' if it coincides with its conjugate.

that's another way to do the exercise, really. if T satisfies your condition then iT will be self-adjoint, and hence maybe already known to have real eigenvalues. so T has eigenvalues of the form real/i, which are purely imaginary. same characterization, via a different route.

D.C. the III

Well now that you mention it, the self adjointess was how I first thought of the question. But I thought the negative sign in front of my adjoint nullified self adjoint being true.

Wait...what.....that last sentence has me confused.,,,if the eignevalues are all real due to self-adjointness how are they imaginary?

robjohn

@Koro: I fleshed out my answer and added it to your question.

leslie townes

19:24

an operator that satisfies T = -T* is not generally also going to be self adjoint. but iT will be self-adjoint, and so you can apply what you know about self adjoint operators to iT, then divide everything by i to learn what that says about T.

if k is an eigenvalue of iT, then k/i is eigenvalue of T. if k happens to be real, then k/i will be purely imaginary.

D.C. the III

All makes sense except for how you got "i" in front of $T$.

leslie townes

no reason except that i wanted to turn something that was skew adjoint into something that was self adjoint. analogous to how if z is a purely imaginary number than iz is a real number.

"multiplication by i" sends the imaginary axis to the real axis.

Koro

@robjohn thanks a lot :-). I'll check out the details soon.

robjohn

@Koro I hope it helps. I have not actually taken a class which covers homotopy.

leslie townes

a good analogy to have in mind is that normal operators are kinda like complex numbers (because they diagonalize to things that behave algebraically like tuples of complex numbers, or more generally, complex-valued functions). operators satisfying S* = S are like complex numbers z with z* = z (real numbers), operators satisfying S* = -S are like complex numbers z satisfying z* = -z (purely imaginary numbers), and the algebraic tricks carry over.

e.g. the suspicion that iT will be self adjoint if T is skew adjoint turns out to be true.

D.C. the III

19:31

This is something for me to ponder and internalize. Thanks for the tip and the help

Koro

@robjohn: how did you make that image?

robjohn

@Koro In Mathematica

D Left Adjoint to U

@Koro I can no come really close to proving twin primes, want to hear it?

Koro

@robjohn Ohh. That's so cool :-).

D Left Adjoint to U

My previous post is unrelated...

Well note that $x$ is a twin prime average if and only if $\gcd(x^2 - 1, p_1 \cdots p_n) = 1$ for $n = \pi(sqrt{x + 1})$ the prime counter function

Silly Goose

19:40

i am reading through baby rudin. is it correct to say that a closed ball around point $p$ with radius $r$ in $\mathbb{R}^1$ is a neighborhood around $p$ with radius $r$ in $\mathbb{R}$ with a particular choice of metric?

in other words, it seems that neighborhood is more general a term than ball. it also seems from some stack answer on this site that a ball is of a particular shape(?) as opposed to a neighborhood being more abstract(?) or not having to be a particular shape. Is the "shape" given rise to by the choice of metric?

Koro

@DLeftAdjointtoU: Have you studied Ireland and Rosen's number theory book?

D Left Adjoint to U

@Koro yes long time ago

never finished

Koro

I'm at chapter 4 in that right now.

Silly Goose

oops i meant open not closed

leslie townes

silly: different books use the term 'neighborhood' differently. there isn't just one use of that term in use.

silly: as one example, for some people, a 'neighborhood' of a real number p would be any subset of R that contains an open interval around p.

Silly Goose

19:44

D:

D Left Adjoint to U

@Koro do you have any questions for me?

I might be a little rusty

leslie townes

which is, as you kind of point out, a broader notion than just a ball, whether open or closed, although open or closed balls of positive radius centered at p would be examples of 'neighborhoods' of p in this sense.

robjohn

@leslietownes I usually see "neighborhood" as an open set, but some texts say "open neighborhood"

Koro

@DLeftAdjointtoU not right now. What I wanted to say was: you seem to know a lot about Number theory, I don't know it yet as much as you do. :(

leslie townes

more generally i think it is fair to say (irrespective of details of particular definitions) that 'ball' has more structure (e.g. determined by a choice of metric) and "neighborhood" could have less structure.

robjohn: yes. in at least one of the pre-"open set" axiomatizations of topology, "neighborhoods" of p weren't required to be open, and some books have carried that over and some haven't.

D Left Adjoint to U

19:46

@Koro i just know some arithmetic functionology

Koro

@DLeftAdjointtoU So I didn't know how to add my input to this one as I have no idea.

Silly Goose

okay i see. so is it safe to associate the metric with endowing shape? and so endowing the set with the quality of being a ball or not a ball

D Left Adjoint to U

Oh i'm in the process of working it out, could be wrong though

leslie townes

silly: in the specific example of rudin, who works with the euclidean metric almost exclusively (maybe exclusively, outside of chapter 2), the "balls" as rudin defines them do resemble what we think of as a structure of a ball.

if you fiddle with the metric, a "ball" as defined via rudin chapter 2 might or might not capture what you would think of as a "ball" in an informal sense.

D Left Adjoint to U

@Koro here's a related post to it. Just requires knowledge of factorial:
https://math.stackexchange.com/questions/4642421/under-what-conditions-is-dfracp-n1-3p-n-1-1p-n-1p-n?noredirect=1#comment9799412_4642421

Silly Goose

19:48

oh i am thinking of ball as whatever the generalization of 3d ball is

to all dimensions

leslie townes

silly: again, it depends on what aspects you are including in "generalization" there. it all comes back to definitions.

Silly Goose

hm i see i suppose i should keep reading on then :P

leslie townes

i think the definitions you get with the euclidean metric are very close to what i intuitively think of 'balls' as being. but with other metrics, less so, or maybe even not at all.

Silly Goose

oh i see

wait well then is the quality of ballness dependent not on the metric, but on the constraint you put on what elements can be contained in your set (which is defined in terms of an inequality perhaps and the metric)

leslie townes

he's defining the general metric space concepts for an arbitrary metric space X, for example, and we're mostly focused on R^n here. for more general sets X, while the notion of metric space ball is still useful, i think you do have to step away from any hope that it will resemble an intuitive ball in R^n.

thankfully, nothing in rudin is going to depend on "the quality of ballness," it will only depend on his definitions.

Koro

19:50

@DLeftAdjointtoU there are atleast two users with number theory hammer on mse.

leslie townes

whatever he says about balls in a metric space will be true, even if those balls do not resemble balls in euclidean R^2 or R^3.

Koro

I'm sure they can help a lot if they get to know about your question @DLeftAdjointtoU

Infact, there are 20 users with number theory gold badge.

Silly Goose

i feel like i am okay with eschewing the notion of an intuitive ball, but i am more so confused about what differentiates a ball from a neighborhood in a case other than Euclidean space (since it seems like open ball is equiv to neighborhood here)? It seems like from what you said it is not the metric because we can use different metrics on $\mathbb{R^k}$ to get balls. Which leads me to think that it is then the constraint on the set of interest that gives rise (or does not give rise) to ballnes

leslie townes

the properties rudin leans on in chapter 2 for proving and defining stuff involving open balls will, ultimately, just depend on the axioms for a metric space. he's not using anything about balls that isn't expressible in those terms.

Koro

@DLeftAdjointtoU ohh

leslie townes

19:55

OK, i peeked at rudin. he does define "neighborhood," but does so in a way that makes it synonymous with "open ball." he says a neighborhood of p is an open ball, of some positive radius, about p.

Silly Goose

:P sorry i should've posted the definitions here

leslie townes

that's not a definition that everybody uses, but it's the one he's using. so for him, neighborhoods are the same thing as open balls.

for other people, a neighborhood of p might be any set that contains an open ball with center p, or any open set that contains p. but not for rudin.

Silly Goose

i guess that is my confusion: why define both concepts at all if they are synonymous for all purposes in the textbook :P

err maybe it will go into it more in a future chapter...

leslie townes

you'd have to ask him about that, and unfortunately, he's dead. but, i don't recall him using the term "neighborhood" very often.

maybe he just wanted two words to choose from, so he doesn't have to say the same exact thing over and over.

Thorgott

neighborhood of a point should mean containing the point in its interior

you typically don't want to require neighborhoods be open

e.g. you wanna talk about closed or compact neighborhoods, which oftentimes won't be open

leslie townes

20:00

reading rudin closely, he doesn't seem to define 'ball' except for euclidean R^n. and his definition there requires him to choose a radius. so maybe he wanted something for more general metric spaces, without having to call it a "ball," and without having to choose a radius.

maybe that motivated his definition.

why he made it synonymous with open ball of some radius in a metric space, god only knows. i guess rudin isn't going to be talking about compact neighborhoods.

Silly Goose

ack well in any case this was helpful

will there one day be unanimous agreement upon all math definitions :D

Koro

Lee is also on mse.

and Hatcher too

leslie townes

ooh, rudin's real and complex analysis is a little better. in a general topological space X: "A neighborhood of a point p in X is any open subset of X which contains p. (The use of this term is not quite standardized; some use "neighborhood of p" for any set which contains an open set containing p.)"

Silly Goose

is that like a grad level book analogue to baby rudin?

leslie townes

look at him, wasting precious print by bothering to include a remark about the use of terminology!

it's why people say "baby rudin." because he wrote another one that's bigger. i guess it's grad level, i don't know.

Silly Goose

20:07

XD gosh would i have enjoyed a remark about the use of terminology

leslie townes

it doesn't seem to be much in use in actual classrooms. i think most people know it only because they are exposed to 'baby' rudin, and then somehow learn that he's got another one.

Silly Goose

oh i see

leslie townes

he also has a functional analysis book, and a book on abstract harmonic analysis, which (like real and complex analysis) do not seem to be standard references for their subjects, at least, not nearly the way that principles is sometimes thought of as a standard for single variable real analysis.

Silly Goose

is functional analysis like the generalization from looking at sequences to looking at functions or is that pretty off base

or like performing analysis on functions as opposed to sequences

leslie townes

PMA is a funny book because people don't use it for multivariable analysis or measure theory, either, while it has chapters on those things.

silly: i'd say it's, aspects of linear algebra and real analysis, but on 'bigger' spaces than R^n, definitely including function spaces.

Silly Goose

20:11

ohh i see

Silly Goose

20:31

a point $p$ being an isolated point of $E$ can be rephrased as there exists a neighborhood $N_r(p)$ such that it is the empty set (i.e. contains no points not equal to p) and $p \in E$?

Silly Goose

20:54

hehe i understood a terse baby rudin proof >:D a victory for the day. this basic topology business is much more fun than cardinality :P

leslie townes

@SillyGoose yes, p is an isolated point of E if there's some r with N_r(p) intersect E = {p}, or equivalently, if p is in E and there is some r with N_r(p) intersect (E \ {p}) empty. (this is almost what you are saying above, but the reference point for the "it" in your "such that it is the empty set" is not clear).

silly: if you want to have a tiny moment of fun with rudin, try to find where he defines "topology," the word that he uses in the title of chapter 2, in that text.

Silly Goose

oh by "it" i am referring to is the neighborhood $N_r(p)$. since it seems like $p \notin N_r(p)$ by definition (positive definiteness of metric combined with r > 0), which led me to believe that $p$ is not a limit point $\implies N_r(p) = empty set$

leslie townes

ok, that's also a little ambiguous, given how rudin has done it. he's fixed X up above, but "all q such that d(p, q) < r" in his definition of "neighborhood" is refering to all q in X.

you could also consider a notion of N(r,p) that was considering only all q in E, but that's not what he was doing.

Silly Goose

heh i shall try to find that

leslie townes

i'll save you some time, he never defines that word, or uses it at all, outside of the title of that chapter.

Silly Goose

21:06

xD

wait so is your point above that it is ambiguous whether the point $p$ which you characterize a neighborhood $N_r(p)$ by is in or not in said neighborhood? (in rudin at least)

leslie townes

in his definition, N_r(p) = {q in X: d(q,p) < r}. so, N_r(p) may have points in it other than p, even when p is an isolated point of E.

the ambiguity (not really made clear by what you screenshotted above, you also need the two lines before it) is what set the "all q such that d(p, q) < r" are understood to be coming from, in the definition of N_r(p).

D.C. the III

I'm given that $U$ is a partial isometry, so I can conclude that $U$ is a projection based on how the partial isometry is defined ($U(x)\| = \|x\|$ for $x \in W$ and $U(x) = 0$ for $x \in W^\perp$. Is it the case then that $U^*$ would also be a projection onto $W$ as well?

Silly Goose

ohh i see so long as $q \notin E$ then $p$ can still be an isolated point if the neighborhood contains such $q \notin E$

leslie townes

p is isolated in E when [p is in E, and there's some r that N_r(p) doesn't contain any other elements of E]. this doesn't mean that N_r(p) itself is empty. but if p is isolated and r is chosen judiciously, then N_r(p) intersect E will be empty.

Silly Goose

gosh i have to read more carefully xD

leslie townes

21:13

dc: a partial isometry need not be a projection. indeed, it might not have any fixed points other than the zero vector. there are senses in which you might think of it as kind of like a projection, but not literally one.

D.C. the III

Ok so how it is deinfed here it would be better to say it is like one then.

leslie townes

it has a domain space W on which it is an isometry, and it has a range space (not named in the above), and it'll be different from a projection when that domain space and that range space are different spaces.

D.C. the III

because I do need $U^*U$ to be a projection before I can jump to orthogonal projection (which is the end game of what I'm trying to show)

leslie townes

i wouldn't spend too much time thinking about what makes something "kind of like a projection", it is better to work with real stuff.

yes, OK. if U is a partial isometry than adj(U) U and U adj(U) will both be projections.

that's an equivalent definition of the concept, even. you wouldn't necessarily need the thing about U(x) and x having the same norm on some space and U being zero on a different subspace.

although the norm condition that you were given does explain a bit why you'd expect adj(U) U to be a projection.

D.C. the III

what's the equivalent definition of the concept? to the concept of projection?

leslie townes

21:18

not "the" equivalent definition, an equivalent definition. i'm sorry, this is distracting you. i don't want to do that.

D.C. the III

all good. I'll focus on "what I know" and think about that comment after

I really abuse these quotations for the purpose of emphasis....

would've been nice if this was the whole space and not the subspace, then I could $U^*U = I$ that bad boy all over....

Oh in my case Leslie, I just read the isometry definition fully and it is defined for the linear operator so domain and range are the same space, but for the generalization this is good to keep in mind. (which comes up in a further chapter)

leslie townes

oh yeah. well in this kind of example it might help to think of even rank 1 operators, like the operator U on C^2 sending the first basis vector e_1 to a unit vector u, and the second basis vector to 0. it's an isometry on the span of e_1, it maps the span of e_2 to 0. its a partial isometry. adj(U) U is projection onto the span of e_1 and U adj(U) is projection onto the span of u.

Obliv

If I'm solving a system of equations that has free variables, does the final form have to be the same no matter how you solve it

leslie townes

more generally adj(U) U is going to be projection onto the space called "W" up above, and U adj(U) will be projection onto the range of U. i haven't explained why that would be, exactly, but that's what happens. understanding that in the 2d example might help for general.

D.C. the III

@Obliv one of the first things learned in a chapter of matrix reduction

Obliv

21:30

I got $x=320+96t-3s, z=186+52t$ but the answer was $z=54+52t, x=100+96t-3s$

are they equivalent

D.C. the III

I don't think you needed that tag Leslie...lol

leslie townes

it may also help to have some terminology. the space W is sometimes called the 'initial space' of the partial isometry, and the range is called the "final space" of the partial isometry. when those two spaces are the same, you get an orthogonal projection onto whatever common space that happens to be. otherwise, you just get something that isometrically maps the initial space onto the final space, and is zero on the orthocomplement.

nxn matrices having all zero entries except one entry that's a 1 are all partial isometries. you get a projection if that 1 is put on the diagonal.

D.C. the III

These are probably facts I'm going to be proving very soon and also the terminology is going to be introduced to them too.

leslie townes

it could be.

D.C. the III

I figure because if for now it is only for linear operators then "initial space" and "final space" don't have much context unless they would be different which of course from what you said will be the case in a broader framework.

Obliv

21:36

ya nvm I figured it out i did something wrong

Obliv

22:25

for $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ this homogeneous coefficient matrix can either $x_2$ or $x_3$ be set as the free variable

like $x_1 = 0, x_2 + x_3 = 0$ so both are correct right

also would $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ mean $x_1 = s, x_2 = r, x_3 = t$ all three vars can be free

Gwyn

23:24

My textbook asks if there exists another continuous function $g:\mathbb{R}\to\mathbb{R}$, different from $g(x)=2x$, such that $\frac{1}{b-a}\int_a^b g(x)dx=a+b$ with $a,b\in\mathbb{R}$ such that $a \ne b$. Using the fundamental theorem of calculus, they say that $2x$ is the only function that satisfy that integral equality. However, I was thinking about $g(x)=a+b$. It is continuous because it is constant, and it is $\frac{1}{b-a}\int_a^b(a+b)dx=a+b$, am I wrong or is the textbook wrong?

The solution of the book is: "Assume $f:\mathbb{R}\to\mathbb{R}$ is continuous and such that for each $a,b\in\mathbb{R}$ with $a\ne b$ it is $\frac{1}{b-a}\int_a^b f(x)dx=a+b$. Then $\int_a^b f(x)dx=b^2-a^2$, letting $F(x)=\int_a^x f(t)dt$ it is $F(x)=x^2$ for each $x \in\mathbb{R}$ and so, since $f$ is continuous, it is $f(x)=F'(x)=2x$".

Thorgott

quantifiers are important

from your quoted solution, I assume the book wants that equality to hold for all distinct $a,b\in\mathbb{R}$

and then your proposed $g$ doesn't make sense

Rithaniel

Anyone here familiar enough with Macaulay2 to tell me if there is some way that I can check to see if an input to a function is a list?

From the documentation I can find, it seems the answer is "no, there's no way. You're just supposed to assume," but that's contrary to all the other programming knowledge I have

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