Timeline for (Closures of sets of) operations in topological groups.
Current License: CC BY-SA 3.0
20 events
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| Jan 9, 2012 at 8:42 | history | edited | François G. Dorais |
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| Jan 8, 2012 at 15:48 | comment | added | Yulia Kuznetsova | @Sebastian: I see now, you were assuming wide quotation marks. | |
| Jan 8, 2012 at 12:40 | comment | added | Niemi | @Yulia: Did you also see my second edit? Anyhow, I have now, in a third edit, removed the terms "probably" and "I guess" from my question, since, yes, I am now convinced that the topology is not discrete (see also the discussion in the comments under answer by Benjamin). | |
| Jan 8, 2012 at 12:40 | history | edited | Niemi | CC BY-SA 3.0 |
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| Jan 7, 2012 at 19:27 | comment | added | Yulia Kuznetsova | I see your edit; you still write that the induced topology is "probably" not discrete. Aren't you convinced now? I think that describing the topology is a too complicated question, so to extend the study to general groups you rather just ask what is the closure. | |
| Jan 7, 2012 at 19:21 | comment | added | Niemi | @Yulia: I have rephrased my example to avoid confusion. | |
| Jan 7, 2012 at 19:19 | history | edited | Niemi | CC BY-SA 3.0 |
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| Jan 7, 2012 at 17:57 | comment | added | Yulia Kuznetsova | To the question in my previous remark, I have an answer: yes, there exists a simultaneous rational approximation to any given set $t_1,\dots,t_m$. This implies that $f_0$ is not an isolated point of $F$, so in fact $F$ has no isolated points. For reference, see page 14 of Cassels, Diophantine approximation books.google.lu/… | |
| Jan 7, 2012 at 17:56 | comment | added | Yulia Kuznetsova | @Sebastian: but you still want to describe this closure in the topology of point convergence, do you? So we have anyway to start with the circle. And this seems not at all easy, see below. | |
| Jan 7, 2012 at 16:50 | comment | added | Niemi | @Yulia and Matthew: I changed the example above, as it might very well be incorrect (see the explanation above). Thanks for your comments. | |
| Jan 7, 2012 at 16:48 | history | edited | Niemi | CC BY-SA 3.0 |
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| Jan 6, 2012 at 20:13 | answer | added | Benjamin Steinberg | timeline score: 2 | |
| Jan 6, 2012 at 16:48 | comment | added | Yulia Kuznetsova | @Matthew: thank you, closedness clear now! Note that in the question $n$ is allowed to be negative. In fact, to conclude with the circle, it remains to separate $f_0$ from all the other $f_n$. And this has to do with simultaneous rational approximations, the area I have only a vague notion of. Is the following true (or its negation): $\forall\epsilon>0 \forall t_1\dots t_m\in[0,1)$ $\exists n\in \mathbb Z$: $\{nt_j\}<\epsilon$ for all $j$, where $\{x\}$ is the fractional part of $x$ (or its distance to the closest integer)? | |
| Jan 6, 2012 at 16:41 | comment | added | Matthew Daws | @Sebastian: Are you sure you get the discrete topology in the circle case? For example, how to show that $\{f_1\}$ is open in the relative topology? what's an open set $U$ in the topology of pointwise convergence with $U\cap F = \{f_1\}$? | |
| Jan 6, 2012 at 16:24 | comment | added | Niemi | The thing with $\mathbb{N}$ is a typo. Of course, $n$ is also allowed to be negative. I have corrected that above. As for the frist comment given by Yulia, I do not know how to give a better answer than "Yes, two identical functions are the same function". The set $F$ is simply the collection of all functions $x \mapsto x^n$ where $n$ ranges over $\mathbb{Z}$. If two function $f_{n1}$,$f_{n2}$ among this set happen to the identical (i.e., $x^{n_1}=x^{n_2}$ for all $x \in X$), then they are of course the same element of the set $F$. So, yes, in your example it would be $F = \{f_0,f_1\}$. | |
| Jan 6, 2012 at 16:19 | history | edited | Niemi | CC BY-SA 3.0 |
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| Jan 6, 2012 at 15:52 | comment | added | Matthew Daws | @Yulia: If G is abelian then each $f_n$ is a homomorphism; so if $f_{n_\alpha}$ limits to $f$ pointwise, then $f(st) = \lim f_{n_\alpha}(st) = \lim f_{n_\alpha}(s) f_{n_\alpha}(t) = f(s) f(t)$ by joint continuity of the group product. Thus $F$ certainly consists of (continuous) homomorphisms. So if G is the circle group, we already see that F is a subset of the functions $x\mapsto x^n$, where now $n$ might be negative. However, I think you can get negative $n$, and so F is not closed (but I don't see a good argument for this right now). | |
| Jan 6, 2012 at 13:42 | comment | added | Yulia Kuznetsova | And could you please explain how you prove your statement for the unit circle (I hope that's what you mean by the "group associated to one-dimensional sphere"). Because to me it seems that the topology on $F$ is not at all discrete and that its closure equals to the set of functions which are arbitrary at irrational points (assume that the circle is parameterized by $[0,1)$), and at rational points $f(x)\in \{nx: n\in \mathbb Z\}$. | |
| Jan 6, 2012 at 13:20 | comment | added | Yulia Kuznetsova | Do you identify functions equal everywhere, even if $n$ are different? E.g. in a group of order two, e.g. in $\{0,1\}^X$ for any $X$, you have $f_{2n}=f_0$ for all $n$, so $F=\{f_0,f_1\}$. | |
| Jan 6, 2012 at 10:00 | history | asked | Niemi | CC BY-SA 3.0 |