@AlessandroCodenotti This should be true. Take any open set $U$ in $X$; $X - U$ is proper closed so any cover of $X - U$ admits a finite subcover. That, together with $U$, is a finite subcover of $X$.
Well, better to parse it as: given any open covering of $X$ pick $U$ out of the open cover and take it's complement - reduce it to a finite covering there.
What powers of z do you see in the expression "z"? Well, you see a z there, which is z^1. Do you see any other powers of z in the expression "z"? No, so all other powers have coefficient 0.
The same idea goes into understanding polynomials in college algebra. For example, when they do long division or synthetic division, if a polynomial is "missing" powers of z they need to put "0" in that column. If you don't see a number written in front of a power of z, that means it's coefficient is 1. If you don't see a constant coefficient, that means it's 0. Etc.
I have a model of cell migration, and I want to calculate what proportion of each type of I cell I want to end up with a certain proportion at the end. So I have red cells, green cells and blue cells. After 24 hours (800 frames), 75% of the cells present should be red, 15% green and 10% blue. Cells enter the domain in 3 different ways.
The first of these is being in there from the beginning. Let's say there are n cells to begin with (so n*k_r of these are red, where k_r is the proportion I'm looking for). The next way is cells entering from outside the domain. The probability that one cell enters is p, the probability that two cells enter is p^2 etc. I do this being calculating a random number at each step, and summing the p^k (plus the probability that none enter) until I get into the right interval.
The last way is through cell divisions. Red cells divide into red cells, green into green and blue into blue. Cells divide once every 10 hours (so approximately every 333 frames)
So I guess what I want is to know calculate how many cells of each colour there will be at the end in terms of the proportion k_r of cells entering that are red
Hello all, anyone can help me with a simple physics (path of sunlight ) question. http://imgur.com/a/nzv65 In the picture uploaded, why is $F_{1}$ the midpoint of the duct with length $l$?
Yes, any closed subspace of a compact space is compact. I agree. I thought you were asserting that any compact subspace is closed. What did "iff" mean?
@DHMO: I looked at part of the first video. It's clear and motivated. The main point is that we actually don't know the true mean of the distribution; we only have our "observed" mean from our data sampling.
huh, seems to be $\sigma_m/m=\sqrt{(1-1/R^2)/(n-2)}$ based on a source I'm seeing. Might reflect the fact that one needs to have at least three data points to have a meaningful uncertainty in a linear fit.
@HarryEvans The radius of convergence can be deduced from the distance of the nearest pole ($2\pi i$ and $-2\pi i$) to the center of the expansion ($0$).
@DHMO Everything that can be proved in ZFC is consistent with those axioms, assuming it can be formulated, which I'm not sure for statements involving $\Bbb R$
@TedShifrin well my wife has also been sick, and just found out she has bronchitis instead if pneumonia. She now has the proper meds and will hopefully be doing better. I am pretty much on the mend, but need to check in on my friend again soon.
@TedShifrin Yeah. Just back from walking the dogs (we picked up a stray at the friend's house I was just talking about on the day before Christmas Eve). It was drizzling a bit.
I don't know (I don't think so but I might be wrong) @DHMO, but I don't have time to think about this right now, I have some uni work that needs to be done
A loan of $ 5100 has to be repaid in two equal installments in 2 years. If the interest is charged at the rate of 4% per annum, compounded annually, find the amount of each installment
@AlessandroCodenotti if you are currently familiar with applications of Baire, I would be interested if you can see how to apply the theorem to this problem: math.stackexchange.com/q/2071544/152424
it already has solutions, but the OP thinks there is a solution that uses the Baire category theorem
Hello all, anyone know a little about the path of light? i need help imgur.com/a/nzv65 In the picture uploaded, why is $F_{1}$ the midpoint of the duct with length $l$?
Hi guys. My book says that a subset $I$ in $\mathbb R$ is said to be an interval, if $I_{a,b}\subseteq I$ for each $a,b\in I$, where we've defined $I_{a,b}=[\min\{a,b\},\max\{a,b\}]$. However, I don't understand why this definition is necessary. Can't we just say that $I$ is an interval of $\mathbb R$, if $I=[\inf(I),\sup(I)]$? Or is this a broader definition?
Oh wait
But then something goes wrong
for open intervals
Is the reason they define it like this, so that they don't have to resort to treating open and closed intervals?
Another thing you could do if you want to be a troll is that instead of considering it as a subset of the real line, just look at it under its subspace topology
So that it is both open and closed in itself
Same for every set
Open and closed will cease to have meaning, and thus we no longer encounter this terminology problem!
@DHMO I really don't like pop-sci like that. What do they think they mean when they say that "only 5% of the universe is normal matter"? Or when they say "atoms are 99.99999999% empty space"? Such vague blabbering, imprecise to the point of being contentless.
@BalarkaSen Saw a talk yesterday, on leavitt path algebras path algebras. It was the second part, and I didn't see the first talk so I didn't get to much out of it, but there was an interesting comment about trying to generalize correspondces of rings and spaces
I guess in the non commutative setting you can't find a (co)functor, which agrees with the classical functor and doesn't send nontrivial algebras to the empty set
I have never really thought much about the correspondence (I havn't thought much about commutative algebra or algebraic geometry) but I found that interesting
I asked if you could maybe try to piece together multiple functors, to try to get a "good" correspondence, and I guess my naive thought wouldn't work, but someone there said that maybe there would be something that could work like that, and he was explaining some of the subtleties on how it might work, but I didn't really understand (and he was explaining it to the speaker) @BalarkaSen
Have you been to those sorts of camps before @ZachHauk
Me neither, but I am sort of curious as to what they do and how the students feel about it, and what they liked about it. I could imagine that one of the best things is being around more like minded people interested in math, even if you were not super interested in those sorts of "problem solving things"
"Problem Solving and competition-style math plays an interesting role at Mathcamp: it can be a piece of your mathematical experience, or your primary mathematical focus, or play little to no role at all" Not what I expected, an really cool of them to organize it this way, I always these things were focused on math competitions
Hope you get in, sounds like a great opportunity to learn, and do, some cool math @ZachHauk
Hi @Astyx. (Sorry, I didn't read it). Also, for anyone reading this: why is a path in $\mathbb R^n$ defined as an $n$-tuple of continuous functions with domain $[a,b]$, for some $a,b\in\mathbb R$? I don't really see the problem with having an open domain? Or to have $\mathbb R$ as the domain.
It is also ok to call a continuous map $(0,1)\to \Bbb R^n$ a path
although usually you also want it to be piecewise differentiable or have some other "niceness" property, so you don't get things like this, which are sort of not something that looks like a path
You can always reparametrize $(a,b) \to X$ to be $(0,1)\to X$, but you can be interested in certain parametrizations, like ones that parametrize the length of the curve, in which case it is nice to have something which isn't just $(0,1)$
I don't think I understand what you mean exactly. But if we have $1/x$ on $(0,1)$, then how is this a path with a beginpoint?
We haven't worked with paths yet btw. I just came across it in my book, so I might be lacking basic knowledge on paths. Maybe I should leave it for now, as I understand why it makes sense to use a closed interval in the definition.
it doesn't have a starting point, if you say "a path is a continuous map $[a,b]\to\Bbb R$" or "$[0,1]\to\Bbb R$" then it must have a starting point, but maps from $(0,1)$ or $(a,b)$ do not need to have starting points
an even worse example would be $(0,1)\to\Bbb R^2$, $x\mapsto (x,\sin(\frac1x)\sin(\frac1{1-x}) )$, which is bounded and continuous but has neither starting nor end points
I would normally call the $(a,b)$ definition a type of parametrized curve, and I would normally think a path has endpoints by conventions I am familiar with, so I would say a parametrized path (type of parametrized curve) would have $[a,b]$ definition. But anyway these are just definition.
