@tb How did you come up with that so fast? I can't imagine ever having such ready access to the repertoire of problems and solutions that some of you guys have.
@tb: thanks for the name of those estimates. I remember that the $n=1$ case is used to prove Liouville's Theorem. I probably have seen them somewhere before because the idea came to mind so quickly.
this bothers me: infinity by definition means uncountable, but Cantor jumps & says he's counted it and the answer is aleph-something - this is not the question I want to ask, just a test to see if someone here is willing to indulge ...
Well, disregarding the (philosophical?) question of whether something infinite can be counted or not, countable is a term with a specific meaning. (See anon's comment.)
counting something means using some kind of counter, whether it looks like integers, or is the volume of water that has flowed past some point, it is still just a means to count . can these different counters for different sets of countable things be compared directly? (loaded question)
@tb Well at any rate, I'm impressed. However, I'm still a bit stuck on this... How can I go from an arbitrary point $(x,y)$ in $\mathbb{R}^{2}-\{O\}$ to a point $(\theta,\sqrt{x^{2}+y^{2}})$ without the use of the $\arg$ function?
That map itself (the left component anyway) is the arg function, so no beans. However, you can use the form $\mathbb{S}^1=\{z\in\mathbb{Z}:|z|=1\}$, and for the left component use $(x,y)\mapsto \frac{x+iy}{\sqrt{x^2+y^2}}$, though I'm not sure what's gained through that.
Note that with the arg function we're using $\mathbb{R}/2\pi\mathbb{Z}$ for the circle. What's wrong with it anyway?
@anon Well, the main issue is that this class is stupid and stifling. We consistently have to pretend that useful and powerful tools don't exist, forcing us to solve problems in convoluted and round-about ways. That is, I don't really have access to such notions in this class. So, I'd have to develop it from scratch.
What's wrong with using complex numbers? Send $z \in \mathbb{C} \smallsetminus \{0\}$ to $\left(\frac{z}{|z|}, \log{|z|}\right)$ and $(\zeta,t)$ to $\zeta e^{t}$.
Those are evidently mutually inverse and continuous functions.
Well if you have $\mathbb{S}^1$ sitting inside $\mathbb{R}^2$ it's not difficult to modify the example given to you. Just take $$\mathbb{R}^2\backslash\{0\}\to\mathbb{S}^1\times\mathbb{R}:(x,y)\mapsto \left(\frac{(x,y)}{\sqrt{x^2+y^2}},\log\sqrt{x^2+y^2}\right).$$
@anon This notation is a bit confusing to me isn't $\frac{(x,y)}{\sqrt{x^{2}+y^{2}}}=(\frac{x}{\sqrt{x^{2}+y^{2}}},\frac{y}{\sqrt{x^{2}+y^{2}}})\in\mathbb{R}^{2}$?
ah, I see your problem now, it is contextual: for indexing (=counting) you happen to use something that looks like integers; you now confuse this counter with the set of integers, an entirely different context; to make it clearer: you have a coffee-mug which you put on papers in a windy room, it isn't a coffee-mug anymore, but a paperweight;
@tb So then $$\left(\frac{(x,y)}{\sqrt{x^2+y^2}},\log\sqrt{x^2+y^2}\right)=\left(\left(\frac{x}{\sqrt{x^{2}+y^{2}}},\frac{y}{\sqrt{x^{2}+y^{2}}}\right),\log\sqrt{x^{2}+y^{2}}\right)\in?$$
make that $\frac{x}{\sqrt{x^2+y^2}}$ in the first component's first component, and $\mathbb{S}^1\times\mathbb{R}$ on the other side of the $\in$, and then yes.
@anon: you & t.b. - by the way I'm not trying to mess about, I'm genuinely trying to understand this stuff, but if you think I'm wasting everyone's time please say so & I'll go away
@slashmais: You can't be wasting our time, since we aren't giving you time :P I can't make out what you're saying for the most part, but you sound like you might be confused about what some or other terms mean. If we could understand you we'd respond.
I love the tone employed by one of our fellow MSEers which always makes me think he is a retiring, multi-prized, famously good math instructor and researcher
@Mariano: while it's not very deep maybe it's worth pointing it out: the identification of the unit tangent bundle of S^2 and RP^3 we discussed recently can be used to give a quick proof of the hairy ball theorem. If there were a non-vanishing vector field on S^2, you could use this to trivialize the unit tangent bundle as S^2 x S^1 but this has a non-compact universal cover while RP^3 doesn't (or look at the fundamental groups).
@anon: (lol, good one): I'll try to be clearer: the context in which something is applied, fully defines that something for that context, and is valid only in that specific context. (it's a coffee-mug if you drink coffee from it; it's a paperweight if it prevents papers from being blown away; a paperweight is not a mug, they cannot be equated) ==> (continued)
it's integers if it is used with respect to the set of integers, it's counters if it is used to count something, a counter is not an integer, integers just happen to be used (like the mug just happen to be used as a paperweight). (continued)
My problem is that I cannot see how the integers-as-counters can be given all the attributes of the integers-as-a-set, they are fully different contexts. All this stem from having the cardinality of the set of integers the same as that of the set of rational numbers: the sameness deriving from using counters and saying that the counters and the set of integers are the same thing, which it is not.
@slash: the integers are the prototypical and canonical example of an index set for counting things. calling elements of other index sets "counters" is misleading (e.g. you can index a collection of objects with an uncountable set). you seem to want the word "count" to be more general than it actually is. the thing in set theory that allows us to compare sizes of sets is the idea of bijections.
fyi, the natural numbers also go by the name "counting numbers"
An obvious proof that the physicists are wrong about the universe being finite is that if it were, there would not be enough room to contain all the Law&Order episodes...
@anon: I may be barking up the wrong tree (wrong audience), my difficulty may be more philosophy-related than purist-mathematical. but still, thanks for your indulgence.
"Perhaps, the term newsletter is misleading and you understand this differently to me." This makes no sense to me if newsletter is used as a generic English word.
No, no. With the phone survey job I wasn't even allowed to hang up first unless I was badmouthed. I was making a joke about how phone conversations often end awkwardly with people who haven't developed a social protocol for them (e.g. me for awhile).
Blah, how do people not confuse "a morphism of schemes, where the morphism is locally of finite type" and "a morphism of schemes, where the schemes are locally of finite type"? They can both be legitimately be phrased as "a morphism of schemes locally of finite type"...
anon: ;) numbers do count: a 'threesome' is three people having fun, a 'twosome' is two having fun, a 'handsome' is ... be leery of people calling you handsome :)
@Raj: I had someone tell me their line was being monitored by the police, and proceeded to explain to me over and over again that my company was at risk for legal action by calling her. Another wanted to sue my company, thinking the Do Not Call list applied to my employer (it doesn't, ha ha). Another kept asking me my full name and phone number and where I lived. Another was a gay male video-gaming couple that apparently filmed harassing the homeless for youtube videos...
... Another time my random dialer got hold of a different market research firm (also, fire department, FBI, the very company I was doing the survey on behalf of, etc). One woman told me Obama was the atheist / muslim / socialist antichrist. Lots of stories.
@Srivatsan Some of these people are really talkative. I once was doing a survey that normally took 30 min (waay too long. and with old people about health care no less. poorly worded and structured, blah blah) and the guy had an entire story for every freakin bullet point on the survey. and there were like, hundreds of bullet points in total. After almost an hour we were halfway done, I was way past shift and I politely terminated the call.
I really liked the CNN surveys. The questions were good and made sense and they took 5 min.
Sometimes I wanted to strangle the people who wrote the surveys. I do not like asking women about how often they apply deoderant and how effective it is, or who "they would prefer to have a Girls Night Out with" (out of a list of inane celebrities)
or what brand of underwear they typically use. shudder
well, there are trends and patterns, but it's hard to see at the individual level. sometimes I would hear about statistics from our studies that weren't quite consonant with my personal experiences doing the surveys.
one time in a briefing my supervisor was talking about how a particular question had so many of the possibilities for a home situation addressed, the only way a respondent wouldn't fit on it is if they lived in a housebout. Lo and behold, on my second call I got a guy in the Navy who said he was currently stationed and living in a navy ship (it was a cell phone survey).
@robjohn Well, I do come back to chat from time to time. :) I am not sure if this is better than spending time in both chat and main; what do you think? ;)
@Srivatsan I talked to Michael Mrozek about any responsibilities and he pointed me to a page describing some things about room ownership, but there seem to be few responsibilities.
@Raj: Asaf said some guy being from Morocco explained his bad English. Gigili made a sarcastic show of the "glass houses" parable and Asaf namedropped and used an offensive gendered insult towards her. (Gigili did not seem to be paying attention at the time; the flag and deletion might have gone over before she even saw the insult.)
I'm trying to see why the function $f_n(x) = \begin{cases} 0 & -1 \leq x \leq 0 \\ nx & 0 \leq x \leq \frac{1}{n} \\ 1 & \frac{1}{n} \leq x \leq 1 \end{cases}$
@BenjaminLim So, this is all off on a tangent. Let me answer your question now.
Take the sequence, 0, 1, 0, 1, 0, 1, ..., obtained by interleaving two constant sequences. Sure, it has a convergent subsequence (e.g., all odd terms, or even terms).
@BenjaminLim Yes. In summary, some parts of the sequence converge to different points in the space, but the whole sequence may not converge.
@BenjaminLim Well, that is quite easily fixed. Take two sequences x_n and y_n with distinct entries converging to 0 and 1 respectively. Interleave them.
@BenjaminLim Yes. That is true. For one thing, one shouldn't expect "having a convergent subsequence" to imply much about the rest of the sequence (if we don't have any additional control).
@Ben: Try your hand at this small exercise: Suppose $x_n$ is Cauchy, and has a convergent subsequence. Show that $x_n$ converges.
@Kannappan: Apparently, the problem that I mentioned (class equation of GL_2(F_p)) is classical. I found this reference: imsc.res.in/~amri/GL2p.pdf, though I have only glanced at it.
@tb Ah, I didn't do it for the badge. I attempted for some time, posted the question, then wanted to try my hand one last time, so I deleted the post...
I'm not sure if this is quite accurate, but the -ment suffix for nouns has something passive about it: payer -> paiement, panser -> pansement, changer -> changement.
@Srivatsan I guessed that because, conjugacy classes of $GL_n(\mathbb R)$ make a classical study. I am not really aware of the full details. I wondered why Artin wanted that as an exercise. : )
