One of my very smart ex-colleagues notoriously wrote real analysis problems he “thought” he knew how to do. It was because of him that we changed the system — not that the change always worked. We made 3-oerson committees write the exams — 2 experts, one non. The non-expert was supposed to hold the feet to the fire. Most didn’t bother.
When I was the non-expert (real or algebra), I worked the exams and raised hell if problems were bad. But most people didn’t care enough … or trusted the experts.
By definition, isn't a set unordered? What would you call an ordered set? Is it assumed a set is unordered until it's qualified with "ordered"? For example in a card game what would you call a hand that must be played in ascending order (e.g. a run like 3,4,5 but not 5,3,4)?
Hi All, I don't understand why there are multiple upper bounds leading to the concept of the least upper bound. Can someone give me an example where there are so many upper bounds and hence the concept of a LUB matters? Thanks
Ok. This is where Rudin uses t = x - (x^2-2)/(x+2) which implies t^2 = 2 + 2(x^-2)/(x+2)^2 and we can get a contradiction. If x^2 - 2 < 0, we have t > x and t < 2. Hence contradiction.
So the answer is that the set of Q such that $x^2 < 2$ has no lub.
Rudin in theorem 1.11: Suppose S is an ordered set with the least-upper-bound property, B c S, B is not empty, and B is bounded below. Let L be the set of all lower bounds of B. Then .......
yeah its like if there was a lake and you want to feel the edge, you need to put your hand into the water first and then slide towards you till you grab the edge
so with LUB you need to go into the upper bounds and then you slide down until you approach the least one
Take the function:
def f(x0:ZZ_N) -> tuple:
x = x0
while True:
y = x**2 + 1
if isprime(y):
return x
x += 1
That's what got the wheels turning in my 🧠.
Let our computer memory integers typed ZZ_N be modeled finitely ie as usual on a modern machine t...
@SillyGoose Well try to imagine maps between them? What might a kernel of such a map be?
@TedShifrin Not so much disagreeing with this, but what would you call $M_{r,s}(k)\times M_{s,t}(k)\to M_{r,t}(k)$? I thought of this family of maps for $r,s,t\in \Bbb N$ as being the 'matrix operations'
I don't know if it forms a structure on its own (although I feel like one could try to set up some bigraded algebra type object, and take as convention that multiplying 'illegal pairs' gives zero, or something - or maybe one includes all the finite matrix sets here into some infinite matrix set or something, I dunno)
I guess technically one is meant to think about these as being actions
Ah yeah I just saw on wiki https://en.wikipedia.org/wiki/Operation_(mathematics)
They introduce language that distinguishes between these types of things. Your message was about what they call 'internal operations', and they leave n-ary operation as $\omega:X_1\times X_2\times \dots\times X_n\to Y$ without assuming $X_i=X_j$ in any case
So I suppose 'operation on a set = internal operation' makes sense
One of the big problems with smoothness seems to be that $x\mapsto \|x\|$ is not differentiable at $0$. But this can be avoided if we compose it with an analytic function of the form $f(x) = \sum a_n x^{2n}$ since $x\mapsto\|x\|^2$ is smooth everwhere. This seems to be one way to handle this, the other being separated from $0$
There is a lot of confusion between what is a number, and what is a representation of a number. It is an extremely ordinary and common thing for students to get stuck on.
It is not a very well phrased question, and I suspect that if it were asked today, it would be closed and deleted, but it is from 2014, and we are generally a bit kinder on older questions. And, again, I think that the asker is clearly struggling to articulate a problem which is very common.
@Jakobian You seem to be extraordinarily dismissive of this problem, which pretty clearly demonstrates (to people who have been teaching for a while) a very common mistake which students make.
Look, I've told you that this looks very much like the kind of question that students who are struggling with the representation of numbers vs the numbers themselves actually ask. My experience with exactly this kind of question, in in-person interactions with students, makes it very clear what the underlying problem is. You have basically responded with "Nuh uh! You're wrong!"
Which... okay, fine. I guess we'll have to agree to disagree.
That too is incorrect. People can voice their opinion in a binary like this, and leave it at that, especially when you are very confrontational, and I don't have the energy for that
Do you believe that one can be confused about something, but the very confusion stops one from being able to state what they are confused about? Cases where the resolution of their confusion actually dissolves the confusion, rather than answering a concrete question?
I figured something like that was what Xander had in mind, and one finds when they teach the same content over and over again, the same types of errors appear, and students ask the same strange broken questions about those same errors. So the teacher/tutor/lecturer slowly learns how to answer weird broken questions - where in particular, the response they get is 'oh wow, that's exactly what I meant to ask about', etc.
I also know that with strongly decreasing frequency I've asked a question that ended up being nonsense, but the response I got was both why I realised it was nonsense, and resolved my hidden confusion that I couldn't express
A question from 2018 I posted here is beyond nonsense lol
I get that view, but I ask questions myself and I know I didn't put any effort into some of them before asking, and I wasn't particularly confused about them either... I just posted them out of laziness. The question looks lazy and not that the OP is confused to me
@Jakobian Sorry, not that the OP is not confused, of course they are. Its more that the question is too open ended, too much there can be taken up to interpretation, even if it is about decimals, its not clear at all, and attempts at clarifying it are really adding your own interpretation to the question, so its better if its closed
I mean that's possible. I do think one can have expertise in interpreting such things though. I.e. some people might be able to better associate a hidden question with the asked semi-nonsense
jakobian is pointing to another kind of idea which is that if for whatever reason you can 'see' a basis of eigenvectors, that is another way to tell, although if the question is "how/why would i be able to do that without calculation" my general answer might be "i don't know"
sometimes for whatever reason it's definitely easier or at least possible to identify eigenvectors without running some general diagonalization algorithm
and if you find enough of them, you know that you have a diagonalization even if you haven't run a general diagonalization algorithm
Question: The functors $S\mapsto S^1, S\mapsto S^0$ of semigroups obtained by adjoining $1$ and $0$ element respectively are defined so that $S = S^1$ if $S$ is a monoid, correct?
its maybe mildly interesting that for general real symmetric matrices, relying on the theorem is always possible in a way that being able to write down "nice" eigenvalues/eigenvectors is not
Do the forgetful functors $G_H:\bf Monoid \to \bf Semigroup^1$ and $G_O:\bf Semigroup^1 \to \bf Semigroup$ have left and/or right adjoints? Here $\bf Semigroup$ is the category of semigroups with semigroup homomorphisms, $\bf Semigroup^1$ is the category of monoids (semigroups with identity) with...
In Vretblad's Fourier Analysis and Its Applications, there is the following exercise.
> Let $f$ be a $2\pi$-periodic function with (complex) Fourier coefficients $c_n, \ n\in \mathbb Z$. Assume that for an integer $k>0$ it holds that $$\sum_{n\in\mathbb Z}|n|^k|c_n|<\infty.$$ Prove that $f\in C^k$.
I'm dumbstruck on this one. The condition above means that the $k$th termwise differentiated Fourier series of $f$ converges absolutely and (I think) uniformly to a function $g$ (maybe $g$ is even continuous, I'm not sure). How do we know that $f=g$?
If I have a map $f:S\to T$ between the two monoids such that $f(1)\neq 1$ then I need to adjoin identity to both to make the morphism unital, so I do need to adjoint it, even if it was a monoid already
psie: can you answer this in the case k = 0? note how the question from the other day (about a specific c_n generated by solving some differential equation) maybe fits into this setup
In the mathematical field of analysis, uniform convergence is a mode of convergence of functions stronger than pointwise convergence, in the sense that the convergence is uniform over the domain. A sequence of functions
(
f
n
)
{\displaystyle (f_{n})}
converges uniformly to a limiting function
f
{\displaystyle f}
on a set
E
{\displaystyle E}
as the function domain if, given any arbitrarily small positive...
Yep. It implies that the function to which the Fourier series converges is in $C^k$
Moreover, there should be some theorem that tells you that since $k > 0$, it follows from uniform convergence that the Fourier series converges to the intended function
I am reading Hogg, McKean and Craig book for Introduction to mathematical statistics 6th ed. There are 2 two definitions there for continuous random variable (R.V): 1- R.V for which space is continuous (for e.g. space is real numbers) are continuous R.V. 2- R.V is continuous if its cumulative distribution function is continuous. Which one is correct or both are equivalent in some way?
If $A\subseteq S$, let $S/A = S/\sim_A$ where $\sim_A$ is the smallest congruence for which $a\sim_A a'$ for all $a, a'\in A$. If $S$ is a monoid, let $\mathfrak{g}(S)$ be its group of units and call $\mathfrak{a}(S) = S/\mathfrak{g}(S)$ the absolution of $S$. Call a monoid absolute if $\mathfrak{a}(S) = S$. Is the absolution of a monoid an absolute monoid?
I'm thinking about iterating $\mathfrak{a}$ operation, but for this I would probably need to know if it even needs iterating. It seems like there should be an example
Say $S = \{0, 1, 2\}$ where $(\{0, 1\}, *)$ is $\mathbb{Z}/2\mathbb{Z}$, $2^2 = 2$ and $\{0, 1\}*2 = \{0, 1\}*2 = \{0, 1\}$ maybe?
consider the relation $\sim_A^{\prime}$ defined by $x\sim y$ if there is an integer $n$, elements $z_1,\dotsc,z_n\in S$ and $a_1,\dotsc,a_n,a_1^{\prime},\dotsc,a_n^{\prime}\in A$ such that $x=a_1z_1\dotsc a_nz_n$ and $y=a_1^{\prime}z_1\dotsc a_n^{\prime}z_n$. this respects the multiplication, is reflexive and symmetric. it is not transitive, but the transitive relation generated by it (so taking sequences of such relations) should yield the congruence $\sim_A$.
now, if $S$ is commutative, then $g(S)$ has the wonderful property of being saturated, i.e. $xy\in g(S)$ implies that $x,y\in g(S)$. then, from the above relation, it is easy to see that $y\sim_{g(S)}1$ iff $y\in g(S)$. so $[x][y]=[1]$ in $a(S)$ iff $xy\sim_{g(S)}1$ iff $xy\in g(S)$ iff $x,y\in g(S)$ iff $[x],[y]=[1]$, i.e. $a(S)$ is absolute.
yeah, and it has this "saturated" or "prime" property
this doesn't really need commutativity i guess, but rather just that an element has a right inverse iff it has a left inverse (not sure if that property has a name)
cause $xy$ having a right inverse implies $x$ has a right inverse and $yx$ having a left inverse implies $x$ has a left inverse, so $x$ is invertible, and vice versa for $y$
sha: p is the range projection of u. an operator always commutes with the projection onto its range (e.g. if u is positive the operators (u + 1/n)^{-1} u converge strongly to the range projection of u)
well, my construction yields a congruence satisfying $a\sim a'$ for all $a,a'\in A$, but it's also generated as an equivalence relation by relations that are composed of $a\sim a'$ for $a,a'\in A$, reflexivity and compatibility with multiplication, hence contained in the smallest such congruence
@ShaVuklia you may be thinking/intuiting that the range projection isn't generally in the C star algebra generated by the operator (which indeed might not contain any nontrivial projections). you do need to be able to take strong limits
we don't necessarily have a finite dimensional spectral theorem or eigenspaces, but it does follow from the fact that pu = u that pu^{1/2} = u^{1/2} (e.g. because u^{1/2} is a limit of polynomials in u)
i wasn't attempting to imply that my approach was the only one, just connecting the question to some ambient operator theory that is used frequently enough to be a common way of thinking about stuff like that
use whatever you'd use to make sense of "u^{1/2}" in general, which you could definitely do using something like stone-weierstrass (meaning, something like identifying 'functions of u' with a function space)
since the range projection of u would commute with u irrepsective of its dimension, 'using' the finite dimensionality would seem kinda odd from some points of view
Let $M$ be a monoid with three elements. Let $G$ be its group of units (elements which have an inverse) and let $I$ be its minimal ideal. Note that if $|I| = 1$, then $M$ has a zero.
Let denote by $C_n$ the cyclic group of order $n$.
If $M = G = I$, then $M = C_3$. Otherwise, $G$ and $I$ are di...
what he does is consider group of units $G$ and minimal ideal $I$ of the monoid
its probably possible to just check which monoids of order four are there on fingers
all monoids for which $|G| = 1$ or $G = S$ are irrelevant, so that there'll be less checking
@leslietownes I'm not sure if I'm understanding you correctly. If $L$ is an operator, and $P$ is the projection onto its range, then we have $L=PL$, and I don't believe that $L=LP$ always holds. E.g., if we take $L:\mathbb C^2\to\mathbb C^2$ such that $Le_1=e_2$ and $Le_2=0$, and if we take $P$ the projection onto $\mathbb C e_2$, then $LP=0\neq L$
if u is not self adjoint then following the same idea you might only get that its range projection is a strong limit of functions of u ustar [or something else that is not guaranteed to commute with u]
this is actually an instance where it makes more sense than usual to avoid chatjax, because when there is more than one literal asterisk in a line, it can be interpreted by the chat as wanting to put something in emphasis, even if the asterisks appear separately in different math modes
@Thorgott Another thing is that if $|S/G| = |S|-|G|+1$ then $S/G$ must be absolute, so we have to only check those monoids for which $|S/G|\leq |S|-|G|$
@TedShifrin I have just added a more complete explanation of $(1)$, I will add how that supports $(2)$, but I am leaving for PT soon, so it will have to wait until I get back.
@Thorgott The case of monoids of order four goes like this. Since $|S/G| = |S|-|G|+1$ implies $G$ is an equivalence class, for $|S| = 4$ we only consider $|S/G|\leq 4-|G|$. Then $|G| = 2$ and $|S/G| = 2$ is the only case. If $|I| = 1$ then $S/G$ has a zero, so is absolute. If $|I| = 2$ then letting $S = \{1, a, b, c\}$ with $a^2 = 1$, $1\sim b$ then $bc = b$ or $cb = b$ so $c\sim b$ as well, contradiction.
So $S/G$ is absolute for monoids of order four
so the monoid I'm looking for must be of order $5$ or above
If $|S| = 5$ then $|G| = 2$ or $|G| = 3$, if $|G| = 3$ then $|S/G| = 2$, the cases of $|I| = 1$ and $|I| = 2$ are like for monoids of order four above