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19h
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
Thanks, yes, I meant the modulus of $y$. It's fixed now.
19h
revised The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
added 2 characters in body
1d
answered The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
2d
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
@G.Sassatelli You're probably right. If so then after rescaling this is the same as the limit of $\lambda_n(B_1(x)\setminus B_1(y))$ as $y\to x$, and now it's clear.
2d
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
I think the statement as written is false. WLOG $x=0$ and $y=e_1$. Then we're measuring the set of points which are a distance $r$ from $0$ but not from $e_1$. This set contains a half ball of radius $1$ centred at $-(r-1)e_1$, so it has measure at least half the measure of the unit ball.
Apr
22
answered Show that there exists an entire function $h$ such that $\lim_{n\to\infty}{h(nz)}=0$ for all $z\ne0$
Apr
21
answered Find $\lim\limits_{n\rightarrow\infty}(n!)^{1/n^2}$
Apr
16
comment Difference between pushforward and differential
@IberêKuntz You can certainly use another map $g$ (instead of $f$, to reduce notation clash) for the identification $T_q\mathbb{R}\cong \mathbb{R}$ if you like, provided that $g'(q)=1$. Then you'll find that, at $q=f(p)$, $f_*(X)(g) = g'(q) X(f) = X(f)$.
Apr
7
answered How do I choose first terms of a Fibonacci sequence?
Mar
12
answered Haar's theorem for the rotation-invariant distribution on the sphere
Mar
10
comment Determinability
If you define $g(x) = \sup\{t\in\mathbb{Q}: x_t=0\}$ then actually $g^{-1}(\{y\})$ is a little nicer than that: whether $x\in g^{-1}(\{y\})$ depends only on $x_t$ for rational $t$.
Mar
10
comment Determinability
Does taking the supremum just over rational $t$ help?
Feb
19
comment How could Collatz conjecture possibly be undecidable?
However I've just remembered that the Collatz problem is not exactly so simple, but at least this is true: the statement "there does not exist a loop in the Collatz problem other than 4,2,1" can obviously never be proved independent, though for all we know it might be independent. While making my earlier remarks I forgot about the other possibility, that there exists an unbounded sequence, which is not so simple. But still the point is that a conjecture might be independent without us mortals ever having the luxury of knowing it.
Feb
19
comment How could Collatz conjecture possibly be undecidable?
@whacka I do apologize for the confusion, I was talking about independence. This confusion of the terms is not so uncommon though, for instance see the remarks at en.wikipedia.org/wiki/…. I certainly was not making the claim that it's impossible to prove statements are independent, but merely something like the following: if $\phi:\mathbf{N}\to\{0,1\}$ is a computable function then the statement "there exists $x$ such that $\phi(x)=1$" cannot be proved independent.
Feb
19
comment How could Collatz conjecture possibly be undecidable?
@user2520938 Yes, exactly.
Feb
19
comment How could Collatz conjecture possibly be undecidable?
The trouble is that if it can be (and it might be!) undecidable without your knowing it. In this case then as you say it must be true, since any counterexample would decide it, but that's not a contradiction, it only means that its decidability would also be undecidable, and similarly the decidability of its decidability would be undecidable, and so on ad nauseam.
Jan
29
awarded  Good Answer
Jan
27
awarded  Yearling
Jan
22
answered $X$ is an infinite set. Prove that $S_X$ does not have proper subgroup of finite index.
Jan
19
comment Finding two numbers given their sum and their product
@HenningMakholm Simple. You know the factorized quadratic will look like $(x-X)(x-Y)$, and $X$ and $Y$ satisfy... oh crap.