@RyanUnger Really? I don't know understand. What's the point in spending that much. She can get educated at a lot less if she wants. Is it just for label?
Ok my question was phrased poorly. Such a space is not a manifold.
Take "manifold" and replace $B^n$ by $B^n\times(0,1]$
You get things with "partial boundary"
user131753
5:14 PM
@AlessandroCodenotti: I think I have figured out an algebraic proof of the problem about which I was talking to you earlier. Are you interested in hearing?
@RyanUnger I don't understand. If you allow locally $B^n$ or $B^n \times [0, 1)$ you have a manifold with boundary. You mean every point has a nbhd which is homeomorphic to $B^n \times [0, 1)$? That cannot happen, take any point $x$ and a $U$ around it such that $U \cong B^n \times [0, 1)$ then take $(0, 1/2) \in B^n \times [0, 1)$ and a nbhd $V$ around it which does not intersect $B^n \times \{0\}$ and pull it back to get a nbhd in your original space back again which is homeomorphic to $B^n$
user131753
@AlessandroCodenotti Unfortunately it is what we were taught.
The estimates I'm working on seem to naturally want this structure. But it'd be easier to say take manifold with boundary and delete part of the boundary
Yeah that happened to me last week, things were getting bad. Today I went to bed at 3AM and woke up at 7ish so I might take a short nap at some point but hopefully I'll be in line for now to mostly get to sleep by 3AM
The convergent sequence of convergent sequences is Cauchy (convergent sequence in arbitrary metric space is Cauchy). Hence,
$\forall \epsilon>0$ ,$\exists k\in \mathbb{N}$, such that $\sup_{n}|x_n^p-x_n^q| < \epsilon/2$, whenever $p, q \geq k$
Again, $(x_n^p)_n$ and $(x_n^q)_n$ being convergent, $\forall \epsilon>0$ we can find $k_p, k_q$ $\in \mathbb{N}$ such that $|x_n^p-l_p|< \epsilon/4, \forall n \geq k_p $ and $|x_n^q-l_q|< \epsilon/4, \forall n \geq k_q $
It looks good. This is a key trick in analysis, you get some estimate assuming some bound on $n$ but then by triangle inequality you eliminate $n$ from the scene altogather
After getting such an $l$, $\forall \epsilon>0$, we can find one $\delta$ such that $|l-l_n|<\epsilon$, whenever $ n \geq \delta$. Choose any $m_0 \geq \max \{k, \delta \}$ [$k$ is from the main proof, that depends on $m$ only].
We get a corresponding $k(m_0)$.
Now, finally, $\forall \epsilon>0$, $\exists k(m_0)= \lambda$ such that $|a_n - l|< \epsilon$ whenever $n \geq \lambda$
@SubhasisBiswas Alternatively, once you've proven $\{l_m\}$ converges to $l$, you can note that $|a_n - l| < |a_n - a_n^m| + |a_n^m - l_m| + |l_m - l|$. For all $m \geq k$, $|a_n - a_n^m| < \epsilon/3$, for all $n \geq k^*(m)$, $|a_n^m - l_m| < \epsilon/3$ and for all $m \geq M$, $|l_m - l| < \epsilon/3$. Combine to get for all $m \geq \max\{M, k\}, n \geq k^*(m)$, $|a_n - l| < \epsilon$. Fix an $m_0 \geq \max\{M, k\}$ once and for all, and let $N = k^*(m_0)$.
The point I made is not about $2\epsilon$; that's fine - that's what you get by combining your initial result about $|a_n - l_m|$
This is a different way to argue that $a_n$ converges to $l$
Try drawing a picture of this scenario; think of $a^m_n$ as a double sequence arranged in a grid maybe. Each row converges, each column converges uniformly.
@SubhasisBiswas Anyway, great work. This is good, you should keep doing problems like this. Pick up Rudin and go through the problems.
My follow up to this problem would be to show that the space of bounded real sequences with the sup metric is complete (every Cauchy sequence converges). You can think about that later
I am vaguely reminded of the fact that there's some length-minimizing flow proof of the isoperimetric inequality (on the plane with embedded curves, I'm a mortal)
I only "know" the Fourier analytic proof of the isoperimetric inequality on the plane to be honest; it's a corollary of the fact that if $f$ is $C^1$, $\int_0^{2\pi} f = 0$, then $\|f\| \leq \|f'\|$.
These are the kind of things Sobolev embeddings are about, right? But for bad Sobolev-regular functions with norms all over the place in different spaces
Sobolev inequalities really have two parts...on the one hand they tell you neat things like you can estimate some L^p norm of a smooth function by L^q norms of itself and the gradient
But it also tells you about continuous embeddings of certain nasty function spaces
Dumb question: What's a good proof that $C^\infty$ is dense in $L^2$? Approximate any $f \in L^2$ by continuous functions outside of an open set of arbitrary small measure by Luzin, this is also $L^2$-convergence, then approximate by a $C^\infty$-functions by Bolzano-Weierstrass?
so you write a paper, then you put your advisor, your undergrad helpers, the guy who runs the lab, the tech who helped you do the experiment, and maybe some other grad studentin your group who helped with the experiment. That's a 6 person paper already
"In 1991, the bilogists Bensimon and Mutz experimentally verified the Willmore conjecture with the aide of a microscope studying the physics of membranes." wot
"In 1991, the biologists Bensimon and Mutz [47] (see also [41]), gave experimental evidence to the Willmore conjecture with the aide of a microscope while studying the physics of membranes"
second part of the problem is show that the equation admits a solution of the form $x = at^2$ for the right choice of $a$. Explain why the picard iteration does not converge to this solution
Is the Extreme Value Theorem, that guarantees that a continuous function defined on a closed interval attains a minimum and maximum value, valid for a continuous function on an open interval?
Here's the thing with generating only primes and semiprimes
You can probably find a decent function that finds a lot of semiprimes in the first few terms but then it slowly decays in it's success rate until you're left with an average success rate
The implications would be huge if anyone found such a function that generated only semiprimes or only primes
