Mathematics

12:01 AM

Can also cut up real bad.

I still cant get it: we cant write d(x_n,\mu) can we? d is defined in a metric space which doesnt contain the limit points of cauchy sequences in it.how can we we think of d(x_n,\mu)

Jonas Teuwen

Perhaps you should just look up the proof and understand how it is done.

Peter Tamaroff

@Pilot Yes, maybe you're right.

user19161

I read the transcript and I don't understand Pilot's problem. What's the problem?

Realz Slaw

oy

Pilot

12:12 AM

I found paper on that topic.Thank you guys for help

Peter Tamaroff

@JasperLoy I think he wants to define the completion of sequential space using equivalence classes of Cauchy sequences.

user19161

@PeterTamaroff Ah, then just read the proof 9000 times until one understands it!

user19161

@Pilot Paper? This stuff is in books, not papers!

Peter Tamaroff

@JasperLoy I think one needs to think about $\{x_n\}\sim\{y_n\} \iff \lim\;d(x_n,y_n)=0$

user19161

If you need to read a paper to get it, you are reading the wrong thing.

user19161

12:14 AM

Unless it is in the newspaper.

Peter Tamaroff

@Pilot Read about sequential closure here It has a bit to do.

Pilot

@ Calm down Jasper please, I read stuff from scientific papers,I like it that way

Peter Tamaroff

@JasperLoy Man, I have some stuff pending of a proof.

It is unsettling me =P

user19161

@PeterTamaroff I answered the low hanging fruit you just answered too with more details =J

Michael Greinecker

@Pilot The kind of scientific papers that "develop" this stuff are like 100 years old...

Peter Tamaroff

12:28 AM

@JasperLoy Which one?

@MichaelGreinecker Man.

user19161

@PeterTamaroff Oh sorry it is the one you edited, the xyz question.

Pilot

it doesnt matter to me,i want to learn

user19161

@Pilot Well done! But reading a book could be more efficient in many cases for well-established material.

Peter Tamaroff

@Pilot Usually, it is better to learn for good books. Papers are usually aimed at different people.

user19161

@PeterTamaroff Like Jonas.

Peter Tamaroff

12:32 AM

@JasperLoy Right.

user19161

@PeterTamaroff Hmm, I think I will try to aim for 4k and then retire. I am only 350 points away.

Peter Tamaroff

@JasperLoy Retire?

user19161

@PeterTamaroff Retire from MSE. Yay!

Michael Greinecker

We should all start to downvote Jasper to keep him on MSE...

Peter Tamaroff

My to-do list is
$1.$ Prove $$\int_0^\infty \frac{\sin x}xdx$$ exists.
$2.$ Show that if $f:[0,1]\to\Bbb R$ is such that $\lim_{x\to a}f(x)$ always exists, the set of points where $f$ is discontinuous is countable. (I have a clear idea how to do this, just can prove one last thing)
$3.$ That question I have on MSE about uniqueness of an ODE solution.
$4.$ Show that if $\sum_{i=1}^n p_i=1$, and $x_i$ $1\leq i\leq n$ are arbitrary, $$\min \left( {{x_i}} \right) \leqslant \frac{{\sum\limits_{i = 1}^{n - 1} {{p_i}{x_i}} }}{{\sum\limits_{i = 1}^{n - 1} {{p_i}} }} \leqslant \max \left( {{x_i}} \r…

user19161

12:38 AM

@PeterTamaroff Did you learn ODEs from Apostol?

Peter Tamaroff

@JasperLoy From Apostol, Spiegel, and a online book from a Caribbean university =P

@MichaelGreinecker Could you help me finish off $2.$?

$$\Huge{\text{Chillax}}$$

Michael Greinecker

@Peter I think so.

Peter Tamaroff

@MichaelGreinecker Thanks.

@MichaelGreinecker I have done the following, suggested by Spivak.

Let $\epsilon>0$.

Consider the set $A_\epsilon=\{a\in[0,1]:|f(a)-\lim\limits_{x\to a}f(x)|>\epsilon\}$

The claim is that $A_\epsilon$ is finite for each $\epsilon$.

Suppose that $A_\epsilon$ is infinite. Then by Bolzano Weierstrass, it must have an accumulation point $\xi\in[0,1]$.

By hypothesis, $\lim_{x\to\xi}f(x)=\ell $ exists.

I must show this is impossible.

Since $\xi$ is a limit point, for each $\epsilon>0$ there are infinitely many $a\in A_\epsilon$ such that $|a-\xi|<\epsilon$.

In particular, since $\ell$ exists, for this $\epsilon$ there is a $\delta>0$ such that for all $x$, whenevr $0<|x-\xi |<\delta$ then $|f(x)-\ell|<\epsilon$.

anon

@spernerslemma The p-adic field $\Bbb Q_p$ is not the same as the finite field $\Bbb F_p$ or any extension thereof. The former can be viewed in three different but equivalent ways: (a) power series in $p$, i.e. $\Bbb Z[x]/(x-p)$ with the $(p)$-adic topology; (b) the metric completion of the rationals $\Bbb Q$ with respect to the metric $d(a,b)=|a-b|_p$ induced by the valuation $|(a/b)p^k|=p^{-k}$ (when $p\not\mid a,b$);

or (c) the fraction aka quotient field of the inverse aka projective limit $\varprojlim \Bbb Z/p^k\Bbb Z$ as a profinite ring. $p$-adic numbers have characteristic zero. Keywords to search for: finite fields, p-adic numbers, metric completion, fraction/quotient field, (I)-adic topology, inverse limits, profinite groups/rings.

Peter Tamaroff

But since $\xi$ is a limit point, there are infiniely many $a\in A_\epsilon$ such that $|a-\xi|<\delta$ yet $|f(a)-\lim_{x\to a}f(a)|>\epsilon$

@MichaelGreinecker That observation is what should show that $\ell$ can't exist, but I'm stuck there.

Michael Greinecker

12:55 AM

@PeterTamaroff I have to think about this for a minute or so.

Peter Tamaroff

@MichaelGreinecker OK. The idea is that there will be infinitely many points away from $\ell$ in a $\delta$ nbhd of $\ell$, which is impossible.

I can't formalize it, though.

@MichaelGreinecker You get the idea, right? Note that $f$ is such that discontinuities are all removable.

Michael Greinecker

@PeterTamaroff Every open $\delta$-neighborhood of $x_i$ contains an $a$ from $A_\epsilon$ since $x_i$ is an accumulation point and a point $x$ such that $|f(x)-f(a)|>\epsilon$. Both $a$ and $x$ can be taken to be different from $\xi$. Since $\delta$ is arbitrary, this shows that $\ell$ can't exist.

Peter Tamaroff

@MichaelGreinecker Where does $x$ come from?

Pilot

1:13 AM

hi once again. I have looked through many defintions of embedding some metric space into another, wanted to ask you as well. Could anyone explain me what is embedding a metric in to another?

Michael Greinecker

@PeterTamaroff Since $\lim_{y\to a}f(y)$ exists, take any sequence $(y_n)$ converging to $a$ not containing $a$ as a term. Then $|\lim_{n\to\infty}f(y_n)-f(a)|>\epsilon$. Since the inequality is strict, we have $|f(y_n)-f(a)|$ for some large enough $n$. If $n$ is large enough, it will also be in the $\delta$-neighborhood. So let $x=y_n$.

Peter Tamaroff

@MichaelGreinecker $|f(y_n)-f(a)|$ what for some large enough $n$?

Michael Greinecker

@PeterTamaroff $|f(y_n)-f(a)|>\epsilon$

Peter Tamaroff

@MichaelGreinecker Right.

anon

@Pilot It's a mapping from one space to another that neither expands nor shrinks any distances by too great of a ratio (i.e. the ratio $d_Y(\phi(x),\phi(y)):d_X(x,y)$ is bounded above and below by positive constants).

Peter Tamaroff

1:27 AM

@MichaelGreinecker So we say

For this particular $\epsilon >0$, given any $\delta >0$, we can choose $x$ and $a$ such that $x,a$ are in a $N_\delta(\xi)$ but $|f(x)-f(a)|>\epsilon$.

Michael Greinecker

@PeterTamaroff Exactly.

Peter Tamaroff

And now I use the triangle ineq to get to a contradiction?

Alexander Gruber

Can somebody explain to me why this question isn't equivalent to @JacobSchlather 's comment?? Since finite groups are Hopfian isn't every epimorphism an isomorphism? math.stackexchange.com/questions/221152/…

Peter Tamaroff

@MichaelGreinecker I think we should've taken $\epsilon/2$... to get $$\varepsilon < \left| {f\left( a \right) - f\left( x \right)} \right| \leqslant \left| {f\left( a \right) - \ell } \right| + \left| {\ell - f\left( x \right)} \right| < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon $$

Bitrex

Maybe some of you real analysis folks could answer a question when you have a chance. If $f(x)$ and $g(x)$ have for coefficients of their series expansions $f_n$ and $g_n$, and their coefficients run over the same indices and are positive and monotone decreasing, is it the case that $f(x) < g(x) \forall x \geq 0 \in \mathbb R \implies f_n < g_n \forall n$?

seems intuitively true..but maybe I'm missing something.

anon

1:49 AM

The converse should be clear, but my intuition says that direction of implication is very plausibly false. (Also you would want $f_n\le g_n$, a weak inequality, I would imagine.)

peoplepower

@AlexanderGruber The hopfian property gives that every surjective endomorphism is an isomorphism.

We're not dealing with endomorphisms.

Bitrex

@anon Thanks. If it's the case that it's false, I'm trying to think of a counterexample. Given my knowledge of analysis (or lack thereof) I might be at this a while :)

peoplepower

@AlexanderGruber You're reasoning is correct, of course, since any surjective function between finite sets is a bijection.

anon

There are some categories of groups where epimorphisms are not surjective. I don't know if finiteness of groups changes that, or if the OP had that in mind at all (doubtful).

peoplepower

2:05 AM

@anon Presumably these maps cannot be homomorphisms in the traditional sense.

anon

Yes, they are homomorphisms.

You just restrict what groups there are in the category, so that some cancellation properties (which is the essence of a morphism being an epimorphism in a category) hold even when surjectivity is absent.

peoplepower

I see the idea.

anon

See for instance arxiv.org/pdf/math/9807117.pdf

(a variety (of groups) is a category of groups with some extra properties satisfied)

MaoYiyi

anyone know a good website of calculus problems by topic?

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$Mathematics$ Mathematics