@robjohn how can I use another e-mail for logging in the site? If that is not possible, I delete my account and create another one with another e-mail. I don't care the points I have and lose.
@robjohn or other solutions, I don't wanna share the e-mail of my brother further.
@Chris'ssis well, the only people that will see your email are mods, but nothing is ever completely deleted from the site. If you delete your account, the CMs and devs can still see the information, even if the mods can't.
@Chris'ssis I don't know if there is a record of old logins left in the history that the mods can see, though changes to your profile, in which emails are also stored are logged.
@Chris'ssis I don't want to go into that incident here, because from what I saw, it was a misunderstanding (and the question "are you mad?" is not the same as "you are mad!"). That being said, you can trust or mistrust whomever you wish. However, before someone can become a mod, even if elected, they have to sign an agreement, and I have yet to see an SE moderator divulge personal information.
Anonymous
I think those people who discuss no math here are mad.That makes me mad :p
@Venus The actual mods are great. I was thinking a bit of the future and of the privacy idea. I'm inclined to trust robjohn, there shouldn't be such problems.
@Venus If you ask me, I think robjohn had already experience working in teams much before joining the site. Maybe he was some kind of leader. To get these skills at this level is terribly hard.
@Venus I mean if I'm annoyed one feels that immediately, but you never know when robjohn is terribly annoyed, he knows how to control himself very well.
Community thinks the other way apparently, if you had seen the flurry of "you have my vote"-type comments below his post before they got deleted @Venus
As suggested on the chat of Mathematics 2014 Moderator Election, I'm opening this thread to give us an opportunity to nominate our choices for moderator.
Please state the your candidate election and set your reason why he/she should be elected.
Maybe we can convince some of them to nominate th...
Could I ask someone something about this exercise: http://math.stackexchange.com/questions/1006120/at-which-p-adic-fields-does-the-equation-have-no-solution ? This question has a bounty of 500 and expires in one hour!!!
@robjohn Could I ask you maybe something about this exercise: http://math.stackexchange.com/questions/1006120/at-which-p-adic-fields-does-the-equation-have-no-solution ?
@skullpatrol Could I ask you something about this exercise? http://math.stackexchange.com/questions/1006120/at-which-p-adic-fields-does-the-equation-have-no-solution
Hey @BalarkaSen Could I ask you something maybe about this exercise: http://math.stackexchange.com/questions/1006120/at-which-p-adic-fields-does-the-equation-have-no-solution ?
@teadawg1337 Oh, you might be right. I was just laughing at how far from reality your thoughts seem to be, at least based on those aspiring mathematicians I know ;)
@Alec: I didn't read your question, but the easiest way I know of seeing that the sphere and torus are not homeomorphic is either using the tools of algebraic topology.
Let $S$ be a set of $n$ integers. Assume you can perform only addition of elements of $S$ and comparisons between sums. Under these conditions how many comparisons are required to find the maximum element of $S$?
I have written an algorithm I found that $n-1$ are required.
How could we show that $n-1$ is a lower bound on the number of comparisons?
@MikeMiller that's not really the crux of the question, I've posted it because it's a symptom of a bigger problem I am having - I don't like/see/agree that punctures mean discontinuous, I am just "told" it
They say puncturing is discontinuous because, well, let's think about the sphere: to get the torus you "pull apart" at the north and south poles, and then glue together the resulting cylidner to get a torus. But the image of two points very very close to the north pole needn't be close at all in the torus when you do this. But I guess you knew that.
I'd like to ask you how you would like to approach the integral below
$$\int_0^1 \frac{(2 e)^{-1/y} \left(e \left(-2^{1/y}\right)+2 e^{1/y}\right)}{1-y} \ dy$$
and then recommend me some tools you'd employ. It's hard to even imagine how to start here.
I like to think about metrics whenever I can, but I see your point. OK, so in this case, look at the image of the north pole, and pick a 'sufficiently small' neighborhood around it. Since this neighborhood won't contain the image of an $\varepsilon$-ball around the north pole, the preimage won't be open. This is just a rephrasing of the above, so I suppose it's still not helpful. Sorry.
I have to keep saying $\varepsilon$-ball because that's how these topologies are really defined...
Yeah it's very easy to "embed" them, if you look in the HR box, I show where that GOES HORRIBLY WRONG considering the torus as defined by taking R^2 and associating points that differ by an integer.
@MikeMiller
Because that discards the metric topology of $R^2$ when you go back
@MikeMiller consider a circle of radius 1/4 centred at (0.5,0.5) in the torus considered as a square of length 1 along each side (and edges associated)
@MikeMiller: Yeah, me too. But in high school it is rather common to denote vectors with arrows, and I'm afraid they'll be confused if I suddenly start writing $v$ instead of $\vec{v}$.
Which contradicts "Simply connected maps to simply connected" by continuous maps @MikeMiller I have shown that A is simply connected and f cont => F(A) is simply connected.
@MikeMiller: I'm introducing them to matrix multiplication before they know about the dot product of two vectors, then I'll prove that $a^T b = |a| |b| \cos (\varphi)$ for $\varphi$ being the angle between the vectors $a$ and $b$. Do you think I should from thereon denote the dot product by $a \cdot b$ or $a \circ b$ or $a^T b$ or even $\langle a, b \rangle$?
So, that's not true. It's true that f maps connected sets to connected sets; it's not true that a continuous function maps disconnected sets to disconnected sets as your example shows). Simply connected things needn't map to simply connected, though, and not-simply-connected needn't map to not-simply-connected.
The projection $p: \Bbb R \to S^1$ is an example of a continuous map whose image isn't simply connected... (and the map we're talking about is just $p \times p: \Bbb R^2 \to T^2$.)
For an example in the other direction, considering $\Bbb R^2 \to \Bbb R$, $(x,y) \mapsto x$, the image of the not-simply-connected unit circle is the simply-connected $[-1,1]$.
For example, you can R^2 with the metric topology, and a space X with the topology {null,X}, then ANY function you like is continuous, given an open set in R^2 maps to either null or X somewhere. That topology is just useless
So if we restrict the projection map to $p: [0,1)^2 \to T^2$, then $p$ is bijetive, and we can define $p^{-1}$, and that one isn't continuous, I agree.
@DanielFischer to do it in any less would not allow every element to be involved in a comparison (bird-I-can't-spell's name principle) and the list is in no particular order. QED
@MikeMiller BTW, what does the quotient map we're talking about actually matter with respect to the question?
@AlecTeal Well, to involve all elements in some comparison, we only need $\lceil n/2\rceil$ comparisons. We need to tie them together. It's totally to be expected that $n-1$ is a provable lower bound, but a rigorous proof is not totally obvious.
Ooooh! @MikeMiller, okay let me prove it: assume f(A) is not simply connected, that means there are two disjoint sets, lets call them U and V such that f(A)\subset U\cup V, f(A)\cap U != null, f(A)\cap V != null, right? Good.
@MikeMiller let .... M=f^{-1}(U) and N=f^{-1}(V), these are both open since f is continuous, I claim they are disjoint
@DanielFischer @AlecTeal How could we prove this? Can we justify it only with pigeon principle? Could we maybe show this using induction or using the the decision tree??
Suppose they are not disjoint @MikeMiller then they have a non-empty intersection, suppose x is in that intersection, then f(x)\in U and f(x)\in V Mike, which contradicts.
@MikeMiller if they're connected .... well... then a set in the shape of washer could map to a circle, simply connected forbids this, but as simply connected => connected, I should be okay.
@MikeMiller okay, supposing that is true (I'll try and prove this in a second) what can I say about the image of the simply-connected set as we make it smaller from that? (this is where I am iffy)
I think you can't say much! For the purposes of your question, here's one you might prefer: if $f$ is a homeomorphism, the image of a simply connected set is simply connected. (That's just because $f$ restricts to a homeomorphism $A \to f(A)$, and simple-connectivity is preserved by homeomorphisms.)
@DanielFischer if you do it in n/2 comparisons, then you have partitioned the input into sets of 2, thus cannot compare all of them. (This is graph theory) there is a nice trick to employ here, a tree on the states (of which there are n vertices) must have n-1 edges.
I have just seen a problem on Euler.net (https://projecteuler.net/problem=493) I'm not understanding what I'm supposed to do. 70 colored balls are placed in an urn, 10 for each of the seven rainbow colors.
What is the expected number of distinct colors in 20 randomly picked balls?
Wait @MikeMiller " The image of a simply connected set is connected. It needn't still be simply connected." right.... but you know that inverse of the quotient we talked about earlier, it's not connected (If that is even relevant! I am so muddled)
@AlecTeal Well, the inverse we talked about earlier isn't even a function, so nothing holds. It's not a theorem that "the preimage of connected under a continuous function is connected", as you see from this example.
@MaryStar every connected graph has a maximal tree, that maximal tree is the least number of things you must do (each edge being an operation) to induce a relation.
What I was saying after that was that if you restrict to some place where $f$ is a bijection - so that the inverse is defined - the 'projetion' $f: [0,1) \to S^1$ doesn't have a continuous inverse $f^{-1}: S^1 \to [0,1)$.
@MikeMiller I disagree with that, it's just not a function with the usual topology on R^2 - I bring this up because there is a topology on R^2 such that the inverse is not only a function but continuous (and dare I say homeomorphism!) which is not intuitive.
@Hippalectryon What is the expected number of distinct colors in 20 randomly picked balls See suppose if I have picked 5 blue ball, 5 red, 6 green and 4 yellow then the distinct colors are 4. Is this right
@MaryStar you are trying to find the greatest member in the set right? You can use the well ordering principle and say "hey, there's a linear order here"
@BalarkaSen He said "some time on Sunday", better play it safe and wait until Sunday is at least UTC-over if not Mike-over lest he extend the ignore another day.
@Hippalectryon Then sir why they are asking that give your answer with nine digits after the decimal point (a.bcdefghij). What ever the answer would be that will be a integer from 3 to 7. Is this right?
Anyway @MikeMiller I hope I'm not being really dumb, but I'm sure there is a topology on R^2 such that inverse of the quotient map that takes R^2 to T is continuous WRT this topology, but not with the metric topology (as established), this means that WRT this induced topology there are seemingly disconnected sets that are actually connected WITH THAT TOPOLOGY
@AlecTeal I'm just having trouble understanding what you're saying. Here's what I think you're saying. We can make the domain of the "projection map" $[0,1) \times [0,1)$. Then we have a continuous bijection $[0,1) \times [0,1) \to T^2$. This we can define the inverse map $f^{-1}: T^2 \to [0,1) \times [0,1)$. With the standard topology on $[0,1) \times [0,1)$, this is not continuous.
@Singh Imagine you have an urn with three balls: two red, one green. You pick two balls. What's the expected number of distinct colors of the two balls ? You will pick one Green(G) one Red (R) with probability 2/3, and two R with proba 1/3. Therefore you expect to have (1*1/3+2*2/3)=1.666666... different colors
Okay anyway by "inverse" I mean "inverse of that map" so that's taking the square back into R^2 by mapping a point to all points that differ by an integer, so (0,0) would go to {(0,0),(1,0),(0,1),(1,1) ....} @MikeMiller
@AlecTeal But they are also points - in a different space. If you map a point of the torus to a set of points of $\mathbb{R}^2$, that map cannot be the inverse of the projection $\mathbb{R}^2 \to T$, wrong codomain.
As suggested on the chat of Mathematics 2014 Moderator Election, I'm opening this thread to give us an opportunity to nominate our choices for moderator.
Please state the your candidate election and set your reason why he/she should be elected.
Maybe we can convince some of them to nominate th...
I'd like to ask you how you would like to approach the integral below
$$\int_0^1 \frac{(2 e)^{-1/y} \left(e \left(-2^{1/y}\right)+2 e^{1/y}\right)}{1-y} \ dy$$
and then recommend me some tools you'd employ. It's hard to even imagine how to start here.
