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12:00 AM
@Chris'ssistheartist It is not too clear when thinking of them using the iterated integral definition.
 
@robjohn Yeap. It's clear when visualizing the series representation.
It's pretty late here. I'm out.
 
Things have sure changed around here
 
They have, @Argon?
 
Many of the people who used to spend hours here (like me, admittedly) are no longer present.
 
That's hardly noteworthy.
Things change ...
=_=
 
12:06 AM
I'm not saying it's bad. It's an observation.
 
user147690
12:22 AM
@robjohn Bad latex for cubed root :P
 
12:38 AM
@AlexClark Oh, okay. I was wondering if there was some meaning I was missing.
 
user147690
Nah haha, I thought I should look up how to fix it, but I was pretty tired, save me the trouble searching maybe xD
 
user147690
I mean I could just /!/!/! it
 
user147690
$\text{}^{\text{}^3}\!\!\large{\sqrt{2x}}$
 
$\sqrt[\large3]{2x}$
 
user147690
Ahh I see, thanks
 
user147690
12:41 AM
$\sqrt[\Huge 8]{12x}$
 
@AlexClark I sometimes make the root index a bit bigger since they turn out pretty small.
 
user147690
Better :)
 
@AlexClark not quite that big :-p
 
1:22 AM
@robjohn Considering that we're integrating $\text{PolyLog}^{(1,0)}(1,-x)$ over the entire positive real axis, the best way I can think of to define the derivative is to differentiate inside the Fermi-Dirac integral to get $$ \text{Polylog}^{(1,0)}(1,-x) = -\int_{0}^{\infty} \frac{\log t + \gamma}{e^{t}/x+1} \, dt .$$
 
1:38 AM
I'm thinking about homotopies and wouldn't mind some reassurance. I have a continuous $H \colon X \times [0, 1] \to Y$ with $(x, t) \mapsto H(x, t)$ that I think I'd like to 'extend' to a continuous function $\hat{H} \colon X \times [0, 1] \to Y \times [0, 1]$.
It seems reasonable to define $(x, t) \overset{\hat{H}}{\mapsto} \big(H(x, t), t\big)$, which "obviously" should be continuous. If this is indeed continuous, I don't see the slick argument that guarantees it, since my topology is so rusty.
 
it suffices to show preimages of basic open sets $U\times (a,b)$ are open. the preimage is $$\widehat{H}{}^{-1}\big(U\times(a,b)\big)=H^{-1}(U)\cap(a,b),$$ and $H^{-1}(U)$ is open since $H$ is continuous
 
Thank you, @anon. I was hoping I wouldn't have to mess around with open sets, but that's not bad at all.
(I probably should have forced myself to think about open sets, being so rusty!)
 
same argument shows if $A\to B$ and $A\to C$ are continuous then $A\to B\times C$ (diagonally) is too
 
Ah, that's a nice result, and was the sort of 'function-based' thing I was trying to think about. Thank you again.
 
more importantly that's an iff; a map to $X \times Y$ is continuous iff each of the maps to $X$ and $Y$ (after projecting onto each factor) are
 
Bib
2:29 AM
Suppose I have a homomorphism of algebras $\alpha \colon \mathbb{C}G \to \operatorname{End}(V)$ where $V$ is a vector space and $G$ is a group. Does $\operatorname{im} \alpha$ consist of the modules of $\operatorname{im} \alpha$?
 
2:49 AM
@RandomVariable what is the integral before differentiating?
 
3:32 AM
how is it possible that a space might not be homeomorphic to the disjoint union of its connected components?
 
@SamuelYusim Take $\{0\} \cup \{1/n \mid 0<n \in \mathbb{N} \}$
You can have things get closer together, so that the connected components topology might be to fine of a topology
Actually let me think...
 
would a homeomorphism of that into $\mathbb{Z}$ not be $1/n \mapsto n$ and $0 \mapsto 0$?
and then you can definitely send that to a disjoint union of connected components
 
$\{0\}$ is an open set in $\mathbb{Z}$ but not in the topology I described
 
oh, duh
okay, then how on earth would one show that this set has the desired pathology?
 
Well, you know the disjoint union topology, and describing the connected components should be quite easy, so show that the subspace topology does not coincide with the disjoint union topology, using whatever methods you would like
 
3:44 AM
oh, sure
 
Some more examples, would be the rationals, the irrationals, and the cantor set.
 
Finally finished understanding pre req required to understand GNS construction proof
 
user147690
How are you doing @Paul?
 
4:00 AM
Doing okay, just moved into my new place today, glad that is over. How about yourself? And how is (was...?) the tropical geometry project? @AlexClark
 
user147690
I am good and Tropical geometry stuff is very interesting. We are hopefully choosing a specific open problem in the field tomorrow, and will work on that for the next three weeks or so.
 
user147690
Lots of polyhedral geometry xD
 
user147690
Uni is back on again, and diff geo looks really interesting
 
Sounds like it is coming along. How did all the preprep (that I gave you a hard time for doing) go, taking all the classes?
I might take diff geo this semester
 
user147690
Taking the four courses and the prep went nicely I think
 
user147690
4:03 AM
Mostly I was just making sure I had all the stuff from last semester down(in the relevant areas) and learning some vocab
 
user147690
I mean I could have done much more, but I spent lots of time on Tropical geometry(despite being very bad at it I found[mainly the geometry xD])
 
So you got the tropical part all figured out :P
 
user147690
Haha yep, studied on the beach
 
user147690
Weirdly enough I expect algebraic methods of physics(3rd year) to have harder algebra than advanced algebra(which is a 4th year course)
 
I would guess you would end up digging deeper into certain parts of algebra, while the advanced algebra is suppose to give a taste of everything and make sure you know some basics
 
user147690
4:14 AM
That does make sense
 
user147690
Good call
 
user147690
Week 7 or so we will be doing Hopf algebras, and I looked at the notes and they seem intense
 
Bib
question: what is "Howe duality"? And where can I find a reference for it?
 
user147690
Just the conditions for a lie superalgebra look intense aswell :P
 
Sounds fun
 
user147690
4:19 AM
What are you studying atm?(not necessarily this minute, but what textbooks if you have any atm)
 
I haven't really done much for a while, with finding a place moving. But I guess I have been doing bits of algebra, to fill some holes, and a little algebraic topology. I guess I also did some research for this problem, which I read a little bit of some of the papers on the subject
It is always fun to run into something by Yves de Cornulier
 
4:34 AM
Oh and for textbooks Algebra Ch0 and Algebraic Topology by Hatcher
 
user147690
Ahhh I really want to read some algebra ch0 for that categorical perspective, but I never have time
 
5:11 AM
@robjohn $ \displaystyle \text{Li}_{s}(-x) = - \frac{1}{\Gamma (s)} \int_{0}^{\infty} \frac{t^{s-1}}{e^{t}/x+1} \, dt$
 
i really cannot do factoring on expressions for the life of me. been trying all night
its jut not intuitive to me
I'm off to bed - hopefully it will click tomorrow.
 
Bib
5:28 AM
How can $GL(V)$ be viewed as a subgroup of $Sp(V \otimes U)$?
 
 
1 hour later…
Bib
6:40 AM
Anyone here comfortable with rep theory?
 
Huy
7:36 AM
@MikeMiller: Still up?
 
"Geometric logic is a syntactic description of a Grothendieck topos" wut.
 
Huy
@BalarkaSen: Do you know about twisted bundles?
 
no.
 
Huy
I should pick my courses according to knowledge in this chatroom. :(
 
haha
Ted or Mike will surely know about this.
 
Huy
7:39 AM
Yeah, but Ted's rather busy these days I think. I guess I'll have to wait till Mike is up again.
 
yeah
 
Huy
I'm scared if I take the QFT course, nobody here will be able to answer. Well, maybe Semiclassical.
I don't want to go into the physics room to ask, they're all mean.
 
apparently there has been some physics discussion building up around here lately, mostly by Semiclassical and Karim and another guy I am forgetting.
so you'll be fine, I guess
 
Huy
Interesting.
Hopefully.
BTW, are you staying in India for your studies or do you have plans to go to a university abroad, @BalarkaSen?
 
presumably the latter, but I am not bothering about that at the moment. still got 2 years in high school to finish.
 
Huy
7:42 AM
Ah, I see.
Any favourites?
 
princeton, maybe. france has a few too, but the mathematicians there are mad.
 
Huy
How mad? :D
 
as mad as Bourbaki :P
 
@RandomVariable That makes sense. Adding powers of $-x$ to the definition of Dirichlet's eta function.
 
on a serious note, I think france has gotten over their algebra-obsession. i think Gromov being in IHES has been a positive influence, although it might have been going off from way earlier.
 
Huy
7:45 AM
I only know Cartan from the founding members of Bourbaki. I know they liked Hilbert's axiomatic method and wanted to do the same for other fields. What else did they do, @BalarkaSen?
 
they liked to view everything algebraically. have you read V. I. Arnold's writings on this?
 
Huy
No.
 
let me link it to you
 
Huy
Sure. Maybe I have, but I don't remember it.
 
Huy
7:47 AM
Ah, yes, that seems familiar.
I definitely have.
 
Arnold might have been exaggerating a bit (he hated french mathematics for some reason), but a lot of it was true.
 
@Alizter they'd be hard pressed to live without themselves...
 
8:19 AM
> To the question "what is 2 + 3" a French primary school pupil replied: "3 + 2, since addition is commutative".
 
8:52 AM
@robjohn do you see an easy way of calculating this one? $$\int_0^{\pi/2} \left(\cot ^2(x) \log (\cos (x))\right)^2 \, dx$$
 
@Chris'ssistheartist Not off hand, but I will think for a bit.
 
@robjohn I only wanted to know your opinion, if it's anything obvious.
OK
 
@Chris'ssistheartist Nothing obvious right now.
 
9:13 AM
@robjohn It has a nice closed form $$\frac{\pi ^3}{24}+\frac{\pi }{3}+\frac{1}{2} \pi \log ^2(2)-\frac{4}{3} \pi \log (2)$$
 
10:00 AM
@Chris'ssistheartist Okay, I don't feel stupid for not seeing something immediately :-)
 
@robjohn It can be viewed as beta function, but this way is still hard. I was just asking, I don't have a magnificent way either.
 
@Balarka So is the point of exercise 26 that one is the direct sum and the other is the direct product?
 
@Khallil How are you
 
Okay, not quite.
 
Hey @Alizter! It's been a while :-)
 
10:15 AM
@Khallil How is warwick
 
Not too bad, how about you?
Warwick's been good fun!
 
I think I will apply to warwick
I am not too sure about cambridge anymore
 
@Alizter No? If you need convincing, ask @Alyosha!
I'm sure you're more than capable for Cambridge, @Alizter! :-)
 
I feel like it isn't for me
I am even considering studying in turkey
Most universities are good universities thats the problem
It's weird because a year ago I was very decided and dedicated
Now I am lost
woops
 
Are you going into lower sixth this September, @Alizter?
It's good to be lost sometimes.
 
10:20 AM
@Khallil upper
I am writing UCAS now
How did you get into warwick?
 
The same way! :-)
 
@Alyosha $H_1(X, A)$ is indeed countable direct sum of $\Bbb Z$. $H_1(X/A)$ is a bit exotic (keyword : Hwaiian earring), but yeah, it's larger than $\prod \Bbb Z$, so you're done.
OK, here's a problem for you (took me a whole afternoon to figure out) : It's known that if $(X, A)$ has homotopy extension property, then $H_\bullet(X, A) \cong H_\bullet(X/A)$. Is it true that if $H_\bullet(X, A) \not \cong H_\bullet(X/A)$, then $(X, A)$ doesn't have homotopy extension property?
 
Yes, I did the thing of thinking it's equal to, not larger than.
 
10:36 AM
@Alyosha off topic : it's only slightly larger than $\prod \Bbb Z$. $\pi_1$ of Hawaiian earring is isomorphic to some quotient of $\varprojlim F_n$. I don't know what's the abelianization of this is, though.
 
Why is that only a little larger?
 
I expect the abelianization of that inverse limit to be not as large as the fundamental group in turn is.
i don't have the time right now, but you can take it as an interesting project to figure out what really the abelianization is.
@Alyosha did you do 22?
 
I think so. Is it not basically obvious?
That is, I did it modulo carefulness.
 
You should stop saying the word "obvious". "Not hard" is a good alternative. :)
Yes, it is not hard, but it's a key to an important tool in homology.
 
Okay. Does 'obvious' sound mean or such?
 
10:44 AM
It's generally frowned upon among the mathematical community.
 
It seems silly if its usage is essentially the same as 'not hard'.
 
For example, "obvious" facts/theorems/proofs might actually have a very deep interpretation and is useful in many contexts.
@Alyosha It's usage is not the same has "not hard", thus the frowning-upon.
Saying something is "obvious" is as if you are refuting it's (however little) usefulness outright.
If you are prepared to listen, I can convince you that that exercise is very useful.
 
I see. Thanks for this.
OK.
 
OK. We will not even need much of the exercise. Recall that Hatcher notes in 2.1.22 that $H_n(X^{(n)}, X^{(n-1)})$ is free abelian on the $n$-cells of the CW complex $X$.
This is the key fact one needs in cellular homology. The goal of constructing cellular homology is to have a homology theory which takes it's values on some chain complex associated with the CW-structure of your space and spits the group.
This means if we can arrange $\{H_i(X^{(i)}, X^{(i-1)})\}$ in a chain complex, we're done.
 
Its values meaning what?
Or is the 'takes its values on' a redundant clause?
 
10:51 AM
The standard singular homology group takes it values on the chain complex associated to the singular structure of your space, not?
 
OK, in that sense.
 
Similarly, our hypothetical "cellular homology" would take it's values on the chain complex associated to the cell structure of your space.
 
Is cellular homology harder to work with than simplicial?
 
Of course not. Cellular structure is easier than simplicial structure.
 
Because if not I don't see why we bother with simplicial. It just seems to be similar but with more restrictions with how we glue.
 
10:54 AM
You can't define cellular homology without defining simplicial/singular homology.
Cellular homology is some sense is "homology of homologies"
(indeed, the generalization of this idea is provided by the beautiful notion of spectral sequences)
 
I mean, surely one still has the notion of a chain in CW complexes?
 
Yes.
:)
 
Then you can build a homology on it independently?
Mm, yes.
 
whoops, mis-delete.
 
OK, anyway. What does this exercise tell you?
 
10:56 AM
@Alyosha no, as those chains are elements of $H_n(X^n, X^{n-1})$.
8 mins ago, by Balarka Sen
OK. We will not even need much of the exercise. Recall that Hatcher notes in 2.1.22 that $H_n(X^{(n)}, X^{(n-1)})$ is free abelian on the $n$-cells of the CW complex $X$.
It's the key to the construction of cellular homology.
 
Oh right, I see.
OK, there is nothing more concrete?
 
what do you mean?
the exercise doesn't tell anything else which is interesting. but there is a lot to do about cellular homology.
we haven't even constructed our chain complex.
 
OK. Is there anywhere nice to read about cellular homology?
I guess May is pretty opaque.
 
I find Hatcher a good read.
It's just in 2.2.
 
I'll go on to the exercises in 2.2 for a while.
 
10:59 AM
You can't do them without knowing the theory in there.
 
I know. By doing exercises I mean backfilling knowledge and doing exercises.
 
ok, right, sure.
make sure you learn the cellular boundary formula thoroughly when you do cellular homology. very important.
personal philosophy (you can ignore this) : I find cellular homology interesting as it, in some sense, computes homology of homologies. Spectral sequence is a generalization of this idea : it contains pages (marked as $E^{n}$) of informations, each page encoding "homology of homology of ... of homology" of your space. This thing keeps a hell lot of data about your space, and is an immensely powerful tool.
I am talking about spectral sequence of a filtered complex (as a special case, a CW-complex with the filters being the collection of k-skeletons), however. There are apparently things called Serre spectral sequences I can't make head or tails of.
anyway, I have to leave. @Alyosha have fun with algebraic topology, and don't forget to do this exercise.
 
Thanks again, bye!
 
11:19 AM
Hello. Is anyone here familiar with series?
and it expires soon. Sorry, I just wanted to attract some attention
I'm very much concerned about this problem
 
11:40 AM
@BalarkaSen both yesterday and today were cool. got a 10 on a test with some wacky binary quadratic form $f$ (although he didn't call it that :P). apparently you had to prove that if $a\sim b\iff f(a,b)=0$ then $\sim$ is an eq. rel. teacher was impressed for some reason.
do you actually think that the proof of FLT is useless?
@PaulP where are you in ch. 0?
 
Hey@Soham
 
You know @Soham I learned a proof today which made me stand and clap
 
@BalarkaSen the way you've stated the exercise, if you assume the well known fact then the thing you're trying to prove follows from truth-table type logic?
 
11:50 AM
Cartesian product of compact sets are compact (using tube lemma)
You know the proof @Soham
 
Read it .. Amazing proof I tell ya
Since you have done ... I guess till rings .. you mind if I ask you a question?@Soham
 
I won't be able to answer, most probably. Shoot.
 
How do rings and groups look like (I mean pictures like in topology.. Not the cayley graphs .. mistake in the spelling I guess)
 
Minus Cayley graphs and subgroup lattices, I don't really know of anything.
 
11:55 AM
Ahh... Then I guess groups are just algebraic stuff , No way of thinking about them
and rings
Though still I might be 200% wrong :p
 
You have topological groups like $SU_2(\Bbb R)$ and all.
 
@BalarkaSen Local degrees seem nice. Anything that looks a little like brached covers makes me happy.
 
In mathematics, the special unitary group of degree n, denoted SU(n), is the Lie group of n×n unitary matrices with determinant 1 (i.e., real-valued determinant, not complex as for general unitary matrices). The group operation is that of matrix multiplication. The special unitary group is a subgroup of the unitary group U(n), consisting of all n×n unitary matrices. As a compact classical group, U(n) is the group that preserves the standard inner product on Cn. It is itself a subgroup of the general linear group, SU(n) ⊂ U(n) ⊂ GL(n, C). The SU(n) groups find wide application in the Standard Model...
 
@SohamChowdhury Even I would think that because I have no idea about how it is done
 
Well, it introduced a lot of new techniques in arith. geo. (which Balarka understands far better than I do).
It was not a direct proof, as you know.
 
12:01 PM
Yes TSC
was involved
But it is no point discussing it .. because we have no damn idea how it works
So what have you been doing @Soham ?
 
schoolwork.
a little algebra.
 
Hmm... I finished my school work just now and I am free
3
Q: L'Hospital rule, exponental ratio

Sarah$$\lim_{x\to ∞} \frac {x^{1000000}} {e^x}$$ could anyone please provide some hits with what result I will end up? After all applyings of L'Hospital rule, I will get $\frac {n} {e^x}$, where $n$ is large number before I got out of the $x$ powers. So, will it be the limit $0$ then? Since the infin...

Look at what is written in the question
 
@SohamChowdhury Nice way to think
@Soham So you have done all the questions from DF?
Till the place you have studied
 
nope. Artin.
 
12:15 PM
Oh.. How is artin??
Because I have gone crazy doing questions
47 question in just subgroups and some 20 odd in centralizers
@Soham you have the pdf of artin?
 
12:46 PM
@SohamChowdhury no, absolutely not. the proof of FLT is of immense importance.
at least, as far as I know.
 
Hello@Balarka
 
@Alyosha no. You know the statement, and you're trying to show if the converse holds. It's not a logic problem - you have to do some concrete topology to solve it.
hi
 
@Balarka What are your views on the proof that the Cartesian product of compact spaces is compact (finite product)
 
it's nontrivial (tychonoff theorem). you needn't restrict to finite products.
 
I liked the proof of finite products a lot for some reason
Though I haven't read about Tychonoff thoerem
But I feel it somewhere uses finite intersection property...
 
12:51 PM
huh? what you're saying is precisely tychonoff theorem.
how come you don't know the name?
 
I am saying that I like the proof of that finite product of compact spaces is compact
Tychonoff theorem is I guess infinite product of compact spaces is compact
 
oh, finite.
sure.
but I'd still call that Tychonoff theorem.
 
Ya I guess so.
@Balarka Does tychonoff theorem use finite intersection property somewhere ...
 
sorry, @Remember, I am a bit busy right now.
 
Fine fine ...
:)
 
12:55 PM
what do you mean by the "finite intersection property" anyway?
 
A collection $\mathscr{C}$ of subsets of X is said to have the finite intersection property if for every finite subcollection
$\{C_1,....,C_n\}$ of $\mathscr{C}$ the intersection $C_1\cap \cdots \cap C_n$ is nonempty
 
ok. I don't know. I have to search through/recall the whole proof to see if it's used anywhere.
Why do you think it's being used?
 
We have this theorem about the finite intersection property saying that the if for every collection of closed sets in a set $X$ have the finite intersection property and the intersection $\bigcap_{c\in \mathscr{C}} C$ of all the elements of $\mathscr{C}$ is nonempty then the space $X$ is compact
 
hmm. makes sense. perhaps it has been used.
i dunno, not willing to recall tychonoff right now.
 
Okay :)
 
user147690
1:24 PM
That moment when you open up some review document that was never mentioned in the first two lectures and there are 35 questions that are beyond your level xD
 
Hello@AlexC
 
user147690
Hey
 
user147690
Just reviewing my advanced alg class atm
 
user147690
and dying
 
And dying?
 
user147690
1:27 PM
Yes, it's very hard straight off the bat, going to be little time off this semester(which is fine since its the first semester I am single haha)
 
Oh..
 
@AlexClark Is the syllabus hard?
 
user147690
Well I was most worried about algebraic physics from the syllabus
 
user147690
But the review book that we are assumed to know everything from goes a little beyond where I was
 
Which things are assumed?
 
user147690
1:31 PM
Mostly things I haven't been able to think about in regard to ring theory
 
Like?
 
user147690
Like exercises that I haven't done, like proving three isomorphism theorems
 
you don't know how to prove the three isom theorems?
 
user147690
I could probably work it out, I am sure I have done the first two for groups
 
user147690
Mainly it is the volume of things I haven't thought about that are assigned exercises
 
1:33 PM
yeah, I think you can work it out too.
what else is there?
 
user147690
Find a subring of this, show that the only ideals are this and that, describe factor rings, radical ideals
 
I guessed that's what you meant, but what you actually wrote was not that.
3
 
user147690
Ahhh I guess it's not so bad @BalarkaSen, I guess I was overwhelmed
 
user147690
By the volume
 
By the way, how do you prove the 'it can be shown' part?
 
user147690
1:36 PM
I am just worried it'll be another semester when I have one lecturer who thinks to himself "This is their only class, better give them a full work load"
 
I just mean "is there a pair $(X, A)$ such that $H_\bullet(X/A) \cong H_bullet(X, A)$ but $(X, A)$ doesn't have homotopy extension property?"
oh, I see that's not what I wrote.
apologies for that, @Alyosha
@AlexClark huge amount of work can be scary, yes.
 
1:54 PM
@Alyosha what do you mean by the "it can be shown" part? that (X, A) has HEP => H_*(X/A) = H_*(X, A)? it's in Hatcher, but he calls them "good pairs".
 
2:54 PM
@Semiclassical, if two OPS share the same recurrence relation, do you know if their weights must have the same support?
 
@SohamChowdhury There is no one chapter I am going through but I am focusing on parts of Ch2 through Ch7 to fill in some holes, and pick up things that were not covered in detail when I actually took algebra. How about your travel through Ch0? Although I think you switched to topology and competition math.... How is that going?
 
that's a plausible claim, but i'm not sure i know how to (dis)prove it
same recurrence relation for OPRL means (almost?) identical jacobi matrices, and from there one has a one-to-one correspondence to weights by Favard's theorem
if i were to make a guess, it'd be that the absolutely continuous part of the weights would have the same support, but perhaps not the singular components
 
i suspect one could write down a linear combination of Chebyshev polynomials that'd be such an example.
a question for you in turn: have you done much work with semiclassical OP weights? (i find it amusing that i'm having to ask about that, given my name :P)
 
3:04 PM
I had to look it up - I'm unfamiliar with semiclassical OPS
 
alright. if i'm remembering right, the simplest definition is that the absolutely continuous part of the weight be of the form $e^{-V(x)}\,dx$ with $V'(x)$ a rational function of $x$
 
3:18 PM
@Huy: If you ping me when you ask a question, I'll probably see it eventually, egnow.
 
Huy
@MikeMiller: I know, yeah.
 
3:30 PM
In the order topology of an ordered set, is an "isolated point" just one without a predecessor or without a successor (or both)?
Well, if it has a last point, that point has no successor but it still might be isolated.
In the order topology of an ordered set without endpoints, is an "isolated point" just one without a predecessor or without a successor (or both)?
Also, if $A$ is a countable ordered set without endpoints, and $A$ has no isolated points in the order topology, must $A$ be homeomorphic to either $\mathbb Q$ or $\mathbb Q\cap\mathcal C\setminus\{0,1\}$ (where $\mathcal C$ is the Cantor set)?
 
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