If $X = \varnothing$, then $\mathcal P = \{\{\}\} = \{\varnothing\}$
So, when we take the case for $\varnothing$, there's nothing in it.
The conclusion that $x \in [x] \to [x] \text{ not empty}$ seems a bit wonky. Was this said in the book?
I guess it can be read as follows. Since there's no $x \in A$, there is no equivalence class that such an $x$ would belong to. If we had an $x \in A$, there would be an equivalence class $[x]$ that was not empty.
I'm not 100% sure of the partition of the null set being $\{\varnothing\}$ or $\varnothing$, but my last message doesn't depend on it, @user19405892
@Danu Sorry, I didn't think to ask. What will you cover in this (and the sequel) class
@ForeverMozart Locally open maps are open, so you just need to check that at each point in $\Bbb R$. Away from $2\pi n i$, it's a local homeomorphism, and near those special points, it looks like $x \mapsto x^2$ (or the negative of that), which are indeed open maps (in that they send, say, $(-\varepsilon,\varepsilon)$ to $[0,\varepsilon^2)$ which is open in $[0,\infty)$).
@ForeverMozart I guess the above was not actually much of a proof, since I didn't prove that the restriction to that neighborhood is open, I just showed that it sends a certain collection of open sets to an open set. But you could either 1) just write down what it maps an arbitrary union of intervals etc to or 2) show that there's a homeomorphism $(\pi -\varepsilon,\pi +\varepsilon)$ to itself that takes the map $\sin$ to $|x-\pi|$, which is open as a map to $[0,\infty)$
The second one is pretty aggressively silly but whatever.
The constant sequence $\{x_{n}\} = 1$, $1$ is a limit point, right? I always get confused about the difference between limit points and accumulation points.
I want to clarify the definition of limit point and accumulation point.
According to many of my text books they are synonymous that is $x$ is a limit/accumulation point of set $A$ if open ball $B(x, r)$ contains an an element of $A$ distinct from $x$.
But from one of the problems in Aksoy: A P...
Is there any way to show that a finite-dimensional normed linear subspace is closed/complete without showing that every norm on a finite-dimensional normed space is equivalent?
@datalava Sure. Let $e_i$ be a basis. Pick a Cauchy sequence $v_m$. Write $v_m = \sum a_{i,m} e_i$. By a version of the triangle inequality, we also see that $a_{i,m}$ forms a Cauchy sequence. Hence they converge in $\Bbb R$ to some $b_i$. Again using the triangle inequality it is then straightforward to see $v_m \to \sum b_i e_i$.
I need to show that the sequence of functions, $\{ x_{n}(t) \} = 1$ is not compact in $L^{1}(0,\infty)$.
I believethat what I need to do is show that it does not have a limit point in $L^{1}(0,\infty)$ under the $L^{1}$ norm: $\int|x_{n}(t)-x(t)|\to 0$. However, I do not know how to show this.
...
@MikeMiller Looks wonderful but now I'm feeling stupid because I can't get $a_{i,m}$ Cauchy... should I be using something like the 'reverse' triangle inequality?
This question is related to another question I asked earlier (feel free to answer that one, too, if you want; I'm not satisfied with what people have said so far).
In addition to the conditions mentioned in the other question, suppose that additionally, $\exists \alpha > 0$ such that $\forall n...
gauge theory leads to (instanton/Seiberg-Witten) floer homology which is closely related to Lagrangian floer homology whose chain complexes form the morphisms in the Fukaya category which is used to build one side of the homological mirror symmetry conjecture
(in particular, there are various flavors of floer homology for 3-manifolds; seiberg-witten floer homology, paradoxically, is more technically complicated than instanton. lagrangian floer homology is just different, though the story is very much the same.)
you probably care more, btw, about classical (or possibly SYZ) mirror symmetry than homological mirror symmetry, both of which to my understanding have a completely different flavor/story to them
homological mirror symmetry is now (soon to be?) known to imply classical mirror symmetry, but i don't think it's how physicists usually approach it
@PVAL I think I have a counterexample for you. Let $\ell_1, \ell_2, \ell_3, \ell_4, \ell_5, \ell_6$ mutually intersect in double but not triple points in a way so that they form a hexagon. $p_i$ be the intersection point $\ell_i \cap \ell_{i+1}$ (indices are mod $6$). Now but $\ell_1 \cap \ell_3$, $\ell_2 \cap \ell_4$, $\ell_5 \cap \ell_6$ are nonempty, so let these points be $q_1, q_2, q_3$ respectively.
Finally introduce another line $\ell_7$ so that $\ell_7$ goes through each of $q_1, q_2, q_3$ making three triple points.
I didn't get very far in it... the first six chapters or so. It was rigorous enough, I could make precise everything they said there, though it was certainly obvious it was written by a bunch of physicists.
I guess, sure, though. Positions of $q_1, q_2, q_3$ in $\Bbb P^2$ is a subset of the space of positions of three points in $\Bbb P^2$. That is given by three coordinates $(X_0: Y_0: Z_0), (X_1:Y_1: Z_1), (X_2:Y_2: Z_2)$. That's $\Bbb P^2 \times \Bbb P^2 \times \Bbb P^2$, @MikeMiller.
And being collinear is a closed condition there. I'll get a closed subset.
@MikeMiller I know :) I just figured out that's probably not something I'd not be able to tell immediately. But yes, I agree with your inherent point that being pedantically precise is sometimes not worth the time.
Given a set of $n$ integers and a fixed starting point, one has to cover all the numbers moving at most a distance 1 from any number previous picked. For example if $n=5$ one solution would be: $\{3,2,1,4,5\}$. Another would be $\{3,2,4,1,5\}$ or $\{2,3,4,5,1\}$. How many solutions are there?
Given my puzzle, If you start at 1, there is only 1 solution. if you start at 2 there are $n-1$ solutions. There seems to be some recursive way of doing it but I can't quite find the pattern
I am interested in effective computations in finding approximate spectral decompositions in some suitable format.
Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can be decomposed as:
...
@DanielFischer Suppose I look at a variety $X$ and a codimension 1 subvariety $V$ of $X$ along which $X$ is smooth. Why is the local ring $\mathcal{O}_{X, V}$ of regular functions at $V$ a discrete valuation ring? I know this is true for $X$ a curve.
But the way I prove it for curves is by embedding $\mathcal{O}_{X, p}$ into $k[[t]]$ where $t$ is a local paramater at $p$. $k[[t]]$ has a valuation.
@BalarkaSen Hmm, so intuitively the codimension $1$ part should translate to a single generator of the maximal ideal, though I would need to write things down to make that precise
This question is related to another question I asked earlier (feel free to answer that one, too, if you want; I'm not satisfied with what people have said so far).
In addition to the conditions mentioned in the other question, suppose that additionally, $\exists \alpha > 0$ such that $\forall n...
@TobiasKildetoft I know that near a point $p \in V$ I can locally write $V$ as hypersurface in $X$. But it is not clear to me how that translates into global.
Hmm, on second thought I think I am confusing the definition of $\mathcal{O}_{X, V}$ with something else.
Never really thought about local rings at subvarieties other than points.
Mm, right, $\mathcal{O}_{X, V}$ is the ring of rational functions regular on some open set of $V$. Not what I was thinking it is.
@TobiasKildetoft In that case what you said gives me the right thing. Locally $V$ is cut out by just a single equation (since $X$ is smooth along $V$). So $m_{V, X} \subset \mathcal{O}_{X, V}$ has to be generated by a single thing.
@JessyCat Probably the assumption that $\int_0^{\infty} t^{\alpha} \lvert x_n(t)\rvert\,dt \leqslant C$ independent of $n$ gives you that the $x_n$ are "uniformly small on the tails", for any $\varepsilon > 0$ there is an $A(\varepsilon)$ such that $\int_{A(\varepsilon)}^{\infty} \lvert x_n(t)\rvert\,dt \leqslant \varepsilon$ for all $n$. Then the uniform convergence on $[0,A(\varepsilon)]$ gives you an $N$ with $\int_0^{A(\varepsilon)} \lvert x_n(t) - x(t)\rvert\,dt \leqslant \varepsilon$
for $n \geqslant N$ and that gets you $\lVert x_n - x\rVert_{L^1} \leqslant 2\varepsilon$ for $n \geqslant N$. For $\alpha > 1$, it's straightforward, I think, but for $0 < \alpha \leqslant 1$, the details may be a bit hairier.
Ah, no, it's straightforward for any $\alpha > 0$.
setting aside the multiplicative factor at the end (which is explicit and therefore easily accounted for) the key difference is that linear term
and while that linear term would seem to complicate things, it's what gave rise to the Fourier series because it implied that (again, setting aside the factor $e^{-x^2}$) the function was $1$-periodic in $x$.
now, at the level of math, this is a lovely identity regardless of what $a$ happens to be. but whether this is a useful resummation depends on what $a$ is
if $a$ is large, then the summation on the LHS (in powers of $e^{-1/a^2}$) will converge slower than that on the RHS (in powers of $e^{-\pi^2 a^2}$). but the opposite is true if $a$ is small. (the crossover point is when $a=1/\sqrt{\pi}$).
that makes a general discussion for the case of $\sum e^{-x^T M x}$ difficult, because whether the resummation is useful or not depends on what $M^{-1}$ looks like
in particular, something like $M=\text{diag}(\epsilon,\frac{1}{\epsilon})$ seems pretty weird.
I want to clarify the definition of limit point and accumulation point.
According to many of my text books they are synonymous that is $x$ is a limit/accumulation point of set $A$ if open ball $B(x, r)$ contains an an element of $A$ distinct from $x$.
But from one of the problems in Aksoy: A P...
I need to show that the sequence of functions, $\{ x_{n}(t) \} = 1$ is not compact in $L^{1}(0,\infty)$.
I believethat what I need to do is show that it does not have a limit point in $L^{1}(0,\infty)$ under the $L^{1}$ norm: $\int|x_{n}(t)-x(t)|\to 0$. However, I do not know how to show this.
...
I am interested in effective computations in finding approximate spectral decompositions in some suitable format.
Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can be decomposed as:
...
