Mathematics - 2022-11-07

Seeker

00:00

I was thinking of starting to learn Abstract Algebra. Is Dummit and Foote still the go to book or have there been any new popular books on Abstract Algebra in recent times?

Xander Henderson

No idea. But Dummit and Foote is a perfectly fine book for an undergrad.

I've heard good things about Carter's Visual Group Theory but I have no first hand experience of the text.

Ted Shifrin

I prefer Artin, but it's more challenging. It's more innovative and shows a lot more breadth of mathematical content.

Dummit and Foote have almost too many exercises.

Xander Henderson

@TedShifrin That is another book that I have heard very good things about.

But my only experience of Artin is Tate's thesis.

Ted Shifrin

My algebra book is more accessible than any of the ones we're talking about, and more restrained. It is controversial in doing rings/field extensions before groups.

Seeker

@TedShifrin @XanderHenderson I will check out all three books. Thanks a lot for the suggestions!

one potato two potato

02:24

@Yai0Phah You know what? Actually, I found that formula during the proof of $[\mathcal{L}_X,\iota_Y] = \iota_{[X,Y]}$.

Ted Shifrin

Better to.prove that formula directly, as I posted to you.

one potato two potato

Yes, I saw it.

one potato two potato

05:11

@TedShifrin I didn't know you have algebra textbook. I just ordered it via my college library book order service.

leslie townes

you can also see it for free in the british museum, it's one of the three things etched into the rosetta stone

3

(this is a joke about ted being old)

rings/fields first is a pedagogically superior approach

copper.hat

seems to be a lot of questions lately that seem to be using mse in lieu of a textbook

leslie townes

what could possibly go wrong

copper.hat

i've been a little snippy

PseudoLooped

It's also that time of year of midterms and such and people panicking because they can't find an answer for one of their HW online (and they don't have Chegg or whatnot)

leslie townes

05:19

i think it must be hard to learn math these days with so many free resources available to mix and match

copper.hat

easily availability has got to remove the impetus to solve

leslie townes

so many misunderstandings on MSE are just, yes, the choice of definition actually matters, no, not everybody in the universe uses the same definition as your instructor, whether this is easy or hard absolutely does depend on how you can relate your current problem to what you already know, and none of that is googlable

copper.hat

i would never have looked for creative solutions to chernoff's problems using the Banach Alaoglu if mse was around.

i can see why different definitions throw folks when they are starting out.

if find theSchwarz lemma very unintuitive.

leslie townes

the stumbling point for so many people is just, the logical structure of this is the point, and the logical structure of this depends on what you know

there's no holy tablet somewhere with the 50 Math Things That You Get To Use etched on it

and if you ask for an explanation of a problem like that's how it is, it will go badly for you

the schwarz lemma is magic

copper.hat

since i am really cranky today, another thing that bugs me is the wandering question. i answer and then "oh, i meant blah", followed by a few iterations of the evolving question.

i find amusing the questions/comments that suggest that the author has just uncovered a deep flaw in mathematics; makes one wonder what they thought all the mathematicians of past were doing all along.

i need a glass of something relatively strong, methinks

Ted Shifrin

05:30

There was a weird convexity question earlier.

Schwartz lemma is about negative curvature :)

copper.hat

the solution to this math.stackexchange.com/questions/4570737/… was what prompted my Schwartz remark.

Ted Shifrin

05:48

Did you see Convexity without Differentiability?

This.

Ajay

06:06

I have the following theorem: Let $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $x_0 \in \mathbb{R}^n$. If $f$ is differentiable at $x_0$, then all of the partial derivatives of the components of f exist at $x_0$, and the standard matrix of $Df(x_0) : \mathbb{R}^n \rightarrow \mathbb{R}^m$ is $f^{\prime}(x_0) \in M_{m \times n}(\mathbb{R})$ in other words, $$Df(x_0)(\textbf{h}) = f^{\prime}(x_0)\textbf{h}.$$

The above theorem says that if $f$ is differentiable at that point, then all of the partial derivatives exist at that point. The converse is not true however. Even if the partial derivatives exist at some point, it does not necessarily mean the function is differentiable.

Jack Ohara

Hello

Ajay

Is there a way to at least find a partial converse to this theorem?

copper.hat

@TedShifrin Thanks! Just taking a look, it think it is just increasing secants.

Ted Shifrin

The standard theorem is $C^1$ (continuous and continuous partials) implies differentiable.

@copper.hat Very subtle, if so. And there’s the weird sorta-converse.

Note that $\lambda a$ isn’t even in the interval :)

Jack Ohara

@TedShifrin Heya Ted , do you have some good reading material about the French railroad metric ? been a while since I took analysis

I did prove tht this is a metric but I would like to understand what are the open sets here and how this related to the eucledian metric space

leslie townes

06:18

the french railroad metric is d(x,y) = infty if x neq y because there's some kind of strike on about something

Jack Ohara

@leslietownes this is why i pinged Ted specifically to avoid such answers

leslie townes

OK. there is a notion of 'equivalence' of metrics, and there is also a notion of a pair of metrics providing the same notion of open sets even if they are not the same as metrics.

copper.hat

@TedShifrin something doesn't look right, but that may be the pinot speaking.

leslie townes

speaking somewhat crudely for a moment, this is sometimes a question about whether balls can be stuffed into other balls.

i don't actually know what the french railroad metric is. i'm guessing d(x,y) has something to do with traveling a path from x to paris and then from paris to y. but if we're talking R^n (as hinted at by "the euclidean metric space"?) i might need to know what paris is.

there is only one notion of open sets induced by a metric, if one assumes that the metric is induced by a norm on a vector space (like R^n) and the vector space is finite dimensional (like R^n)

is this strike thing induced by a norm?

Ted Shifrin

@JackOhara I’m afraid I don’t know even what the metric is (at least by that name).

Jack Ohara

06:29

$d(x,y) = \begin{cases} \|x-y\|, & \text{if $x,y,0$ are collinear;} \\
\|x\| +\|y\|, & \text{otherwise} \end{cases}
$

Ted Shifrin

@copper.hat It may, but it looks wrongly to me, and my martini was 5 hours ago.

Jack Ohara

it is this one Ted !

you have to go to the origin to measure the distance

so if the points are not colinear you have to do some detour of sort

Ted Shifrin

Not quite. If x and y are on the same side of the origin, that’s wrong (in the collinear case).

Jack Ohara

what do you mean by same side of the origin?

Ted Shifrin

I’ve not seen this before, that I recall.

Jack Ohara

06:32

okay i will keep my search !

thanks anyway sir

Ted Shifrin

If the origin is outside the line segment $[x,y]$.

Jack Ohara

yes but from what I understood that would fall in case 2

Ted Shifrin

The small balls around $x$ are just line segments parallel to $x$, unless $x=0$. Very not comparable to the usual topology.

You said collinear … you didn’t refer to ordering

Jack Ohara

it is one strange metric

let me try some examples and come back

Ted Shifrin

Take $y$ a positive multiple of $x$ versus negative multiple .

copper.hat

06:35

I added a counterexample to the convex question.

Ajay

@TedShifrin So, if the function exists in a neighbourhood around $x_0$ and is continuous on that neighbourhood, then $f$ is differentiable at $x_0$?

Ted Shifrin

I’m just saying your saying you have to go through the origin isn’t right as stated.

copper.hat

back to scratching my head with the Schwartz lemma.

Ted Shifrin

That’s not at all what I said, @Ajay

Ajay

Huh, i'm confused...

Ted Shifrin

06:37

@copper That statement was lunacy. He wants to deduce $b>a$ from convexity and …

You left out the most important criterion, Ajay.

Ajay

You do mean to say that if the partial derivatives of a function exist and are continuous at some point, then the function is differentiable at that point. Right?

copper.hat

@TedShifrin oops, i didn't read properly. i think the OP may be correct.

i need to adjust my answer. should not drive & derive.

Ted Shifrin

Yes … but your previous sentence didn’t mention partials!

@copper My second comment gave a stupid counterexample.

copper.hat

indeed, but i am curious if one assumes $b \neq a$.

Ted Shifrin

Me too.

copper.hat

06:49

even with $b \neq a$, i believe the statement is false, i think it depends on the signs of $b,a$ as well. however, i am done for the night! good night @Ted.

Ted Shifrin

Gnight!

Ajay

Wait ted!

I still have a few things to clarify

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $x_0 \in mathbb{R}^n$. Write $f=(f_1,...,f_m)$. Then if the function $\frac{\partial f_i}{\partial x_j} : \mathbb{R}^n \rightarrow mathbb{R}$ exists in a neighbourhood around $x_0$ and is continuous on that neighbourhood, then $f$ is differentiable at $x_0$?

Since the Jacobian matrix of $f$ at $x_0$, denoted by $f^{\prime}(x_0)$, is $$\big( \frac{\partial f_i}{\partial x_j}(x_0) \big) \in M_{m \times n}(\mathbb{R})$$ provided that all of partial derivatives are continuous.

I think this works because we can consider the functions $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ because $f$ outputs vectors in $\mathbb{R}^n$ we can write $f = (f_1,...,f_m)$ where $f_i : \mathbb{R}^n \rightarrow \mathbb{R}$ for $i = 1,...,m$. Since each of these functions $f_i$ are scalar-valued, we can compute their partial derivatives.

Do let me know if i'm correct or not.

Well nvm for now, have a good night's rest :)

PM 2Ring

08:04

@copper.hat Related:

607

Q: Exit strategies for "chameleon questions"

I'm not sure if there's already an existing term for this, so I'm inventing my own. (tl;dr: I call them "chameleon questions" because they change every time you submit or edit an answer. If you're already intimately familiar with the phenomenon, please skip past the first set of bullet points to...

one potato two potato

09:14

@leslietownes I know and I love that joke series.

Goku

I caught myself eating paper twice

one potato two potato

I know that. That's an old-fashioned memorizing technique.

one potato two potato

09:46

12

A: Show that the zero set of $f$ is an orientable submanifold of $\Bbb R^{n+1}$.

Let $$\omega=\sum_{i=1}^{n+1}(-1)^{i-1}\frac{\partial f}{\partial x_i}dx_1\wedge\cdots \wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots \wedge dx_{n+1}$$ Since $0$ is a regular value of $f$, $$df\wedge \omega= \sum_{i=1}^{n+1}\left(\frac{\partial f}{\partial x_i}\right)^2dx_1\wedge\cdots\wedge dx_{n+1}...

This argument cannot be generalized to an arbitrary smooth map $f:M\to N$ between smooth orientable manifolds $M$ and $N$.

one potato two potato

10:58

Let $f:\Bbb R^2\to\Bbb R$ be a smooth curve given by $f(x,y) = \sqrt{1+x^2+y^2}$. Define a smooth vector field $\nabla f$ on $\Bbb R^2$ by
$$df = {xdx+y dy\over\sqrt{1+x^2+y^2}} = {4\over\sqrt{(1+x^2+y^2)^3}}\langle\nabla f,\cdot\rangle,$$
where $\langle\cdot,\cdot\rangle$ is a standard inner product in $\Bbb R^2$.

To find the vector field $\nabla f$, I let $\nabla f = f_1{\partial\over\partial x}+f_2{\partial\over\partial y}$ for some smooth functions $f_1,f_2$ on $\Bbb R^2$. Then
$$df\left({\partial\over\partial x}\right) = {x\over\sqrt{1+x^2+y^2}} = {4\over \sqrt{(1+x^2+y^2)^3}}\left\langle\nabla f,{\partial\over\partial x}\right\rangle = {4\over \sqrt{(1+x^2+y^2)^3}} f_1$$
So $f_1(x,y) = {1\over 4}x(1+x^2+y^2)$ and similarly, $f_2 = {1\over 4}y(1+x^2+y^2)$. Is this a correct argument?

ILikeMathematics

13:17

If you are given something like this

Is there a way you could quickly tell the function?

In this case it's $f(x) = (x^2 + 1)(x + 2)$ btw, the $(x + 2)$ part is deduced from it being $0$ at $x = -2$, so what's left is the $(x^2 + 1)$ part

How would you go about finding that out?

Xander Henderson

@ILikeMathematics Not without more information.

If you know that the function is a polynomial, then there are some things you can do---the structure of polynomials is pretty rigid.

For example, if you assume that it is cubic, and you know four points on the curve, then you can deduce the coefficients of the polynomial.

For a kind of high level approach: en.wikipedia.org/wiki/Lagrange_polynomial

At a more high school / precalc level, your polynomial function is given by $p(x) = ax^3 + bx^2 + cx + d$. Each point you "plug in" gives you a linear equation in four variables, e.g. $0 = p(-2) = -8a + 4b - 2c + d$. Plug in four points, and you get four linear equations in four variables, which is solvable using elementary means.

ILikeMathematics

Thank you

copper.hat

14:19

@PM2Ring :-)

ILikeMathematics

Do you have any tips on drawing the tangent to a point on a function?
I heard sketching the osculating circle around that point and joining the center and the point, giving you the normal from which you can construct the tangent is sometimes more accurate

Xander Henderson

@ILikeMathematics For what purpose?

I mean, what you are describing is essentially Descartes' approach, and is perfectly fine.

ILikeMathematics

@XanderHenderson Reading off the slope; determining the derivative

Xander Henderson

@ILikeMathematics Again, why? We have better technology, now, such as the derivative.

What is the end goal?

ILikeMathematics

@XanderHenderson Well, on an exam, you might just be given the graph and you need to determine the derivative at a certain point

Xander Henderson

14:27

On such an exam, I would rely on the instructor to have given you some direction about how to address such a problem.

Personally, I wouldn't assign such a problem, as I don't really see the point of it (though I might ask about the sign of the derivative at some point, I suppose).

ILikeMathematics

@XanderHenderson I thought of quickly finding the function and taking the derivative, then plugging in the x value, but in the end, it seems like just drawing the tangent is faster if the function isn't a polynomial of very nice form

Xander Henderson

@ILikeMathematics Are you trying to get an approximation of the derivative, or do you want to actually know what the derivative is?

ILikeMathematics

@XanderHenderson Most likely only a good approximation is expected for such a problem

Xander Henderson

Again, I would ask your instructor what their expectations are. As I don't really see the value of the question, I have no idea what might be expected in terms of an answer.

ILikeMathematics

Alright, thanks

leslie townes

16:58

copper: is "howeve" how the young kids say however these days? :D

copper.hat

:-) i think that is last gen now

Akiva Weinberger

I just learned about the Post correspondence problem

leslie townes

haha his name was even Emil Post! that's too perfect

Akiva Weinberger

Basically you have a bunch of dominos with a string on the top and bottom, and you want to arrange these dominos (repetitions allowed) so that the same string appears as you read across either row

leslie townes

i have the most pressing quandary in etiquette. my father-in-law has an alphabet A with at least two letters, and my in-laws are bringing two lists of words over A to thanksgiving dinner

Akiva Weinberger

17:10

lol

I'm curious what the smallest example is whose smallest solution is, say, over 1000 dominos long

It's an undecidable problem which means that the possible lengths of solutions grow extremely fast in terms in the size of the input

Curio

Consider the following statement.
Integral between 0 and 2pi of [f(x)]^2 is a real number (not infinity) IF AND ONLY IF Fourier serie associated with f has mean square convergence to f.
Is it true?

Akiva Weinberger

I don't remember

I thought it worked iff the sum of the squares of the Fourier coefficients converged

(assuming f is real, anyway)

leslie townes

the "only if" is definitely true.

yeah, the question is just if what akiva said is equivalent to the above. it probably is because of orthogonality? i dunno

Akiva Weinberger

In $PCP[s,w]$, $s$ is the size (number of pairs) and $w$ is the width (longest word involved)

(PCP = Post Correspondence Problem)

$PCP[2]$, where there are only two dominos, is apparently known to be solvable

Ted Shifrin

17:44

@leslietownes Can you translate this into an alphabet I’ll understand?

Akiva Weinberger

people.ksp.sk/~kuko/pcp

@TedShifrin It's an oblique reference to Post's Correspondence Problem, which I was talking about

(and which is explained in the link I just shared)

leslie townes

ted: there's some computational problem, "post's correspondence problem," and for hilarious reasons it is due to someone named emil (not emily) post

Akiva Weinberger

I don't get it

Oh, author

leslie townes

for people who (unlike ted) did not go to high school with emily post, i should say that emily post was known for writing about etiquette in the early 20th century

Ted Shifrin

I have lost my curiosity.

Akiva Weinberger

17:48

That reference went completely over my head

Ted Shifrin

Yes, Emily Post I got.

Akiva Weinberger

Apparently the problem with dominos $\binom{10}0$, $\binom0{001}$, $\binom{001}1$ is still unknown

PM 2Ring

"A Post canonical system, also known as a Post production system, ...". Of course, Googling "Post production system" is unlikely to get you stuff about Post canonical systems.

Not to be confused with the fictional architect, Wilbur Post, who owned a talking horse.

Akiva Weinberger

No one can talk to a horse, of course!

leslie townes

18:04

there's also mike post, who wrote the theme songs for various tv shows, including law and order and (my personal fave) the rockford files

Xander Henderson

@AkivaWeinberger Unless, of course, that horse is the famous Mr Ed.

@leslietownes Rockford is such a great exemplar of what it is. Gosh, I love that show.

leslie townes

and so many wonderful guest stars. and some of the locations you can still see more or less unchanged.

Akiva Weinberger

leslie townes

i like to imagine that my southern california life is just an episode of the rockford files.

free bill posters

Xander Henderson

@leslietownes Indeed.

robjohn

18:21

Bill Posters will burn in hell

Ted Shifrin

howdy @robjohn

PM 2Ring

At last, they're working on improving the workflow to deal with plagiarism:

19

Q: Feedback request: Proposed new flagging and moderation workflow for plagiarism

A while ago one of the mods from SO requested help in dealing with plagiarism on SO. While we didn’t have an answer at the time, we have been working with the moderators to better understand this issue so that we could come up with a plan to help tackle this problem. For more context please see t...

discussion moderation plagiarism flag-reasons

robjohn

@TedShifrin hey, Ted. How goes? It is slated to rain a bit here for the next few days.

leslie townes

we've had drizzle on and off all morning.

Ted Shifrin

18:42

You mean we have to refine our methods of stealing others' answers?

@robjohn Yes, apparently here too, tomorrow. Bad weather is never good for voter turnout :(

Maybe it'll stop those who want to vote out democracy.

robjohn

19:13

@TedShifrin yes, that is true.

won't stop my family.

Goku

Heavy rain here too

robjohn

We've had less than 0.1" so far, but we were slated for 2" at one point. I don't know what the forecast is now.

Goku

Underwater integral challenge when?

MathLearner

19:39

Hello! Can someone please help me?

Let $c:I\to\mathbb{R}$ a regular curve with curvature $k > 0, \forall t\in I$. <br>
Then, $k(t)=\frac{||\dot{c}(t)\times\ddot{c}(t)||}{||\dot{c}(t)||^3}$, where $\dot{c}$ and $\ddot{c}$ are first and second derivative of function $c$. <br>
Ok, so this is the context. Now, my proof starts with: <Br>
>Let $s$ = natural parameter of curve. Then $\frac{dc}{dt} = \frac{dc}{dt}\frac{dt}{ds}$.<br>
Let $h$ be the inverse function of $s$. So, $h(s)=t$. Then, $\frac{dc}{ds}(h(s))=\frac{dc}{dt}(h(s))\frac{dh}{ds}$ [...]

copper.hat

20:01

@TedShifrin i finally added a $b<a$ counterexample to that convex question from yesterday.

geocalc33

just curious. does a function satisfying the following exist? $$ \sum_{\text{p}~\text{prime}}f_p(x)=\prod_{n\in \Bbb N} g_n(x) $$

Mathphile

20:30

@geocalc33 I would write a question on MSE on that

Overtherainbow

Could someone give me a hint to this question:

Given a real v.s. X an d C an open convex subset of $X$ such that $0\in C$. I define $p(x)=\inf\{ t^{-1}: t>0, tx\in C\}$. I want to show that $x\in C$ iff $p(x)<1$.

Then $\Leftarrow$ implication was easy. Now I want to show $\Rightarrow$ but somehow with contradiction I don't see what to to. I wanked to pick $p(x)=1/t'$ such that $1/t'\geq 1$ so $1\geq t'$. But then I don't see where to continue

Thorgott

your inequality is reversed

also, you may not be able to write $p(x)=1/t^{\prime}$

but $p(x)\le1/t^{\prime}<1$ for some $t^{\prime}>0$ s.t. $t^{\prime}x\in C$ and that's all you'll need

Overtherainbow

hmm I don't see why this is enough

So I want to show it by contradiction, isn't this the best way @Thorgott

copper.hat

20:47

@Overtherainbow if $x \in C$ then there is some open set $U$ such that $x \in U \subset C$ and so $\rho(x) <1$.

Ted Shifrin

@copper.hat Yes, I saw. I also brought the confusion to the attention of the guy who'd written a perfectly good answer to the question we expect it to be.

copper.hat

:-)

of course, i completely misread the question originally, thanks for catching that!

Ted Shifrin

I don't think the OP meant my interpretation. I think he meant the obvious one. But who knows.

Thorgott

@Overtherainbow I wouldn't do it that way

ZaWarudo

Consider $F(y)=\int_1^{+\infty} \frac{\sqrt{\log(xy)}}{x^2+1}dx$. The function $f(x,y)=\frac{\sqrt{\log(xy)}}{x^2+1}$ is $C^1((1,\infty)\times(1,\infty))$, so the integral function is $C^1((1,\infty))$ and hence it is $F'(y)=\int_1^{+\infty} \frac{1}{2y\sqrt{\log(xy)}(x^2+1)}dx$. My book asks for $\lim_{y\to+\infty} F(y)$ and $\lim_{y\to+\infty} F'(y)$. Since $f$ is $C^1((1,\infty)\times(1,\infty))$

I believe that the limits should commute with the integrals, but I am not sure about this. Is this still valid for improper integrals? I am not sure because the improper integral is a limit itself, hence if it is valid I am inverting the order of limits and I don't think this is valid in general.

geocalc33

20:55

@Mathphile Okay I will

Overtherainbow

@Thorgott would you do it in copper's way or in a different one?

@copper.hat why do I know that such a $U$ exists?

copper.hat

I thought you wrote that $C$ was open?

if so, then $s x \in C$ for some $s>1$, just by continuity.

Overtherainbow

Ah right it is open. But I don't get why we immediately deduce that $p(x)<1$

Thorgott

copper is talking about a different direction of the argument than I

Overtherainbow

@Thorgott okey but could you also explain me yours?

copper.hat

21:13

i thought you said you were working on the => direction?

Overtherainbow

@copper.hat yes, so I assume $x\in C$ and I want to show $p(x)<1$.

I agree with you that since $C$ is open there exists $U$ open such that $x\subset U\subset C$ . But I don't get the connection to $p(x)$

geocalc33

@Mathphile math.stackexchange.com/questions/4571504/…

just in case your interested

copper.hat

if $sx \in C$ for some $s>1$ then $\rho(x) \le {1 \over s} < 1$.

it would help to draw little diagrams of the situation imo

Overtherainbow

But for what do you need this open $U$ then?

copper.hat

how did i conclude that there is some $s>1$ such that $sx \in C$???

Overtherainbow

21:24

Doesn't this follows since $C$ is open?

Otherwise I don't see how you conclude it, you say from continuity but continuity of what?

copper.hat

which direction are you trying to prove?

Overtherainbow

$x\in C$ implies $p(x)<1$

copper.hat

forget the continuity remark, if there is some $s>1$ such that $sx \in C$ then can you see that $\rho(x) < 1$ by definition?

Overtherainbow

yes because then $1/s<1$ and $p(x)\leq 1/s<1$ right?

copper.hat

yes. this is what you ae trying to show, right?

Overtherainbow

21:27

right so I need to show that there exists such an $s$ then I would be done right?

copper.hat

yes, but that is straightforward since $C$ is open, right?

Overtherainbow

So I mean intuitivly yes since there is a small open ball around $x$ contained in $C$ and then I can multiply $x$ buy a real number $s >1$ such that $sx$ still remains in this open ball so inparticular $sx\in C$

copper.hat

ok, now back to continuity.

Overtherainbow

But wouldn't my argument be done after my last comment?

copper.hat

consider the function $f(t) = tx$. Since $f(1) \in C$ and $C$ is open...

well, intuition is nice for inspiration, but it does not pass muster in aproof, which is what you are trying to do, right

Overtherainbow

21:32

ah so the continuity argument should verify my intuition comment?

copper.hat

that is the goal, yes

Overtherainbow

@copper.hat Okey now back to your comment. $f(1)=x\in C$ and $C$ is open. So there exists a ball $B(f(1), r)\subset C$ right?

But since $f$ is continuous $f^{-1}(B(x,r))$ is open in $\Bbb{R}$

copper.hat

as i wriote above since the is an open $U$ such that $x \in U \subset C$ then there is some interval $(a,b)$ with $1\in (a,b)$ such that $f(t) \in C$.

in particular, $b>1$.

Overtherainbow

ah so your open interval $(a,b)=f^{-1}(B(x,r))$?

copper.hat

sure

Overtherainbow

21:42

ah and so I can pick $b>1$ and get that $f(b)=bx\in B(x, r)\subset C$ and I have found my $"s"$

copper.hat

exactly

Overtherainbow

makes sense thanks!

copper.hat

the Minkowski functional is a very useful idea

Overtherainbow

what is the minkowski functional?

copper.hat

the $\rho$.

defined by the set $C$.

Overtherainbow

22:12

ah

copper.hat

23:37

@TedShifrin The OP changed the question to add $a,b \ge 0$.

Ted Shifrin

Are we still trying to prove $b>a$?

That OP doesn't understand basic language/logic. He now clarifies that he wants to prove that $b>a$, and yet tells the other responder that he had shown exactly what he wanted.

Transcript for

$Mathematics$ Mathematics