vulpes the question is at least arguably missing context but yes, if i had to guess what to do, it would be to find f'(x) and use the formula for that to indicate where f'(x) is positive and where f'(x) is negative
because the graph of f is particularly simple, it is very possible to answer this particular question without calculus, but calculus provides the general machinery that would answer questions of this type
@VulpesInculta Do you need to? Not at all. Indeed, as @leslietownes points out, there are ways to do this without calculus that are probably simpler: the function is a parabola, given by $f(x) = 2(x^2 - 4)$. That is your basic parabola, shifted down four units and scaled vertically by a factor of two. It is symmetric about it's vertex, which is at $(0,-4)$, so it decreases on $(-\infty,0)$, and increases on $(0,\infty)$.
But you can use the first derivative, instead, if you like.
yeah i was thinking, in the context of california state standards you could ask this question after people did 'graphs and transformations' in algebra 2
@leslietownes Yeah, and it is a major emphasis of the current precalc curriculum at UCR, which is also being rolled out at a number of CCs where former UCR grad students were hired.
(There's a grant of some kind.)
I really wish that folk emphasized transformations more in instruction. It is such a nice concept for understanding what the hell is going on, and it generalized in some really nice ways (e.g. in linear algebra).
@SineoftheTime The notation $5 + \sqrt{4}$ means $5+2$. This is unambiguous. Comments about branches of the square root are only going to confuse people.
There are contexts in which one might need to consider branches of the square root in order to evaluate $5+\sqrt{4}$, but in any such context, that will be spelled out from the start.
@think_meaning_buildß That would be "further context".
Though I would argue that it is misleading to then say "the" square root.
The question should read "What are the two integers which described by the notation 'five plus radical four'?" or something like that.
Hi. Does anyone know about a good blogging website where I can write posts that natively support MathJax (or LaTeX), while also getting subscribers to the blog (like a newsletter)?
Let $\mathcal L$ be the Lebesgue $\sigma$-algebra and $\overline{\mathcal B}(\mathbb R^d)$ be the completed Borel $\sigma$-algebra with respect to Lebesgue measure $\mu$. Showing $\overline{\mathcal B}(\mathbb R^d)\subset \mathcal L$ is straightforward; the other direction is trickier I think. My book does the following (I'm paraphrasing).
> Let $A\in\mathcal L$. Without loss of generality, assume that $A\subset (-K,K)^d$ for some $K>0$. Then $\mu^\ast(A)<\infty$ and for every $n\geq1$ we can find a countable collection $(R_i^n)_{i\in\mathbb N}$ of open rectangles such that $$A\subset\bigcup_{i\in\mathbb N}P_i^n,\quad\sum_{i\in\mathbb N}\operatorname{vol}(P_i^n)\leq\mu^\ast(A)+\frac1n.$$
> We assume that the boxes $P_i^n$ are contained in $(-K,K)^d$ (the intersection of an open rectangle with $(-K,K)^d$ is again an open rectangle). Set $$B_n=\bigcup_{i\in\mathbb N}P_i^n,\quad B=\bigcup_{n\in\mathbb N}B_n.$$
> The author goes on to show that $A\subset B\in\mathcal B(\mathbb R^d)$ with $\mu(B)=\mu^\ast(A)$ and similarly, that there exist a $B'\in \mathcal B(\mathbb R^d)$ such that $B'\subset A$ and $\mu(B')=\mu^\ast(A)$. Thus there are Borel sets $B,B'$ such that $B'\subset A\subset B$ with $\mu(B\setminus B')=0$ and this shows $A\in\overline{\mathcal B}(\mathbb R^d)$.
Question Why can we assume without loss of generality that $A\subset (-K,K)^d$ for some $K>0$?
I'm reading Measure Theory, Probability and Stochastic Processes by Le Gall. In Proposition 3.7, the author proves that the Lebesgue $\sigma$-algebra $\mathcal L$ is the same as the completed Borel $\sigma$-algebra $\overline{\mathcal B}(\mathbb R^d)$ with respect to Lebesgue measure $\mu$.
Showi...
@psie i'm not 100% sure i understand the source of the difficulty. however one has defined overline{B}(R^d) it should not be too much work to verify that it is a sigma algebra (i.e. closed under countable unions). any subset A of R^d is the union over n of A intersect (-n,n)^d, so to show A is in some sigma algebra (any sigma algebra, not just the one you're focused on) it would suffice to show that each A intersect (-n,n)^d is
i haven't rendered that latex or clicked into the question you posted on main however so i may be missing something :)
@leslietownes oh, ok, that's an aha moment there for me :) thank you leslie! I think I understand now. The "without loss of generality" is more a statement of "it suffices to show". Thank you!
The author is basically showing that A intersect (-n,n)^d is in overline{B}(R^d), yes, and then since the sigma algebra is closed under countable unions, we have that A is in overline{B}(R^d). Makes sense.
yeah, and in general it is not an 'if and only if' kind of thing, e.g. one might be able to define some sigma algebra on R^d where some set A is in it despite none of the sets A intersect (-n,n)^d being in it. but the borel/lebesgue/whatever example you're focused on, that does not happen
Let $U, W \subseteq V$. Show that $(x + U) \cap (y + W)$ is either empty or of the form $z + U$ for $z \in V$.
Assume that $x + U = y + W$. I've shown before that then $U = W$ and so this becomes $x + U = y + U$. Now assume that $x + U \neq y + W$. Assume that there is a $z \in (x + U) \cap (y + W)$. We will show a contradiction. Since $z - x \in U$, $$x + U = x + (z - x) + U = z + U$$ and since $z - y \in W$, $$y + W = y + (z - y) + W = z + W.$$
Since $x + U \neq y + W$, it now follows that $z + U \neq z + W$ and so $U \neq W$. But how does this help?
There is some difference between $V/U = \{v + U \mid v \in V\}$ (set of cosets) and $V + U$ (sum of subsets) yet at the same time they feel very similar
Well, the elements of $V/U$ are contained in $V + U$
Let $U, W \leq V$ be subspaces. Consider the restriction of the canonical projection $\pi_U\mid_W: W \to V/U$. Then $\operatorname{Image}\left(\pi_U\mid_W\right) = (U + W)/U$. Why is it not $W/U$?
@ILikeMathematics the elements of $V/U$ are subsets of $V+U$, if you will (although I don't think that's necessarily a helpful perspective), but they're not elements of $V+U$
@Thorgott So we can generally say that $V/U = (V + U)/U$, right? Since $V/U = \{v + U \mid v \in V\}$ and $(V + U)/U = \{v + u + U \mid v \in V, u \in U\} = \{v + U \mid v \in V\}$
@Thorgott Assume $U$ is not a subset of $W$. We have $\pi_U|_W: W \to V/U$ and we need to express the image somehow. We can say $(W + U)/U$ since $U$ will be a subset of $W + U$. Also, this will be the same as the set $\{w + U \mid w \in W\} [= W/U]$.
[] because we aren't really allowed to say that but that's what we want
@leslietownes it sounds like the author is trying really hard to convince people to consider Cantor cube $\{0, 1\}^\kappa$ for $\kappa >\aleph_0$ instead of the Cantor set. Who knows why or what for
> In the history of mathematics, there are many instances of exotic mathematical objects which have found extensive application in physics
oh, haha. i wouldn't assume anything on the basis of that
springer has all of the incentives that paper mills do, receive hard currency, provide published articles
i have no idea what their operations are but i'd expect that outside of some internal "top tier" of stuff they do, the oversight is minimal, maybe limited to reviewing who is on the editorial board
i don't want to sound too elitist, haha, most academic papers including my own basically amount to nothing in the grand scheme of things
but this paper is lacking any hallmarks that a person has any regard for their audience, i don't even see anything set out as a theorem, let alone offered as a purported proof
its just word salad with citations to "real" things
Let $d\in\mathbb N$ be fixed and let $a_j,c\geq0$ be reals. What is $$\lim_{n\to\infty}\left(\prod_{j=1}^d\lfloor na_j\rfloor\right)\frac{c}{n^d}?$$The floor function there makes me unsure about what the limit is or even where to begin thinking about what it is.
This is a translation of a Russian journal And knowing how it works over there I am not sure if it was under review My first papers only needed a "review" from my advisor
An unspecified preprint by E.K. van Douwen, naturally Here is the list of papers: https://www.math.buffalo.edu/~sww/0papers/van_douwen_eric_k.html maybe you'll find something relevant around the year of the publication of the paper
not to harsh the vibe, but one time somebody told me and a coauthor that they had proved something and without asking for a proof we took their word for it and said they had done it in a paper
I am looking at an example of an $F'$-space which is a subspace of an $F$-space but not an $F$-space, in the article "On $F$-spaces and $F'$-spaces" by Dow
it's hard to explain so I'll just refer you to the write up here that I made
the only citation I could access was "An Extremally Disconnected Dowker Space" by Dow and van Mill
it revealed a lot, I've started to understand how this example works
but now I think either 1) there's a mistake or 2) I am misunderstanding something
the article I've been able to access proves that a $P$-space $X$ can be $C^\ast$-embedded in $E(2^\kappa)$ which is fine to me, but I have some problems
maybe the weight of $\beta X_\delta$ and $X$ where $X_\delta$ is the $G_\delta$-modification of compact Hausdorff space $X$ are somehow nicely related, but I don't know
I was thinking that perhaps my bound $w(\beta X)\leq 2^{w(X)}$ was too strong?
either way I have no access to the article to tell me why $E(2^\kappa)_\delta$ embeds as a (nowhere dense?) subspace of $E(2^\kappa)$
> Theorem Lebesgue measure on $\mathbb R^d$ is invariant under translations: for every $A\in\mathcal B(\mathbb R^d)$ and every $x\in\mathbb R^d$, we have $\lambda(x+A)=\lambda(A)$. Conversely, if $\mu$ is a measure on $(\mathbb R^d,\mathcal B(\mathbb R^d))$ which is finite on bounded sets and invariant under translations, there exists a constant $c\geq0$ such that $\mu=c\lambda$.
> Remark For every $a\in\mathbb R\setminus\{0\}$, the pushforward of Lebesgue measure $\lambda$ on $\mathbb R^d$ under the mapping $x\mapsto ax$ is $|a|^{-d}\lambda$. This is an immediate application of the theorem.
I'm trying to understand the remark and how it follows from the theorem. We have that the pushforward is $\lambda(f^{-1}(A))$ where $f$ is the dilation by $a$, but I don't see how the theorem applies here. Any help is really appreciated.
without thinking too deeply, maybe they are implicitly using that the pushforward is finite on bounded sets and invariant under translations and using some model case (e.g. the unit cube) to identify the value of c guaranteed by the theorem
@leslietownes yeah, I think you're on the right track :) I buy that the pushforward is finite on bounded sets since Lebesgue measure is (I think), but why is it translation invariant? Is it because $f^{-1}(A+x)=x+f^{-1}(A)$?
I'm just guessing here.
@psie well actually, the preimage of a bounded set has to be bounded too. Hmm.
well, are you just guessing? could you prove it? would the translation invariance of lebesgue measure imply that x + f^{-1}(A) has the same measure as f^{-1}(A)?
i agree with your implicit contention that f and f^{-1} are linear maps
i'm not sure i understand the remark as the best route to understanding the result that it claims, but a theorem like that reduces the comparison of translation invariant measures to the problem of computing the measure of any single set of nonzero measure
apparently van Douwen proved that weight of $\beta (2^{\omega_2})_\delta$ is $\omega_2\cdot 2^{\omega_1}$, so that I can $C^\ast$-embed my space $X$ into $E(2^{\omega_2\cdot 2^{\omega_1}})$ but I'm not sure if its any better
@leslietownes regarding your third question, it would, however, for the pushforward to be invariant under translation, we want $\lambda(f^{-1}(A+x))=\lambda(f^{-1}(A))$. And I'm not so sure how one verifies that $f^{-1}(A+x)=f^{-1}(A)$ for $f$ being given by $x\mapsto ax$ where $a\neq 0$.
@VladimirLysikov the article by Woods helped me get to an article of van Douwen where he calculates weight of $\beta X$. The bad news is that it's $\omega_2^\omega$ and its consistent with ZFC that it equals to $\omega_2$. So I might have by accident posed a hard set theory problem
And the details of method of van Douwen of embedding $P$-space into $E(2^\kappa)$ is still unknown to me as well
