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12:03 AM
@leslietownes Dude... that's not cocaine. It's Tide.
 
psie all joking aside wikipedia is fine for exactly that kind of use, just do not think of it as a place for double- or even single-checking hypotheses of theorems relating to the fourier transform, or even formulas about it
 
and remember to smoke crack
 
Quickly skim off as much cream as you can.
 
postal dude smokes crack, why can't you
 
 
2 hours later…
2:25 AM
@TedShifrin are you curious about $\text{RSpec}$ notation as well? I could just ask about it on the site
 
Sure, I’d be interested in an explanation.
 
3:16 AM
@TedShifrin Apparently, the "$R$" is short for real, because its called the real spectrum, an analogy for the spectrum for the formally real fields
 
 
11 hours later…
2:26 PM
@Balarka The notion of a group acting on a thing can be generalized in two ways: 1. the notion of a monoid acting on a thing, 2. the notion of a groupoid acting on a thing. There is a common generalization in the notion of a monoidoid (i.e. a category) acting on a thing. To be concrete, a category $\mathcal{C}$ acts on a category $\mathcal{D}$ if you have a decomposition $\mathcal{D}=\coprod_{x\in\mathrm{Ob}(\mathcal{C})}\mathcal{D}_x$ (equivalently, a functor $w\colon\mathcal{D}\rightarrow\mathrm{Ob}(\mathcal{C})$) and to every morphism $f$ in $\mathcal{C}$ we associate a functor $f.\colon
 
Let $\lambda$ be a Hausdorff space with certain properties and let $\tilde{\lambda}\to\lambda$ be a normal covering space with abelian deck group $G$. By certain property of $\lambda$, there is an associated $\Bbb Z$-module $T(\lambda)$. So $G$ acts on $T(\tilde{\lambda})$, making the latter into a module over the group ring $\Bbb Z[G]$. Any $G$-module $B$ determines a bundle of twisted local coefficient over $\lambda$.
Why is the last statement true? what does that even mean?
derekhsorensen.com/docs/sorensen-local-coefficients.pdf I think this note is related but still I can't interpret the statement
 
2:54 PM
@Thorgott this ought to be a special case of the Grothendieck construction for discrete fibrations or whatever lol
 
 
1 hour later…
4:11 PM
This question just closed in real time for me. Xander takes no prisoners
 
 
2 hours later…
5:55 PM
@Thorgott This is too much category theory for me, I am afraid
 
fair, it's not very insightful anyway
just a funny obversation
 
6:14 PM
@Thorgott I prefer to say that the cat of ringed spaces is the Grothendieck construction of the pseudofunctor $\mathbf{Top}^{op} \to \mathbf{Cat}, X \mapsto \mathbf{Sh}(X)$
oh you did mention Grothendieck constructions yourself, I see
I don't think discrete fibrations is enough. The fibration $\mathbf{Sh} \to \mathbf{Top}^{op}$ doesn't have discrete fibers
 
@onepotatotwopotato Horrible; $\lambda$ cannot be a serious notation for a space. Anyhow, the idea is that if $X$ is a (nice) space, $\rho : \pi_1(X) \to H$ is a representation, take $E := (\tilde{X} \times H)/\pi_1(X)$, where the action is happening diagonally.
Here, $\tilde{X}$ is the universal cover of $X$. Then, $E \to X$ given by forgetting the second factor is a covering map.
 
hi @Balarka!
 
6:42 PM
@Lukas some times ago you were asking for recommendations to study some set theory, did you end up doing that in the end?
 
@Alessandro I didn't get very far
 
A transitive set is defined as a set $A$ wherein $x \in A \implies x \subset A$. What can we say about sets obeying the converse?
 
@LukasHeger sad
 
@EE18 there is no sets obeying the converse
 
I may teach a set theory course next year
 
6:45 PM
They are too big?
 
@EE18 sets are not allowed to contain themselves
@AlessandroCodenotti sounds cool
 
Not that they're too big, they just don't exist
 
I hope you include topos theory, lol
 
I will teach a course for sure, but it's not clear yet whether it'll be model theory or set theory
It won't have any topos theory in either case though
 
they don't exist in ZFC*
 
6:46 PM
there's even a topos-theoretic description of forcing
 
Why can't sets contain themselves?
 
We can't have $A\in A$ because of the axiom of foundation
 
More specifically, why can there not exist even a single set that contains itself
I think that is the one axiom Halmos does not talk about lol
 
Axiom of foundation implies that there exists $x\in \{A\}$ disjoint from $\{A\}$
that is $A$ and $\{A\}$ are disjoint, that is $A\notin A$
you see, some other axioms are involved as well, but main thing here is axiom of foundation
There exist models of $ZFC\bar{\ }$ for which $R\neq V$ where $R = \{x : x\notin x\}$ is the Russel class and $V = \{x : x = x\}$ is the universe class
where $ZFC\bar{\ }$ denotes $ZFC$ without the axiom of foundation
in other words, if foundation fails, then there can exist sets $x$ for which $x\in x$
 
Got it, thank you Jakobian
See also Naive set theory for the mathematical topic.Naive Set Theory is a mathematics textbook by Paul Halmos providing an undergraduate introduction to set theory. Originally published by Van Nostrand in 1960, it was reprinted in the Springer-Verlag Undergraduate Texts in Mathematics series in 1974.While the title states that it is naive, which is usually taken to mean without axioms, the book does introduce all the axioms of ZFC set theory (except the Axiom of Foundation), and gives correct and rigorous definitions for basic objects. Where it differs from a "true" axiomatic set theory book...
Seems Halmos omits only this foundation axiom (though I have as yet not seen him define Axiom schema of replacement unless it's tacit somewhere i've missed)
 
6:53 PM
Axiom of foundation is not really needed for anything
its reasonable to not assume it
 
is replacement axiom equivalent to existence of the empty set?
 
but its also reasonable to assume it, we don't lose anything by assuming axiom of foundation
@EE18 no
 
Where does the existence of the empty set enter then? Axiom of infinity?
 
axiom schema of comprehension implies that the empty set exist (in addition to assumption from logic that there exist a set)
 
ah ok
so one must assume a set exists (and i guess this is the entire purpose of the axiomatic method)
 
6:56 PM
You take a set $x$ (its guaranteed to exist from logic)
Then you take a set $y$ such that $z\in y \iff z\in x$ and $z\neq z$ by axiom schema of comprehension
 
When you say guaranteed, do you mean in the sense described here (math.stackexchange.com/questions/4867823/…)?
 
the existence of a set is guaranteed by the axiom of infinity
 
maybe but I'm considering foundations weaker than that of ZFC
existence of a set is a convention from logic
its similar to the assumption that all universal algebras are non-empty
I'm just struggling linguistically to describe this
 
Can you explain the construction/argument then Lukas? Axiom of inifinty as given to me says there exists a set containing the empty set and all its successors. But if we use the axiom to infer that a set exists (so that we can use comprehension/specification to extract the empty set therefrom), isn't this all circular?
 
The axiom of infinity can be stated in a way that does not involve the empty set
 
7:04 PM
just like in above, you should assume a set exist
 
@Jakobian I don't agree on this. If you're doing model theory, then you can define things as you like, but ZFC is commonly used as a metatheory. If your metatheory can't show that a set exists, you have a problem
@EE18 just replace the part in the axiom that says $\varnothing$ by $z:\forall x: x\notin z$
 
Got it, so, to be sure, you would define the axiom of infinity as "there exists a set which contains the set that does not contain any other set and which contains every successor of this set"?
 
Correct
 
I feel like I must be missing something. What I wrote above is basically equivalent to saying the following: "Axiom of Infinity. The empty set exists. The set which contains the empty set and all of its successors also exists."
 
In axiomatic set theory, the axiom of empty set is a statement that asserts the existence of a set with no elements. It is an axiom of Kripke–Platek set theory and the variant of general set theory that Burgess (2005) calls "ST," and a demonstrable truth in Zermelo set theory and Zermelo–Fraenkel set theory, with or without the axiom of choice. == Formal statement == In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: ∃ x ∀ y ¬ ( y ∈ x ) {\d...
 
7:12 PM
@EE18 yeah, it's true that the axiom of infinity pretty much includes the axiom of the empty set by definition
you could also postulate another axiom that explicitly says that a set exists or something like that
 
You don't need another axiom
 
"This formula is a theorem and considered true in every version of set theory. The only controversy is over how it should be justified: by making it an axiom; by deriving it from a set-existence axiom (or logic) and the axiom of separation; by deriving it from the axiom of infinity; or some other method."
it appears there is controversy
 
If you're making it an axiom of logic, it's still an axiom
 
but not of set theory
 
yeah
 
7:21 PM
Another question for the experts here:
oops maybe not. will think a bit more before asking
Oh ya OK, I remember now
Wikipedia's ZFC page says: More colloquially, there exists a set X having infinitely many members. (It must be established, however, that these members are all different because if two elements are the same, the sequence will loop around in a finite cycle of sets. The axiom of regularity prevents this from happening.) The minimal set X satisfying the axiom of infinity is the von Neumann ordinal ω which can also be thought of as the set of natural numbers.
Specifically, I want to call attention to the fact that it claims we need axiom of regularity to prove there is no looping
Why is this so? Without regularity Halmos walks me through proving 1) no natural is a subset of any of its elements and 2) every natural is a transitive set.
The supposition that the sequence loops is the supposition that some natural $n$ is the kth successor of another, $n'$: $n = n'$. Thus we have $n' \subset n$ in particular while $n \in n'$ by construction of the successor. Doesn't this contradict 1)?
 
7:37 PM
@LukasHeger yeah I was just saying words, I never actually learned how the Grothendieck construction works
 
@EE18 yes you don't need regularity for this. But its a quick way to see it
sorry
you do need it to see that $I$ has no "loops"
but the set $\mathbb{N}$ which is the smallest inductive set, you don't need foundation to see there's no loops in $\mathbb{N}$
Inductive set is set that axiom of infinity tells you existence of
 
ahhhh ok got it
so axiom of infinity as stated is "there exists a successor set" *you call this $I$ and without regularity, this may have loops
had we stated "there exists a minimal successor set" then we'd have no such problems
 
not sure what successor set is
those are called inductive sets
 
successor set defined as $\emptyset$ in the set and each element of the set means a successor exists too in the set
sorry, Halmos's naming for it, didn't realize not standard
 
well maybe it is somewhat standard, I am not a set theory expert
but yes, if there is a set such that $x\in x$ then by taking an inductive set $I$ and $I\cup \{x\}$ we obtain inductive set with loops
but regardless of if such sets exist, the smallest inductive set, that is $\mathbb{N}$ by definition, doesn't contain any loops
 
7:53 PM
touche, thank you!
one sees why having a math professor is probably prudent...
much appreciated as always Jakobian
 
I'm not a professor, I only have a Bachelor
 
What I mean was someone much more knowledgeable to whom you can go with questions :)
 
That's certainly almost essential, I would say
 
8:28 PM
Halmos at the beginning of his proof of the recursion theorem:
My specific question: am I correct that he is at the very least being sloppy when he keeps saying "at most one" in the last few lines in the attached picture? That is, he should be saying "precisely one" ("at most one" is not enough; if there are no $x$ such that $(n,x) \in u$ then $u$ is not a function)
perhaps that there is one should be obvious to me from the construction though
 
@RandomVariable: Hey, there. Haven't seen you in a while. Could be my fault, as I have been swept away by RL a lot these days.
 
that there's at least one such pair is clear by definition, no?
cause $(0,a)$ is in all the sets, hence in the intersection $u$
and then iteratively all it's successors
 
8:44 PM
coming back to this just as i finished proving it to myself :)
 
EE18: i would be hesitant to use the word 'sloppy' about this, but he is emphasizing only part of what he is definitely using to say that the resulting thing is a function
 
yes i just had to use a basic induction argument
just surprised he didn't mention it, but i guess that's the style of the book as he says in the preface
 
one might infer that in context he expects the reader to see that there is at least one, or maybe he just wasn't thinking about it and didn't write it the way that you or i might have
 
that's well taken. far be it for me to call someone like Halmos sloppy after all
or from me i guess
 
the reason i wouldn't call this "sloppy" is that it implies an ideal that he is falling short of, and a book that genuinely attempts to follow that ideal is probably (a) a lot harder to read, and (b) not actually any more rigorous
i do agree that if you just paste the paragraph in from nowhere i do kinda wonder why he specifically phrases it that way twice
let's dig him up and ask
 
8:49 PM
"charmingly morbid"
 
@leslietownes in the name of education
 
yes
due to your skill in exposition during life, you have been condemned to haunt the earth forever and answer our expository questions
 
he died for our educational sins
 
"In these memoirs, Halmos claims to have invented the "iff" notation for the words "if and only if" and to have been the first to use the "tombstone" notation to signify the end of a proof,[7] and this is generally agreed to be the case. The tombstone symbol ∎ (Unicode U+220E) is sometimes called a halmos.[8]"
did not know either of these facts
 
9:11 PM
haha that's stretching the usual meaning of the term "sometimes" but ok
how much better would it have been if he had claimed to have originated both of those things in his book but he obviously hadn't
 
@leslietownes You won’t have me to dig up. You’d better ask ahead.
 
I'd say it's just not the usual meaning
 
like dr evil's dad claiming to have invented the question mark
 
>I'd say it's just not the usual meaning
after a search in dictionary, I'd say it is actually usually what it means
it appears that the author of the cited phrase didn't include the word "rarely" (or similar)
thus it seems like they might be talking about it being actually in use
 
Hi!!
 
9:19 PM
it's koro!
 
(stuck at showing that these intersect.)
Hi Leslie!!
 
Use Jordan curve theorem
 
how?
 
A "soft" question. What strategies did/do you use to remember math you read or learn?
 
\alpha divides D^2 in 2 parts.
and so does \beta.
 
9:24 PM
I always seem to forget these things. Right now the Peano axioms are burned into my brain. By next Sunday they won't be.
 
@Koro So why does alpha divide D^2 in 2 parts?
 
so I thought that if \alpha were an arc of S^1, then there would be nothing to prove as it would touch north or south pole (hence \beta).
 
Don't you want to extend $\alpha$ to a map into a bigger ball $2D^2$
 
Shh @Jakobian
 
ok
 
9:29 PM
ok if I consider \alpha to be from I to R^2, then will it help?
 
Jordan curve theorem is about embedded circles in R^2. What is your embedded circle?
You have to give an argument for why alpha divides D^2 in two regions. You haven't given one yet.
 
I guessed it should be \alpha[0,1] but it is not apparently because \alpha is 1-1.
 
That's an arc, not a circle. The statement of Jordan curve theorem does not apply.
 
Also, Halmos asks me to do the following: Prove that if E is a non-empty subset of some natural number, then there exists an element k in E such that k \in m whenever m is an element of E distinct from k. Going to think about this one on a walk but in terms of the why we're doing it. Is it basically presaging that the naturals are well ordered?
 
\alpha divides seemed obvious to me and I assumed it to see next step postponing the proof for the division.
 
9:35 PM
@EE18 minimal element
 
On the contrary it is the most nonobvious step.
 
can someone help me here
1
Q: Why are this two stochastic processes independent?

user123234 Let us consider a filtered probability space $(\Omega, (\mathcal{F}_t)_t, \Bbb{P})$, and let $M$ be an $((\mathcal{F}_t)_t, \Bbb{P})$-continuous martingale. Let $T_t:=\inf\{s: \langle M\rangle_s>t\}$. Let $(\Omega', (\mathcal{F'}_t)_t, \Bbb{P}')$ be a probability space supporting a brownian moti...

 
And once you give an argument the rest is clear
 
@BalarkaSen yeah, I don't understand how to get the circle.
 
Draw some pictures and stare at it.
 
9:37 PM
And that's why I got stuck on how to apply the Jordan curve theorem.
 
isn't well ordering the statement that there exists a total order wherein every nonempty subset has a least element?
 
@EE18 yes
 
i think you and i are saying the same thing, ok got it
thank you, helps me with the context
 
$n$ is ordered by $\in$
@BalarkaSen and since $D^2$ is Hausdorff we can replace arcs by paths too, if need be
 
@BalarkaSen I join \alpha(0) and \alpha(1) by a path outside the disk.
 
9:42 PM
That's a non-issue
@Koro Very nice.
 
this path+ \alpha[0,1] gives a circle.
 
But does that help you conclude D^2 is disconnected by \alpha?
 
In here its not an issue. I'm just pointing out the statement is a little more general
 
Unnecessary information, @Jakobian. Arcs and embedded paths are synonymous in this side of the world.
 
isn't arc = embedded path
 
9:45 PM
@BalarkaSen hmm yeah.
 
I use them interchangeably at least. I have no idea where Hausdorffness comes from
 
Similarly, \beta will disconnect D^2.
 
the proof that path-connected implies arc-connected is slightly non-trivial and requires Hausdorff assumption
 
@Koro No, wait. You did not prove alpha disconnects D^2.
You just gave me some Jordan curve in R^2. So what?
 
If $\alpha, \beta$ are just paths, then we can replace them with arcs, then use Jordan. So this is slightly more general, what we are proving
 
9:47 PM
Ok. So I get two connected open sets $U_1$ (bounded) and $U_2$ (unbounded) in R^2.
 
Oh, I get your point. Yes, of course I agree. Koro started with embedded paths, though.
 
And the circle is boundary points of U_i's.
 
I'd just say "because it's a manifold" than ponder endlessly about Hausdorffness
 
So $D^2 \cap U_1$ and $D^2\cap U_2$ should do.
 
Why is the circle boundary points of $U_i$?
 
9:49 PM
by the JCT.
 
You mean the circle you constructed.
 
I don't see how you can prove it more easily for manifolds. The proof of this requires existence of arcs in nice enough spaces, and for this we have topological characterization of the interval
 
@Jakobian Connected iff path connected in manifolds...
 
@BalarkaSen yes.
 
And once you have a path you can just easily approximate it by a PL path then remove singularities and make it embedded
It's a standard exercise.
Good to think about if you haven't thought about it before
 
9:52 PM
I had an exercise like this in my multivariable calculus class. Thanks for reminding me
 
@Koro OK. But this proves $\alpha$ disconnects $D^2$, and not that $\alpha$ disconnects $D^2$ into two pieces.
Because apriori $D^2 \cap U_1$ can have multiple connected components. Can this happen?
 
@BalarkaSen hmm. Still, if $\alpha$ is a path you want an arc in image of $\alpha$. So this won't work
 
@Koro no that won't do. I skipped the points on the circle on which I applied JCT.
 
yeah I still don't see how we could use that this is a manifold to reduce the case of paths to arcs in this case
 
@BalarkaSen this can't happen in pictures atleast but I don't know the proof.
:(
 
9:57 PM
It can happen in pictures, @Koro.
 
@Jakobian I actually needed this fact recently, even though I don't know how to prove it
 
Suppose $\alpha$ intersects $\partial D^2$
 
ok
but how can it have more than 1 components?
 
Suppose $\alpha$ touches $\partial D^2$ at two points, other than the endpoints.
 
@AlessandroCodenotti If $\alpha$ is a path, you consider an image of $\alpha$, and the topological characterization of interval $[0, 1]$ in terms of non-cut points
 
10:00 PM
I see.
 
Then you consider, iirc chains by compact balls? And their intersection, to exhibit an interval
There is a proof in van Mill with a slight error, but if you fix it, then it works
 
so how to get rid of this?
 
@Jakobian I meant path-connected implies arc-connected is very easy for manifolds. But yes, in this case you'd have to do something like: suppose $\alpha : [0, 1] \to X$ with an arc between $x, y$. Then define $\tau_t \in [0, 1]$ to be the last time $t_0 \geq t$ such that $\alpha(t) = \alpha(t_0)$. Then write $\alpha'(t) := \alpha(\tau_t)$.
 
Jeremy Brazas has a note on this fact
 
Or something to this effect
It's the standard way to generate a loop-erased Brownian motion
 
10:05 PM
I see. I needed this fact to prove that the closure of a connected, uniformly locally connected, open subspace of a Peano continuum is itself a Peano continuum
 
i mean, it shouldn't be hard to reduce this question to the PL case anyway
though i dont see why we would want to do that
 
@Koro You don't, because this case occurs.
 
You work in the space $\text{im}(\alpha)$ which is compact and metrizable (as long as the image is Hausdorff), so we aren't actually in any bizarre Hausdorff space
 
It's not true that $D^2 \setminus \alpha$ has two components. You'd want to work in a slightly larger disk, as Jakobian suggested.
 
ok, thanks. I consider R^2 - \alpha-circle.
it has two components.
 
10:09 PM
That's connected.
No, wrong
 
@BalarkaSen I bet that when trying to check that this works you'll end up with an argument equally as tricky or even trickier
 
@Jakobian I realize now that I needed the recursion theorem to make this precise
to make this n++++ and the claim made in our discussion precise
 
@Jakobian I believe it.
 
after all if this worked out easily then there wouldn't be need for weird topological characterizations of the unit interval to prove this
 
EE18: you always did need the recursion theorem to make this precise [gif of spaceman about to execute an infinite chain of other spacemen]
 
10:11 PM
haha :)
 
@Jakobian That's not a weird characterization
 
@BalarkaSen I meant \alpha circle there, which is the circle that I constructed to apply Jordan curve.
 
Math chat needs GIF support
or maybe that would cause things to devolve
 
Use a different notation, @Koro.
 
so maybe not
 
10:11 PM
Knowing that a continuum must have at least two non cut points I think it's very natural to ask which continua have exactly two
 
@AlessandroCodenotti Sorry. I was thinking from Balarka's perspective
 
Let's use $\Sigma$ for that circle.
 
if you use the word "continuum", you're nothinking naturally
 
I don't think you need that characterization of the interval (which I'm aware of), actually.
 
10:12 PM
But I won't check this
Loop-erasure, which is the obvious idea, should work.
 
I never see one. Those ideas exist, but I never seen one actually follow with it
 
$R^2- \Sigma$ has two components.
 
I think its just overconfidence
 
Most of the time we just avoid it. In the case of Brownian motions, we loop erase the random walks and take limits than loop erase BM directly.
@Koro Agree. Then?
Can you now show that $\beta$ and $\alpha$ must intersect?
 
\beta does the same.
@BalarkaSen no.
 
10:16 PM
Don't argue symmetrically for $\beta$.
Can you show $\beta$ and $\Sigma$ intersect?
Just draw a picture
 
I see from the picture that they must intersect.
 
@AlessandroCodenotti I agree. Same for triads
 
But from the same picture, I also see that \alpha and \beta intersect and that is without using jct.
:(
 
You proved R^2 - Sigma has two components, one bounded and another unbounded.
Which one does (0, 1) and (0, -1) lie in?
 
upto this point, I don't see how the current situation is weaker than the initial situation.
@BalarkaSen ok, they lie in different U_i.
 
10:22 PM
It depends on the other arc you chose outside the disk to close $\alpha$ and create $\Sigma$.
 
@AlessandroCodenotti Which continua have exactly three non-cut points was actually one of my questions to my topology professor. That's when he recommended me Nadler as a book
 
(But yes, they do lie in different U_i)
 
yeah, that's why I deleted the message. :)
 
I haven't read Nadler to this day
 
inside, outside were irrelevant here.
 
10:23 PM
(not in full anyway)
 
image of \beta is connected.
but still $\beta(0)$ and $\beta(1)$ lie in different components.
no idea. I'll think about it some other time.
 
@BalarkaSen You've become as bad as Ted :D
 
Hi @Ted :)
 
Hi, a Balarka :)
 
10:48 PM
@BalarkaSen Proof that loop erasure of a path exists?
 
Yeah.
Discretely of course it's totally trivial.
 
maybe conveniently, in this article they only mention it exists a.e.
 
Then one just takes a limit in the sense of distributions
Right, precisely
 
oh okay. So you don't think that this works anymore?
in general that is
 
I think it will work with some effort
 
10:52 PM
I have a feeling there exists a curve without loop-erasure
 
Let me know if you find an example
That would be morbidly interesting
@Jakobian Nice paper
SLE's are so beautiful, if only I understood them
 
@BalarkaSen how would you make Hilbert's curve loopless, for example?
 
11:08 PM
interesting question
I feel that you'd have to avoid traversing certain quadrants every time you erase a loop
The union of the vertices of the $2^k \times 2^k$ mesh are exactly the multiple-points, right?
 
other than that Hilbert curve is injective
points of self-intersection (I assume that's what you're asking) of the Hilbert curve, are precisely pairs $(a, b)$ such that either $a$ or $b$ is a dyadic rational
So $A\times [0, 1]\cup [0, 1]\times A$ where $A = \{\frac{r}{2^k}\in [0, 1] : r, k\in\mathbb{N}\}$ are dyadic rationals
See here
there's also an explanation how the Hilbert curve can be thought of as a map $\{0, 1, 2, 3\}^\mathbb{N}\to [0, 1]^2$ from the Cantor set to the square
each element of the sequence tells you which quadrant you are in
 
Makes sense
 
so this is similar to the standard map $\{0, 1\}^\mathbb{N}\to [0, 1]$ which maps a sequence to a number using binary expansion
 
11:24 PM
Yup
 
I didn't know hilbert's curve has such an intuitive interpretation, actually. Very neat
5
Q: Can a set of non self-intersection points of a space-filling curve contain an arc?

MichalConsider a continuous surjection $f:[0,1]\to[0,1]\times[0,1]$. It can be proved that set of self-intersection points must be dense. In the Hilbert curve, the set of self-intersections are points (a,b) such that either a or b can be written as $\frac{m}{2^k}$ for some integers $k≥1$ and $1≤m<2^k...

 
@Jakobian Of course not. But loop-erasure doesn't mean you erase all the intersection points.
 
Its just that this was the only other interesting post I found
 
Ah OK
 
If you consider only sequences of $0$ and $2$, won't that be a loop-erasure of a Hilbert space mapping $[0, 1]$ to the diagonal?
 
11:38 PM
Oh, smart, maybe. I haven't thought too hard, but sounds immediately correct.
 
oh well it was a good try
@Jakobian curve*
maybe the set of curves $[0, 1]\to [0, 1]^2$ with loop erasure is meager
I mean... trajectories of Brownian motion are a.e. nowhere differentiable too
 
11:55 PM
This sort of reminds me of a smooth version of this problem I had encountered in trying to prove the Whitney-Graustein theorem by hand
You can demarcate a set of intervals in $[0, 1]$ each of whose endpoints map to the same point in the target by the path. In the smooth setup this was finite. Then try to rearrange things so that these intervals are disjoint.
Obviously, if you could do this, then by collapsing these intervals you'd get an embedded arc in the image.
Nestedness is not a problem either, but rather overlaps are.
 

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