Mathematics

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?

Dylan Moreland

@Srivatsan Hey there.

@Srivatsan what do you assume about $f$?

6:15 PM

@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.

6:17 PM

@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.

6:23 PM

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?

6:25 PM

(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

6:30 PM

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 :)

6:39 PM

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

Matt

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.

Matt

6:56 PM

@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)

7:12 PM

@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 :-)

7:25 PM

@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.