It's called pleurisy, it's kind of this sharp pain in a lung that worsens with breathing. Sometimes it's a symptom of other things but it's the sort of thing that can just happen. Do be cautious, as you should always be about health, but it's not necessarily cause for alarm
@Dartek12 For example the series $$ \sum_{n=1}^{ \infty } \frac{1}{n^p} $$ diverges for $ p \ge 1 $ while converging for p<1 despite having very similar forms.
Vrouvrou. I tried to simplify those lim sup and lim inf one by one but it appears I have done it wrongly. I think I don't have the requiste real analysis background to answer your question
The original proof (given here) needs only a minor modification to avoid the flaw pointed out by Timothy Chow in the comments.
Instead of placing the spheres with their centers on the plane, place the spheres tangent to the plane, such that the projection onto (rather than intersection with) the plane gives us the circles.
Also, @Semiclassical, I have a much easier proof for the 3D version (equivalent to a four-dimensional thing but can be thought of as being three-dimensional)
Otherwise, define a function $f:\Bbb R^3\to\Bbb R$, such that for every sphere, the value of $f$ at the center of the sphere equals the sphere's radius.
in the first iteration, we remove (0,1/3) in the second iteration, we remove (3/9,4/9) and (6/9,7/9) in the third iteration, we remove (12/27,13/27) and (15/27,16/27) and (21/27,22/27) and (24/27,25/27) etc
And this is easy to see, because for any cone and for any point $x$ on the line through the middle of the cone, $f(x)$ equals the radius of the largest sphere centered at $x$ that's contained inside the cone.
(This follows from the fact that this is linear.)
So, at the vertices, $f(x)=0$. This determines a plane, QED.
Not quite as awe-inspiring as the "view it as the $n$-dimensional projection of an $n+1$-dimensional situation" proof shown in the video, but (in my opinion) much more concrete and easy to verify.
It's so strange that $\Bbb R$ is contractible, despite being infinite. It kinda violates my physical intuitions. I guess it's a consequence of the fact that the concept of "uniform continuity" doesn't exist in topology.
@Semiclassical The relationship between the two proofs is that, in the projection one, instead of a function $f(x)$ determining size, you have a function $g(x):=\frac1{f(x)}$ determining how far away it is in the $n+1$-eth dimension. And then the desired hyperplane is the solution to $g(x)=\infty$.
@Secret forget about all that would dare to say a different thing. It is precisely like that: many years of very hard work, days and nights one after another of research.
It's contracting everything to $0$. If it was uniformly continuous there would be a bound to how fast you could contract, so you wouldn't be able to contract stuff near infinity.
Take any three circles in the plane of distinct radii. For each pair, there are four common tangents; draw the two outer ones. They'll intersect at some point. Repeat this for each pair; the theorem is that the three points lie on a line.
DogAteMy: This has the sound of Pascal's theorem in projective geometry, although it's different. I'm sure this theorem is in Pedoe's book that I refer to frequently.
Similarly, for any four spheres in space of distinct radii, you can take each pair and draw the cone enveloping them; it will have a vertex somewhere. Repeating for each of the six pairs, you'll get six points; the theorem is that they'll lie on a plane.
Similarly for any amount of dimensions; for $n$ dimensions, you take $n+1$ spheres, and end up with lots of points lying on a hyperplane.
In essence, you view it as the 2D projection of a 3D situation. The circles of different radii become spheres of different distances from the projection point, and the tangents become tubes.
Yeah, the thing in nonstandard topology (like nonstandard analysis, but topology!) is, given a point $p$ in a space $X$, you take the hyper-version $X^*$ (I forget what it's called). And then the monad is the intersection of all $F^*\subseteq X^*$ for $F$ an open set containing $p$.
I can immediately think of a darker version of that, and I'm a bit disturbed by that.
Hmm. I think I'll go to this talk simply because of the scattering diagram connection, but the chance that I'll actually understand any of it seems pretty low: math.umn.edu/seminar/…
Hello. I'm confused about a minus sign in do Carmo's book on curves and surfaces. On page 155, formula (4) is deduced by taking the determinant of formula (3), but a minus sign seems to disappear. Where has it gone?
