For a sequence $\{f_n\}$ of measurable functions with common domain $E$, show that the following functions are measurable: $\inf \{f_n\}$, $\sup \{f_n\}$, $\lim \inf \{f_n\}$, and $\lim \sup \{f_n\}$
Here is my proof:
It suffices to show that $\sup \{f_n\}$ and $\lim \sup \{f_n\}$ are ...
Problem: Let $I$ be an interval and $f: I \to \Bbb{R}$ be increasing. Show that $f$ is measurable by first showing that, for each natural number $n$, the strictly increasing function $x \mapsto f(x) + x/n$ is measurable. Here is proposition 4 in my book: A monotone function that is defined on an interval is measurable.
In light of proposition 4, what is the point of constructing the sequence $\{f(x) + x/n\}$ of strictly increasing functions, or am I misunderstanding something?
@BalarkaSen I have an idea that I think is wrong and I was wondering if you can see wy its wrong. Consider any smooth map $f:T^2 \to S^3$. My idea is that $f$ is always homotopic to a constant map. The idea is since $T^2 = S^1 \times S^1$ consider $f \lvert_{\{p\} \times S^1}$, this should be a map $S^1 \to S^3$ and since $S^3$ is simply connected, this is homotopic to the constant map. In a path-connected space, all constant maps are homotopic.
So then, enforce that for all $p \in S^1$, $f\lvert_{\{p\} \times S^1}$ is homotopic to $q \in S^3$, then $f$ is itself homotopic to a constant map.
I think this has to be wrong, but I'm not sure why
Maybe the problem here is that the homotopy for each $p$ is different. And you have to stitch them together in a smooth way.
@BalarkaSen Really?? We don't need any further restriction on $f$? I was thinking its probably right if $f$ is injective. But I was worried that if $f(x) = f(y)$ then there could be 2 different homotopies applied to $\{x\} \times S^1$ and $\{y\} \times S^1$ and so when you try and put all the $S^1$ homotopies together to get one for $T^2$, you don't know how to move the point $f(x) = f(y)$. There are, in some sense, 2 ways you want to move it.
@BalarkaSen what would happen if $x>0$ and $\exp x = 1 + x + \dfrac {x^2} 2$? Can we get a contradiction somehow by considering points in the middle of $0$ and $x$?
@BalarkaSen I remembering doing that one, we had to prove already that $S^n, n > 1$ is simply connected. If I recall right, I argued that any such map is homotopic to one transverse to the north pole. The only way to be transverse to the north poll in this case is to not intersect it. Then the image of the map lies in a single chart, so you can go to $R^n$, homotope htere because its contractible, then map back to the sphere
Hahaha I misspelled pole as poll and Im not fixing it
@BalarkaSen We have approximation theorems though. So you should be able to apprixmate any $C^0$ guy by smooth maps, and I'm not sure but I guess thats good enough
@LeakyNun You're bounded below by $(1+x/n)^n$. This has terms that are always positive, and the first three terms are $1+x+\frac{1-1/n}{2}x^2$. Thus $1 + x + \frac{1-1/n}{2}x^2 \leq e^x$ for all $n$.
You should be able to play out the binomial theorem to calculate the power series of $e^x$, if you really wanted to. The essential point is Stirling's asymptotic.
What do you mean you can't do anything from there?
Problem: Let $I$ be an interval and $f: I \to \Bbb{R}$ be increasing. Show that $f$ is measurable by first showing that, for each natural number $n$, the strictly increasing function $x \mapsto f(x) + x/n$ is measurable. Here is proposition 4 in my book: A monotone function that is defined on an interval is measurable. In light of proposition 4, what is the point of constructing the sequence $\{f(x) + x/n\}$ of strictly increasing functions, or am I misunderstanding something?
Why are unruled wirebound notebooks so hard to find? It's like mankind has forgotten how to use a pen, and everyone needs handholding by means of lines on paper!
I don't feel strongly about anything, not even on what I'm informed are more pressing social issues, but this one gets on my nerves.
I need to hang in there for a few more days - finish my homework. but i'm oon her end of life directive, and i have to make a ddeciion to withhold food and hydration
Let S be an uncountable set with a non aleph cardinality and A be a countable subset. Then S-A is infinite dedekind finite
Note that S-C where C is finite will not make it infinite dedekind finite, but its cardinality is strictly smaller than S
Likewise, S-B where B is a countable subset of A will also not make it infinite dedekind finite, despite cardinality will still be reduced
For S-D where D is a cofinite subset of A, it is infinite dedekind finite because S-D contains only infinite dedekind finite sets and finite sets
And therefore, cardinal arithmetic becomes a lot more sensitive to the structure of the set
(Actually, I should have said: Let A be the maximal countable subset of S)
Since some dedekind cardinalities themselves can be linearly ordered if they originated from considering a chain of dedekind finite subsets, and aleph cardinalities are always well ordered by definition of Hartogs number, it follows there might be sets with mixed cardinalities such that as elements are being added or taken away, the dedekind and aleph cardinalities should respond to that independently
This will be explored in more detail using some generic $\aleph_1$-dedekind finite set
Consider the following $\aleph_1$-dedekind finite set with the following structure:
$S = A \sqcup B$
where $A$ is maximally countable, and $B$ is is infinite dedekind finite
Let $C = A \cap B$. It follows that the intersection is finite. Suppose otherwise, then $B$ will contain a countable subset thus contradict dedekind finiteness
In particular the intersection $C$ can be any arbitrary large finite number, thus we can have a family of $C$ with cardinality ranging over all possible natural numbers
Given the current frequency I (basically to the eyes of most) spam chatrooms across SE (because the number of weird users are falling, and the number of users who are not interested in my rambles are increasing much rapidly than I expected, result in all these spontaneous discussion type messages to basically becomes spam) there should not be a problem to keep that room alive for every week
Mathworks had failed me (or succeeded?) due to the unexpected spike in users, converting it from a room that is purely used for my math rambles, into a more formal room for themed mathematical research ideas (which is why a lot of cleanup occured in there recently)
And therefore, this new room I am going to create, should live its intended purpose
O btw, given the fact that the number of ordinal users are increasing, it means there will be less people will spontaneously engage with my thoughts and ideas. But that is not a problem, because the SE server 'funded' this and it knows how this is to be solved
[Intuition] : Is there any reason that the curves look like the way they do? Take for instance an ellipse, why is it a closed curve and looks like an oval?
We name it an ellipse because of the shape of the curve. As for why it is closed, it is because in parametric form, there are two parameter values that give the same (x,y) value
so a topological embedding is taking a topological space putting it in a bigger one and it has the property to have the subspace topology on the other hand an immersion is just putting the object into a space but cant have the subspace topology right?
Now my question is. I can see why embed objects within a bigger space so it gets its topology and do stuff
im in applied math and i next semester i can complete my degree with emphasis on pure math. Since ive taken courses pure theoritic.BUt i need one more .which is logic. but if i take logic plus my regual program courses .Ill have 6 courses on total
I was going through the proof of Malgrange-Ehrenprein theorem in chapter 8 of Rudin Functional analysis .. the author writes '$\sigma_n$ Haar measure on $T^n$ that is lebesgue measure divided by $(2\pi)^n$' .. if I am not mistaken $d\sigma_n = \frac{1}{(2\pi i)^n}\frac{dz_1 \cdots dz_n}{z_1 \cdots z_n}$ .. but why is the author calling it lebesgue measure?
@KonformistLiberal In fact, this group is isomorphic to $H^1(X; \Bbb Z)$, because any map $\pi_1(X) \to \Bbb Z$ factors through the abelianization $H_1(X)$, so your thing is just $\hom(H_1(X), \Bbb Z)$
(More generally, $[X, K(G, n)]$ is a group where $K(G, n)$ is the Eilenberg-MacLane space with $\pi_n = G$. The idea is to realize $K(G, n)$ is a homotopy-associative, homotopy-identity admitting H-space.)
But of course that's easier to see with $G = \Bbb Z$, $n = 1$ because $S^1$ is even a group
@BalarkaSen The question I linked previously now has an answer. It turns out the solution is "obvious." There is a well-known fact that tensor fields are independent of extension (cf. Lee). Simply show that "d" as defined is a tensor field, then we're in business.
@ManolisLyviakis I think that having a thorough understanding of Euclidean/classical geometries will give you a greater appreciation of the latter subjects. On the other hand, in most cases, you don't need to know Euclidean geometry fluently, for the most part.
@Antonios-AlexandrosRobotis i just find euclidean geometry rather dull and raw in the sense it seems more like an iq test than the other geometries when u can bend the rulles sort of speak
@ManolisLyviakis oxi de ksero :/ Alla ama einai steile mou email: [email protected].
But, yeah. The nullstellensatz is a great theorem. I think that Euclidean geometry seems best absorbed through solving problems as puzzles. It can be a nice way to spend an hour here or there.
In the comments of my question math.stackexchange.com/questions/2538844/… someone told me, there CAN be different a's with the same value and $z^n-1$ can indeed be divided by $(z-x)^2$. I'm am cofused because it's the opposite of my task.
I am trying to show that a strictly increasing function $f$ over an interval $I$ is measurable. I know that $f$ is an open map or, equivalently, since $f$ is invertible, $f^{-1}$ is continuous, but I can't see how to use this fact, if it can be used at all.
Can anybody help me understanding some notation? It is related to algebra and I'm supposed to show that $\phi(g)(x)=gx$ is a group homomorphism, however I'm not sure what it means for a map to have "2 parentheses" like in the above example.
Hi all! I am a newbie and I am trying to learn calculus by myself, and I noticed that $\frac{d}{dx}x^2=2x$ and that $\frac{d}{dx}x^3=3x^2$. Is it true that $\frac{d}{dx}x^n=nx^{n-1}$ for any $n,x\in\mathbb{N}$?
They probably use $\phi(g)$ because they're thinking about $\phi$ as a function $G\to\text{Aut}(G)$ that sends $g\mapsto \phi(g)$ which is the map defined as above
The full problem is: Let $G$ act on the set $X$ and define the map $\phi: G \rightarrow \text{Sym}(X)$ by $\phi(g)(x)=gx$. Show that $\phi$ is a homomorphism of groups. (where Sym(X) is the group of all bijective maps from X to itself with multiplication given by composition of functions.)
Got a question which I’m not sure how to categorize:
Suppose I’ve got a closed convex set in R^3 defined as $f(x)\geq 0$ where $f(x)$ is some polynomial in $x,y,z$.
I can project this set on to a plane through the origin and the image will be a convex set as well. From this I can infer some inequality the original convex set must satisfy.
(For instance, suppose my initial set is $x^2+y^2+z^2/4\leq $. Then the projection onto the xy-plane is the set $x^2+y^2\leq 1$, and every point in the original convex set indeed satisfies this bound.
I just read this certifiable tutorial which is indeed informative:
However, I couldn't go further after the following sentence:
The values of x1 and x2 are chosen such that the elements of the S are the square roots of the eigenvalues.
My question is how did they calculate $x_1$ and $x_2$?...
What I’m wondering is how many such two-variable inequalities one needs in order to define the original convex set. If $f(x)$ were some arbitrary smooth function I’d guess you would need infinitely many, but in the case of polynomial $f(x)$ it seems plausible that only finitely many are needed.
@Cardinal I see two things of note.
First off, if you’ve got two linear equations in two unknowns then the generic scenario is that there’s a unique solution (x1,x2)
hey if $R \subset S$ , then S^n is a R-module. Is it a R-free module, I think it isn't, but then does it have any special property compared to usal f.g modules?
@Semiclassical This is a very nice question! I don't have insights, unfortunately. Just to clarify: the closed convex set you are considering is $\{x \in \Bbb{R}^3 \mid f(x) \ge 0 \}$? I wonder if this ties in with sums of squares and algebraic geometry.
But they claim that the solution is not a pair (x1,x2) but the ratio x2/x1. For this to be the case, the two equations have to really be the same equation up to an overall factor. (And that does seem to be the case from their coefficients.(
@Semiclassical If you like questions like the one you posed, you should do a google search on sums of squares and positivstellensatz. It's somewhat related and there is a lot of current work being done on it, both from a pure and computational aspect.
@Semiclassical That's a cute question; I suspect the answer is going to be complicated. Even for algebraic sets instead of semialgebraic sets, the issue about how many polynomial equations cut out a given variety is a subtle one. (Eg, twisted cubic)
(And cut out in what sense? Set theoretically? Transversely? Something in the middle?)
I just read this certifiable tutorial which is indeed informative:
However, I couldn't go further after the following sentence:
The values of x1 and x2 are chosen such that the elements of the S are the square roots of the eigenvalues.
My question is how did they calculate $x_1$ and $x_2$?...
