I have a question; inputs welcome. Suppose $f : \mathbf R \to \mathbf R$ be such that $f^{-1}(a)$ is dense in $\mathbf R$ for all $a$. (In fact, if necessary, you may assume that $f(x+q) = f(x)$ for all rational $q$ and real $x$. This means that $f^{-1}(a)$ is closed under translation by rational numbers for all $a$.) Then does it follow that I can right-invert $f$ continuously? That is, I want a continuous injection $g : \mathbf R \to \mathbf R$ such that $f(g(a)) = a$ for all $a$.

Should I be reminded of some standard idea/theorem at this point?

hi @Dylan.

  Hi

You can certainly get a continuous g:Q->R, but I don't see that you have any right to expect it to work if you extend it continuously to R->R.

@HenningMakholm Why does such a continuous $g: \mathbb Q \to \mathbb R$ clearly exist?

@Srivatsan Hey there.

@Srivatsan what do you assume about $f$?

@ZhenLin yes (Sorry, I have shitty internet connection here and was offline for some time)

I don't know what to expect as the answer. There is no real topological restriction on the codomain; we only know that every preimage is large.

@Srivatsan Enumerate Q and select the values of g(x) one at a time, subject to the condition that g must be Lipschitz with some fixed contstant, say 1. For each number you have an entire interval to choose g(x) values from, and the result is Lipschitz and thus continuous by construction.

@tb Nothing more than I said: every preimage is dense and is closed under translation by rationals.

@Srivatsan sorry, I was offline; can you please tell me about two injections between $[0, 1] \cup [2, 3] \cup [4, 5] \cup \cdots$ and $[0, 1]$ please?

@Daniil Take any injection from R to (0,1) and restrict it to your domain.

@HenningMakholm Aw, I see. I'm not fully sure this works, but that is neat.

@HenningMakholm yeah, basically a binary encoding would do, but what about an injection from the latter to the first?

[I hope the question is clear. :)]

@Daniil An injection from [0,1] to [0,1] cup blah? The identity function works.

If you want a bijection, use Cantor-Bernstein.

Yeah, I do want a bijection. Cantor-Bernstein states that if $X' \subset Y$ and $X \supset Y'$ then $X = Y$, right?

($X'$ and $Y'$ are isomorphic to $X$ and $Y$ respectively)

tb/Henning: I have to leave right now; I will be back in a while. I will think about what Henning said for $g: \mathbb Q \to \mathbb R$ in the mean time.

So I understand how Y= $[0, 1]$ is embedded in X=[0, 1] cup ... And for the $X' \subset Y$ we can use the binary encoding of the X

@Daniil It's a bit stronger than that. It states that if there exist injections X->Y and Y->X then there is a bijection X<->Y.

@Daniil Remove $0,1$ and the sequence $1/n$ from $[0,1]$. Then map $(n,n+1)$ to $(1/(n+1),1/(n+2))$ and map the end points of the intervals to the points that are missing.

The problem wiyth chopping [0,1] up is you get pieces like [0,a), [a,b), [b,c), ...

nice one, you take all the end points separately and of course, there's countably many of them.

@Daniil I'm not entirely sure which "binary encoding" you're talking about -- how about 1/(1+e^x) or (1+atan x)/2pi ?

@HenningMakholm but it does not give us the specific bijection, does it?

(1+atan x)/2pi sould be (pi/2 + atan x)/pi, of course.

Well, I messed up the numbering a bit, but the idea should be clear.

@HenningMakholm By binary encoding I mean taking a number and representing it with only 0s and 1s in the binary system. So, for example, the number 5,4 (which is 101,100 in binary) is mapped to 0,3001101. 3 there represents where to put the comma in the number.

And 1 is also included, since there is 0,99999....

I am sorry if this is a bit unclear and thanks for the help!

@Daniil Well, 3 is not a binary digit. What does 1024 map to? (Of course this can be made to work, but the details are gritty).

1024 is 0,0 000000000 1

I think the problem is solved

@Daniil

So you have the union $\bigcup_{n=0}^\infty [2n,2n+1]$. Map the interval $(2n,2n+1)$ to $(1/(2n+2),1/(2n+1))$, map $0$ to $0$ and map $n$ to $1/n$ for $n \geq 1$. This gives a bijection to $[0,1]$

@QED It is.

@tb yeah, thanks.

@Daniil Is that not also the image of 0.00000000001_2?

@HenningMakholm Image of that number would start with 0,1.. because the comma is between first and second digit in that number.

Actually, nevermind, that was a horrible idea.

And to see that the map has to be somewhat nasty, observe that there can't be a continuous bijection from the union to $[0,1]$ because this would permit you to write $[0,1]$ as a countable union of disjoint closed sets (actually intervals) and this is impossible.

@tb Oh, I have a chapter about that latter in my book :)

yeah they have a different topology

so there is a bijection, but not a continuous one.

@QED It's not quite that obvious... It's a theorem of Sierpiński (essentially an application of Baire's category theorem)

See also here for some nice arguments

Does anyone know if there is a question like "Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous"? I have a proof I'd like to have checked by you but obviously if it's already there I won't post it.

I searched for "compact support" "uniformly continuous".

@Matt I couldn't find a question that this would be a duplicate of.

(and I don't remember having seen anything similar)

@robjohn: Concerning Stirling: I have no problems with your substitution and what you now identified as Lagrange Inversion Theorem. What I don't understand about your argument is how you pass from the local analytic representation of $x$ as a power series of $u$ to a global expression of the integral.

@tb Good, thank you.

eerg math.stackexchange.com/questions/95602/…

yes possible, but doing so is going to be extremely tedious

not sure why someone would ask this

@tb That is because when $u>>0$, $x'=u\frac{1+x}{x}\sim u$ and when $u<<0$, $x'=u\frac{1+x}{x}\to0$.

@tb So the integral vanishes quickly away from $0$ since we are integrating against $e^{-nu^2/2}$

@tb The important part of the asymptotic expansion happens near $u=0$.

Perhaps I should add that, too.

@tb Does that make sense?

@robjohn I agree with that intuitively and morally. :)

@robjohn Yes it does definitely make sense. However, I was unable to bound this to get the estimate $n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^{n} e^{r_{n}}$ with $|\frac{1}{12n} - r_{n}| \leq \frac{1}{120n^2}$, for example.

(but I didn't try for too long and I don't want to bug you too much)

@tb Please don't worry about bugging me. I am glad you are discussing this with me. I can improve my answer better responding to your concerns than trying to figure out what is bothering Didier. His last comments were more direct, but still a bit vague.

"Prove that every geometric figure on the Eucledian plane, which contains a piece of a line has the same ordinality as a line". Can anyone give me a hint on how to prove this formally?

*has the same ordinality as that piece of line, actually

@robjohn I think that your mentioning of Lagrange should have taken care of most of the concerns that seem to be more or less implicit in the last three comments. I still have to think about why $x(u)$ doesn't blow up too badly. Unfortunately, I have to leave now for a while, but we can come back to this a bit later.

@tb When $u>>0$ is $u^2/2=x-\log(1+x)$ implies that $x\sim u^2/2$ and when $u<<0$, $x\sim e^{-u^2/2}-1$.

@tb note that when $u>>0$, $x-\log(1+x)\sim x$ and when $u<<0$, $x-\log(1+x)\sim-\log(1+x)$

@robjohn Oh, right! This sound like what I was missing. I was staring too much at the graph of $x - \log{(1+x)}$. I'll consult my scribbling later and tell you if I'm convinced. See you later!

@tb I have to leave in 15 minutes and won't be back until after dinner. I will have to write up these things when I get back. Thanks for the comments and concerns :-)

@Daniil What is "ordinality" in this context? It is clear that the cardinalities are the same, because a line segment has the same cardinality as the entire plane, namely 2^alephnull.

@robjohn Thanks for your explanations. So, see you tomorrow (from my perspective). Have a good day!

@HenningMakholm uh, sorry, it should by cardinality, of course.

Am I totally out of line for downvoting a Project Euler question that did not disclose its PE origins until it was pointed out in comments?

Hmpf. I'm writing and writing and writing and writing....

I feel that I wrote about twice the original post, and I'm not close to finish.

sigh

@N3buchadnezzar, what's up?

Wondering if I should return a necklace I found at home, it belongs to my ex-girlfriend.

Hi guys.

@N3buchadnezzar Send it by mail.

How much time since the breakup and how expensive is the necklace?

Also, did either of you move on?

Well, its complicated... But we are friends, I do not know wheter she has moved on or not.

If you're friends, and it's complicated... return the damn thing.

I want to give it to her, but I do not know why. It is a cheap necklace, but I spent quite some time finding it.

Okay =)

If it's hers, isn't that reason enough?

Problem is, if she wants it

How so? It's not as if it would be a gift -- it't already hers, just misplaced.

Henning, I'm awful at just adding stuff. I am piling on the post a lot of things, which I don't think is a good idea. I think I will instead wait for his reply.

But it is a heart, sigh. I do not know if Girls would you know... from previous companions.

@AsafKaragila Fair enough. Presumably you're keeping a backup of your efforts so far.

@HenningMakholm Of course. I wrote soooo much.

but it`s hers, and I do not want it so. I guess I will give it to her, even though it might be awkward.

@N3buchadnezzar If you find a wallet in the street, do you or don't you send it to the address on the ID card within? Surely you don't try to estimate whether or not the owner would want to keep it.

The current post is about 500 words, I wrote 700+ in addition.

There is no equation for love

$\cos(\pi\heartsuit)=?$

@N3buchadnezzar As you describe it, it's not a matter of love -- simply one of returning lost property.

It is totally. Just returning propperty

@HenningMakholm Obviously he got a thing for her.

@AsafKaragila ... xkcd.com/55

ask her to marry you

@HenningMakholm Obviously... :-)

if shes says yes you've won, if she says no snap the necklace and throw it at her feet

How very dramatic.

It is like dust that has already settled, It hurts whirrling up all the all dust all over again.

I miss the time we spent together, the necklace reminds me of that. I do not miss her, just what we had. So yeah, it is wierd..

Well, put the damn thing in an envelope, with a note reading "found this while cleaning; I believe it's yours". Stick a stamp on it. Drop it in a mailbox. Problem solved.

I before e cept after w

@N3buchadnezzar Burn it in the fire of a thousand torches. Glaze at the ambers eating through the necklace while sipping cheap whisky, and let the flames fill your mind.

I will return it, because I am bigger than this. Mind over heart. It is hers, my emotions is redundant =)

Cool

Mrs Talisker $\heartsuit$.

Hey Jonas

Ilya was looking for you earlier

Why

what's upw with this crap JSeaton comment math.stackexchange.com/questions/95602/…

@DylanMoreland So I could write an answer :-P

@AsafKaragila My impression was that the OP wanted an answer of the form { (x_i) | .... }, i.e. without any initial \prod.

Good enough.

@HenningMakholm Mine wasn't, but I will add that anyway.

@AsafKaragila Isn't the RHS of your condition trivial in any case? If you take J=Ø, then is always true.

Obviously the contents of $J$ and the choice of $U_i$ make the difference.

Huh? In fact, the longer I look at your answer the less I understand it.

Hang on. Me too. :-D

I'm in mid-editing though.

So I'll cut the irrelevant parts out.

Okay, I give up.

test

@Henning, After chewing on what you said, I now think that:

Every Lipschitz map $\mathbb Q \to \mathbb R$ can be uniquely extended to Lipschitz function over $\mathbb R$.

I agree so far. But will the extension have the right-inverse property you seek?

Aha, no clue about that! It seems I lost track of the problem completely =)

But in any case: can you elaborate your construction of right inverse from the rationals? I do not see why that should work either.

So, let Q = { q_1, q_2, ... }. Suppose we are shooting for Lipschitz constant 1.

Right.

Does the (removed) indicate that you figured it our yourself?

Duck! I had a counter-example in mind; that doesn't work anymore either...

But I would like to see a proof (sketch). :)

Perhaps it would be time to take it to the main site?

Of course, I presume we will define g inductively.

@HenningMakholm I could do that. :)

[I hope I wasn't demanding too much time/attention from you. It's just that I assumed that the construction doesn't work for Q and wanted to point it out to you.]

It's not as if I have anything better to do.

But the answer box gives me better freedom to write long paragraphs.

Not to mention the possibility of upvotes ...

Quite true. =)

But does this mean you figured out the answer to the actual question? Or still only inverse from the rationals?

I can phrase the question accordingly.

Still only rationals, sorry.

The actual question being "Does f necessarily have a exist a continuous right inverse R -> R"?

@HenningMakholm That's fine. :) I appreciate it nevertheless.

I have to be out briefly; I will post it in a while.

ehh?? math.stackexchange.com/questions/95626/… vs math.stackexchange.com/questions/95606/…

are we having kids from the same class with some bad teaching?

@QED No, it's the same OP.

That's even more confusing..

My interpretation is that after the first question he figured out that he actually don't understand what f: A->B means.

@N3buchadnezzar Any particular reason to think it would be true?

That is what I am asking, I know they are equal

How do you know that?

aka $$ \pi \int_{0}^{1} \left[ \sin(\ln x ) \right]^2 dx \, = \, \pi \int_{0}^{1} \left[ \sin(\ln x ) + \cos(\ln x) \right]^2 dx = \frac{3}{5} \pi $$

@Henning: Three different calculators, and my textbook

So that answers your question, or what?

@robjohn Did you see the remarks of Didier on your Stirling-thing?

I have no idea how to prove they are equal, without evaluating them separately.

does evaluating them separately prove they are equal?

Evaluating them separately would appear to be a perfectly cromulent way to show they are equal.

I guess I will just keep bashing my head against the last integral then =)

$$\int_{0}^{1} \left(\left[ \sin(\ln x ) \right]^2 - \left[ \sin(\ln x ) + \cos(\ln x) \right]^2 \right) dx = 0$$

how about just compute this one integral

Note that sin(lnx)+cos(lnx) = sqrt2 sin(lnx-pi/4)

(or something like that)

And then using a substitution might work, or noting symmetry or something like that.

Thanks =)'

test

@Henning, posted. =)

@Henning: I posted a similar answer to that cardinals thing. Mostly to put a reference to a possible case where the continuum is indeed decomposable to two smaller cardinals.

I wonder if whoever gave the exercise is aware of such consistency result ;-)

@Daniil: In what class were you given this question? Topology or something like that?

@AsafKaragila Yes, I thought you might want to weigh in there.

Obviously :-)

Four more questions tagged as cardinals and I get my stinkin' badge :-D

Robert posted an answer. Quite surprising conclusion.

@Srivatsan And even before I got my thoughts in order to write my partial one.

By the way, Srivatsan, "the preimage of every point is dense" $\equiv$ "the fibers are dense".

@AsafKaragila Well, I understand only one side of the equivalence unfortunately =)

But thanks for that.

Fibers are the preimages of single points.

OK. I suppose there should be an abstract definition of fiber as well...

There's a less trivial use of "fiber" = "preimage of point" in fiber bundles in topology.

Hm. The answer is simple, yet I don't have the feeling of fully understanding it.

test

tæst

When did we decide to reopen the mispronunciations thread?!

Ello.

Howdy.

Hi all

Hi Srivatsan!

What's up?

Not much apparently.

Quite.

Going to watch the end of ghost in the shell...

Good movie.

"If you search for Erdos, you will get Erdos, not Euler." (Aryabhata)

Heh

Hammerhead!

@AsafKaragila A nitpick: the mispronunciation thread is the big list question; that is still closed. The reopened one is the pronunciation thread.

@Srivatsan The big list is also being voted to be reopened. Check the mod tools page.

@AsafKaragila I see, then it's just my ignorance as usual. :)

@Srivatsan Over a billion Indians, not all can be as bright as trust_god... ;-)

Yes, I am yet to find those helpful lighthouses he keeps talking about.

:-P

Seriously, dude?

@Srivatsan I always see myself as a lighthouse using light beyond the visible part of the spectrum...

Hm, those two reopen votes -- they are recent, right?

In a not well-orderable subset of the EM spectrum, I assume?

@HenningMakholm Of course :-)

It's like two hit and run reopen votes. No explanation why they want the dead post resurrected.

@Srivatsan Perhaps they were paid off by Chandru?

@AsafKaragila Quite likely. Perhaps he fancies his own name posted as well. =)

@Matt I'm not too surprised. Your taste of movies tend to diverge from philosophical-existentialistic-nihilistic-fatalistic narratives.

@AsafKaragila And yours converge to that?

@Matt It has a subsequence that does.

Or a subnet, since I am not well ordered, and sequences may not get to every point ;-)

The train to Hammervile is boarding! Last call!

@AsafKaragila I'm already hammered... I mean I have already hammered.

I'm passing on that; I don't understand why it is too localised.

This is not suited to the site to begin with.

@Matt Have you hummered?

@AsafKaragila ?

Thwack.

Is this a slow day or what?

@Matt Never mind :-P

@HenningMakholm - what makes you think so?

@HenningMakholm Quite. The last three days were awfully slow; I just got a lot of reputation for that 1+1=2 question.

@HenningMakholm For me it is. Although slow might be an understatement.

At least I gave two answers today.

Is this also too localised? My gut feeling is yes.

I don't know.

Probably.

I am going to bed, it's late and tomorrow I have to go to Jerusalem by bus again (my advisor stopped going there every week, so I cannot use him for the ride)

Actually, I am not very sure: the question is posed generally, unlike the thread that was recently closed.

Night, Asaf. Don't let the sets haunt you.

If only I could.

Technically, it's bed time here, too.

Srivatsan, you'll get a batch of 40 new votes in about five minutes. How fast do you think it'll take you to finish them all? :-D

40 votes from what?

Much like I enjoy voting for closing and deleting questions, Srivatsan enjoys the voting on questions...

However, it's much easier to spend your 40 votes in an hour than to find new targets to close.

@Matt Each SE site has a limit of 40 up/downvotes per user per UTC day.

@HenningMakholm Sounds very reasonable. : )

I enjoy mocking people, in particular the Hindu God of Computer Science... Srivatsan. ;-)