Do you know the construction of the vitali set?
Just assume that $\mu(S)\neq 0$ and use the vitali set construction on $S$.
So you see the only subsets of $\mathbb{R}$ with this proberty are null sets.
A rough sketch for the vitali set:
Define on $[0,1]$ an equivalence relation via $$x\sim y ...
Does this answer make any sense?
I mean: it shows only Vitali construction and not how it is to be applied to S.
I mean we can't say a set of points satisfying a certain condition of points, in this sentence, I just replaced locus with its definition. It makes no sense :'(
62047071 yes I exactly agree. But why locus of points again. Sorry if I am bothering you.
> the set of the points that satisfy some property is often called the locus of a point satisfying this property. The use of the singular in this formulation is a witness that, until the end of the 19th century, mathematicians did not consider infinite sets. Instead of viewing lines and curves as sets of points, they viewed them as places where a point may be located or may move.
I don't know if that Wiki quote's quite correct, but it's certainly true that we didn't have a rigorous definition of sets, particularly infinite sets, until Cantor.
> Your strength as a rationalist is your ability to be more confused by fiction than by reality. If you are equally good at explaining any outcome, you have zero knowledge.
Ok so I thought about it and came to this conclusion : ( as we know locus is a set of dots that have the same characteristic, so for example ' locus of points 1cm away from dot A means that in our locus in our set of dots, these dots they are all 1 I'm away from point A
In my opinion except 'of' that had to write which : like the locus which dots are...
"Let $S \neq \emptyset$, $S \subseteq R$ such that
i) $S$ is closed under subtraction ii) $S$ is closed under multiplication
Then $S$ is a subring of $R$."
When I use something like this, I know that I first have to show that $S$ is nonempty and a subset of $R$, but once I've done that can I then assume the other properties of a ring to show that the above two things hold?
Like if I needed to assume associativity to show that $S$ is actually closed under subtraction.
Inspired of this question : What is the largest known semiprime which can be expressed with a short description , such as $$3^{227}-2^{227}$$ ? Powers , factorials , primorials and sums are allowed.
But the product of two huge known primes is of course not , since the problem is then trivial.
Aditionally, the smaller prime factor should be relatively large (say at least $10$ digits) to avoid Wagstaff-primes or similar stuff.
@copper.hat depends on why the question was closed. If it was a duplicate, then I can easily see voting on answers. If it was a poor question, and likely to be deleted, then it is a bit odd.
Hello. I know that the statement $\liminf\limits_{n\to\infty}|\frac{a_{n+1}}{a_n}|>1\Rightarrow \sum_{n=0}^{\infty}a_n$ diverges is false (here $(a_n)_{n\in\mathbb{N}}$ is a sequence of nonzero real numbers) but I can't see where the following proof of the false statement, which I had come up with before encountering a counterexample to the statement of the theorem, goes wrong.
Let $L=\limsup\limits_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$; then there exists a subsequence $\left(b_k\right)_{k\in\mathbb{N}}=\left(\left|\frac{a_{n_k+1}}{a_{n_k}}\right|\right)_{k\in\mathbb{N}}$ such that $\lim\limits_{k\to\infty}b_k=L$ so if we take $\varepsilon:=\frac{L-1}{2}$ we have that there exists $K\in\mathbb{N}$ such that $|b_k-L|\leq\varepsilon$ for every $k\geq K$ hence $b_k\geq\frac{L+1}{2}>1$ for every $k\geq K.$
So if $k>K$ we have $|a_{n_k}|=\left|\frac{a_{n_k}}{a_{n_{k-1}}}\right|\cdot \left|\frac{a_{n_{k-1}}}{a_{n_{k-2}}}\right| \cdot \dots \cdot \left|\frac{a_{n_{K+1}}}{a_{n_K}} \right|\cdot |a_{n_K}|>\left(\frac{L+1}{2}\right)^{k-K}|a_{n_K}|=A\left(\frac{L+1}{2}\right)^k$, where $A:=\left(\frac{L+1}{2}\right)^{-K}|a_{n_K}|$ which implies that
$\lim\limits_{k\to\infty}|a_{n_k}|\geq\lim\limits_{k\to\infty} A\left(\frac{L+1}{2}\right)^k=+\infty$ i.e. $\lim\limits_{k\to\infty}|a_{n_k}|=+\infty$ hence $\lim\limits_{k\to\infty} |a_{n_k}|\neq 0$ thus $\lim\limits_{n\to\infty}a_n\neq 0$ and therefore the series $\sum\limits_{n=0}^{\infty}a_n$ cannot converge, as desired. $\square$
@copper.hat thanks. Let me reformulate. What I am trying to understand is why my proof of $\limsup\limits_{n\to\infty}|\frac{a_{n+1}}{a_n}|>1\Rightarrow \sum_{n=0}^{\infty}a_n$ diverges is false
@lorenzo the $\limsup >1$ only says something about a subsequence. take a sequence that is zero everywhere except every now and then ${1 \over 2^n}, {1 \over 2^{n-1}}$. then the $\limsup$ is $2$ but the sum converges.
half plane is a better term in the context of a plane.
When optimising a linear function over a polygon one should only check the vertices right? This can be seen geometrically by imaging the image of the polygon in 3d space
@copper.hat I know, but if a subsequence doesn't converge to $0$ then the original sequence shouldn't converge to $0$ too so the series shouldn't converge because the necessary criterion for convergence ($\lim\limits_{n\to\infty}a_n=0$) is not satisfied; that was the idea behind my proof and I still can't see where it failed.
@Shinrin-Yoku a compact convex polygon can be written as the convex hull of a finite number of points. then it is straightforward to show that the $\max$ (or $\min$) of a linear functional over the set is the same as the corresponding on the finite number of points.
@Shinrin-Yoku a $\min$ or $\max$ need not be attained, for example the functional $(x,y) \mapsto x$ has no $\max$ or $\min$ on the half plane $x+y \ge 0$.
I have a question about the conditional expectation of $X$ given $Y$.
Let $Y$ be a discrete RV with values in $E$ then we define $E'=\{y\in E:\Bbb{P}(Y=y)>0\}$. There we can compute $\forall X\in L^1(\Bbb{P})$ and $\forall y\in E'$ $$\Bbb{E}(X|Y=y)=\frac{\Bbb{E}(X\Bbb{1}_{Y=y})}{\Bbb{P}(Y=y)}=:\...
Hello. Are open simply connected sets on C imply they are Contractible
I am reading a proof, where the set is only required the set to be open, simply connected subset of C and then they say "Since every closed curve is nullhomotop then ..."
But i thought thats attribute only to contractible spaces
Infact, we said, that Simply connected: Every two curves with the same beginning and end points are homotopic. Contractble: Every closed curve is nullhomotopic. So if i have two curves, for example closed and another curve that consists of one point $p_o$ on the closed curve. i can pick the begining and end points of the closed curve to be $p_o$ so if a space is simply connected, it imply contractble, but appearently this is not the case, what am i missing?
Your definition of contractible is most definitely wrong.
Contractible = the identity map is homotopic to a constant map.
Your definition of contractible is actually the definition of simple connectivity. Something's all messed up. Your definition of simple connectivity is not the usual one, but you can easily prove they're equivalent.
@Mad ... See above. Any closed curve can be broken into two pieces. Reversing the direction of one of them gives you two paths with the same beginning/end points.
matheducators.SE asked if there was even a point in typing prose to mathematics. I through this chat see it can be abused and put in a mixee with all the other budhijeevi content.
You could answer this question: "Is there a mixed type surface whose mean curvature vector field is smooth and does not vanish along the curve of type change?"
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.
Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence relation
$$\begin{cases} f(0) = 2 \\ ...
Hi, could someone explain why the edit got rejected on this question? The answer is clearly wrong if you don't add the "without boundary", and it actually confused me and took me a bit to understand that it was missing... math.stackexchange.com/review/suggested-edits/1847147
Is it because it is a 'minor' edit on a 2010 answer?
Nah, can't agree with the "for many people", most people think of the general case of possibly nonempty boundary
In any case, I think it's rigged and the rejects weren't thought through properly x') oh well, I guess the next random person who'll end up reading that question will have their 10min of "existential crisis" for not understanding x')
well, irrespective of what people think of, a lot of textbooks would disagree at the level of the definition of "manifold"
the 'with boundary' case is a little harder, and maybe the person who rejected the edit thought that someone familiar with manifolds with boundary would understand the intent of the post without the "without boundary" qualification
The 'with boundary' case is plain wrong though Also, if I understand this website's guidelines (I hope I do, after all this time here), then it should be understood that people asking questions and people looking for answers are not necessarily gurus in the domain, and sometimes small things can make you feel stupid x') so a simple 2-word clarification was not hurting anyone, quite the contrary
@AnthonySaint-Criq with all respect, your “most” is vastly overinflated. A few, some, maybe.
In that context … a compact or closed manifold is without boundary unless specified otherwise. That said, you could add a comment to emphasize your point.
What can I look into when calculating $$f(n)=\left|\left\{S \mid S \subseteq \{2,3\ldots n\}, \sum_{x\in S}\frac 1{x^2}=\frac 12 \right\}\right|$$ for a given $n$
Do you know the construction of the vitali set?
Just assume that $\mu(S)\neq 0$ and use the vitali set construction on $S$.
So you see the only subsets of $\mathbb{R}$ with this proberty are null sets.
A rough sketch for the vitali set:
Define on $[0,1]$ an equivalence relation via $$x\sim y ...
