Timeline for What is the quotient (pseudo)metric $d_\sim$ and how do I identify the infimum of possible sequences in this instance?
Current License: CC BY-SA 4.0
27 events
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| Jan 20, 2021 at 13:59 | comment | added | Robert Frost | You say I don't really need to quotient as the quotient can be identified with $X$. I want to show $g=x+\frac{21}{64}\cdot2^{\nu_2(x)}$ converges on transfinite iteration. For some $x:\lvert x\rvert_3=1$, the infinite limit of $g^n(x)$ satisfies $\lvert x\rvert_3=3$. Fine, so apply $f^{-1}$ and I'm back in $X$. Therefore I can just use $j(x)=f^{-1}\circ \lim_{n\to\infty}g^n:X\to X$. Does your $h$ help me with the proof that $j$ converges on iteration? The claim that it does converge is equivalent to the Collatz conjecture. | |
| Jan 11, 2021 at 16:29 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jan 11, 2021 at 15:58 | comment | added | LSpice | @samerivertwice, I agree: $(f \circ h)(x) = \frac{2^{x + 1}}3 = 2^{x - \log_2(3) + 1} = h(x - c + 1)$ for $x < c$, and $(f \circ h)(x) = \frac{2^x}3 = 2^{(x - \log_2(3) + 1) - 1} = h(x - c)$ for $x \ge c$. | |
| Jan 9, 2021 at 20:34 | comment | added | Pietro Majer | Topologic conjugacy, as it is written in the last line. | |
| Jan 9, 2021 at 20:04 | review | Suggested edits | |||
| Jan 10, 2021 at 1:19 | |||||
| Jan 7, 2021 at 20:10 | comment | added | Robert Frost | Thank-you. This will provide a good month or two of food for thought for me. I am much obliged to you. | |
| Jan 7, 2021 at 17:44 | comment | added | Robert Frost | Yes, that's it. That is the ultimate motivation of this question... I hoped this question would either be it, or be a stepping stone towards it. The convergent sequences or open sets I'm interested have the recurrence relation $x_{n+1}=x_n+21\cdot2^{\nu_2(x_n)-6}\cdot3^{\nu_3(x_n)}$ | |
| Jan 7, 2021 at 17:25 | comment | added | Pietro Majer | So you mean the quotient topology on $Z/\!\sim$ for the topological space $Z$ (as a subspace of $\mathbb R$) and the equivalence relation, as defined in your post here? | |
| Jan 7, 2021 at 17:23 | comment | added | Robert Frost | Simply $Z/{\sim}$ in the quotient topology as per math.stackexchange.com/a/3810857/334732 which may be stronger than the quotient pseudometric. I.e. the standard absolute value metric from $\Bbb R$ reduced by the equivalence relation $\sim$. Or maybe it's no stronger and it coincides with the pseudometric topology in this particular case? | |
| Jan 7, 2021 at 16:44 | comment | added | Pietro Majer | ok but which quotient topology? | |
| Jan 7, 2021 at 15:54 | comment | added | Robert Frost | I'm sorry to push for more - please feel free to ignore. This is to do with the motivation behind this question: Consider the $n$-indexed sequence $\dfrac{2^{6n+2}-1}6=\frac12,\frac{85}{128},\frac{5461}{8192}\ldots$. This converges in $\Bbb R$ to $\frac23$ and in $\Bbb Q_2$ to $-\frac16$. In the trivial pseudometric here it trivially converges as it's the identity sequence. But it also converges to $[\frac12\sim\frac23]$ in the quotient topology (non-pseudometric). Any pointers how I might attempt to derive/construct this quotient topology and show it is Hausdorff or nontrivial? | |
| Jan 5, 2021 at 16:21 | comment | added | Pietro Majer | You don't really need to quotient, as your quotient set $Z/\sim$ can be identified with $X$. Alternatively, you may even consider $f$ on all rationals, and quotient on the $\mathbb Z$ action, so that equivelence classes are the whole orbits of $f$. Any class then contains exactly one element of $X$. | |
| Jan 5, 2021 at 15:54 | comment | added | Robert Frost | It seems LSpice overlooked the condition in the first sentence: whose $3$-adic valuation is either $−1$ or $0$. This constrains the equivalence classes to have cardinality two. I'm just checking you didn't overlook the same, and whether your answer still stands notwithstanding this? | |
| Jan 5, 2021 at 14:56 | history | edited | LSpice | CC BY-SA 4.0 |
Light proofreading and TeXing
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| Jan 5, 2021 at 13:29 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jan 5, 2021 at 12:16 | comment | added | Pietro Majer | Don't worry, it is perfectly normal that we sometimes miss something, and we have the right to ask any question --even because sometimes "trivialities" hide not-at-all-trivial facts. | |
| Jan 5, 2021 at 11:39 | comment | added | Pietro Majer | Yes, it was meant to be a proof (sorry if it wasn't clear). Having every orbit dense is a conjugation invariant (recall that $f=hgh^{-1}\Rightarrow f^m=hg^mh^{-1}$ for all $m$), and the translation $x\mapsto x+c\mod 1$ on $\mathbb{ T:=R/Z}$ does have every orbit dense if (and only if) $c$ is irrational (for the orbit of $x$ is $x+c \mathbb Z\mod 1 $), so every orbit of $f$ is dense too. | |
| Jan 5, 2021 at 10:30 | comment | added | Pietro Majer | PS: The quotient (semi)distance $\tilde d$ on $\tilde X$ for a quotient $\pi:X\to\tilde X$ is the maximum (semi)distance on $\tilde X$ that makes $\pi$ a $1$-Lipschitz map. This leads to the construction of $\tilde d$ via chains of jumps from a class to anothe (which I would call the construction rather than the definition of $\tilde d$ ). | |
| Jan 5, 2021 at 9:04 | history | bounty ended | Robert Frost | ||
| Jan 5, 2021 at 9:03 | vote | accept | Robert Frost | ||
| Jan 4, 2021 at 23:28 | comment | added | Pietro Majer | 3) in this semi-distance, $1/2$ is close to the right-end, for $f(3/4-\epsilon)= 1-4\epsilon/3$ and $f(3/4+\epsilon)= 1/2-2\epsilon/3$. So nothing changes if one starts from a circle, identifying the endpoints of $[1/2,1]$, instead using $[1/2,1)$, which makes $f$ a homeomorphism. This allows to use the tools of homeomorphims of the circle. I wrote $f$ as a time-1 flow to compute its rotation number and find a conjugation with a rotation: only at the end I realized the conjugation was so easy. | |
| Jan 4, 2021 at 23:22 | comment | added | Pietro Majer | On the PS: To me, it was not obvious, and I wasn't even sure that the distance was trivial. How I proceeded, it's nothing special, but here it is. 1) It seemed that $X$ and $Y$ do not play a special role in studying these objects, and that it is healthier to have $f$ as a self-map, to make iterations. 2) one works better on the interval, than on the quotient. | |
| Jan 4, 2021 at 23:02 | comment | added | Pietro Majer | I think so (quotient being Hausdorff) but I do not see immediately. | |
| Jan 4, 2021 at 22:58 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jan 4, 2021 at 22:33 | comment | added | Robert Frost | P.S. was this fairly instantly obvious to a better mathematician than I, or did it take some degree of thought? It seems fairly obvious the way you put it about every orbit being dense - must imply the distances between them are zero. | |
| Jan 4, 2021 at 22:32 | comment | added | Robert Frost | Thank-you most kindly. This will take some time for me to digest properly to my lack of skill but I think the gist is that the pseudometric is $d_\sim=0$. Does this necessarily push through to the quotient space not being Hausdorff? I assume not because the quotient space is richer than the quotient metric. | |
| Jan 4, 2021 at 22:07 | history | answered | Pietro Majer | CC BY-SA 4.0 |