me too lol, I would have probably offered that you can construct topologies without referencing bases by saying 'the smallest topology so that some collection of functions of interest defined on the space are continuous' , and this thing exists because the intersection of topologies is a topology
so this does not require any notion of basis to define and it is often actually realized by a lot of topologies we care about
including so called 'large' topologies as referenced in the post
there are definitely weird topologies out there, not gonna defend topology, but, couldn't for the life of me figure out what that discussion was all about
Suppose that A is an algebra on X and that $\mu$ is a measure on A. It is given that $\pi^*(E\triangle F)=0, F \in M$, the set of all $\pi^*$ measurable sets (I know that M is a $\sigma$ algebra). Then, it is to be shown that $E\in M$.
I'm having difficulty showing that.
I can show that $E\triangle F\in M$.
$\pi^*$ is an outer measure w.r.t. $\mu$.
So we have $E\triangle F, F\in M$. Can it be shown now that $E\in M$?
TrystwithFreedom: Every topology has a basis (being the topology itsefl) that generates the topology.
Suppose that $\mathfrak A$ is an algebra on $X$ and that $\mu$ is a measure on $\mathfrak A$. It is given that $\pi^*(E\triangle F)=0, F \in M$, the set of all $\pi^*$ measurable sets. Then, it is to be shown that $E\in M$. $\pi^*$ is an outer measure on $X$ w.r.t. $\mu$.
(Note: $E\triangle F:= (...
Suppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$).
Thus we have $B = A+ \xi e_{1}^T$, where $\xi$ is a $n \times 1$ column ve...
in atiyah-macdonald intro to commutative algebra, there is a remark that I want to make sure im not misunderstanding, (For $A$ a Noetherian ring, $M$ an $A$-module) "For non-finitely generated modules the functor $M \rightarrow \hat{M}$ is not exact: the good functor, which is exact, is $M \rightarrow \hat{A} \otimes_{A} M$ and the two functors coincide on finitely generated modules", I get the statement minus 'good functor' , that isnt a technical term in category theory right?
i guessed he just meant this is the 'better' of the two functors because its always left exact
Suppose that $\mathfrak A$ is an algebra on set X. Let F($\mathfrak A$) denote the smallest $\sigma$ algebra that contains $\mathfrak A$. Then for any subset $A\subset X$, there is a set $R\in F(\mathfrak A)$ such that $A\subset R$ and $\pi^*(R)= \pi^*(A)$. How do I show existence of such R?
Here $\pi^*$ is an outer measure on X.
I thought about using the set $S= \{E\in F(\mathfrak A): E\subset A, \pi^*(E)\ne \pi^* (A)\}$.
Assuming on the contrary that the statement is wrong, S is non empty because it contains X.
Then, I thought of using Zorn’s lemma but it didn’t work. Any chain (partial order inclusion) has X as upper bound so there is a maximal element of S.
But X is also maximal and that doesn’t give me what I want.
But X is also maximal and I don’t know how to use this to prove the statement.
Of course, Zorn’s lemma is not required here to conclude that X is maximal.
you need to say how $\pi^{\ast}$ is defined, it cant just be any random outer measure because then your statement is false. Take the algebra $\mathcal{A} = \{X , \emptyset \}$ which is also the smallest sigma algebra it generates, let $X$ be $\{1,2,3,... \}$ and let $\pi^{\ast}$ be equal to the counting measure. Then $\pi^{\ast}(\{1 \}) = 1$ and there is no $R$ in $\{X, \emptyset \}$ such that $\pi^{\ast}(R) = 1$
typically the way this is proved is by using a definition like $\pi^{\ast}(A) = \inf \{ \sum \mu(B_j) : A \subset \bigcup B_j ; B_j \in \mathfrak{A} \}$, it basically comes right out of the definition
It can be said so. Basically I want $\mu$ to have these three defining properties: 1) $\mu$ is non negative, 2) $\mu$ is sigma additive, 3) $\mu(\emptyset)=0$.
Can maybe someone look at this question? https://math.stackexchange.com/questions/4517798/how-do-i-compute-the-area-of-the-surface-where-z-x2y2-and-0-leq-z-leq-1
Suppose that $\mathfrak A$ is an algebra on $X$ and that $\mu$ is a measure on $\mathfrak A$. Let F($\mathfrak A$) denote the smallest $\sigma$ algebra that contains $\mathfrak A$. Then for any subset $A\subset X$, there is a set $R\in F(\mathfrak A)$ such that $A\subset R$ and $\pi^*(R)= \pi^*(A...
@porridgemathematics This "goodness" is only kinda philosophy, and I would not agree with that. Today we have derived completions which is believed to be "the" correct completion.
A few weeks ago, I wrote a program to plot the number of primes in the intervals $[n^2-n, n^2+n]$ for natural $n$. Each interval contains $n$ odd numbers, and according to the prime number theorem, the number of primes in each interval is asymptotic to $n/\ln(n)$. We can give crude bounds of $n/\ln(n)(1+u/\ln(n)$ for $u \in \{-1,-1/2,1/2,1\}$
Is there a logical formula in the language of linear operators that distinguishes $\omega$ and $\omega+\zeta$
One thing that you could try is expressing "there is/isn't an infinite descending sequence", but can that be expressed in the language of linear orders?
ok, i was thinking something like, exists S, S subset [linearly ordered set] and forall y, y in S -> exists z, z in S and z < y. i presume i can't use "subset" here.
I wonder if there's some sort of "topological first-order language" - then we can ask whether the sphere and the torus are logically indistinguishable. One thing I know is, the proposition "For any two disjoint closed sets, if the complement of each one individually is connected then the complement of their union is connected" is true of the sphere but not of the torus...
Well, I suppose that would be second-order, actually. But I feel like we need to talk about sets for topology. EDIT: There's something called 'pointless topology' (funny name) where we focus on just the relations between the sets and not their elements...
I wonder what sort of language we could set up to allow that? (Things like "The space is simply connected (meaning every image of a loop is the boundary of the image of a disk)" strike me as 'higher-order', somehow.)
the idea there is that because of the linear independence of the (e_i)'s you can deduce from that stuff that a_j (lambda - lambda_j) = 0 for all j, and this implies a_j = 0 except for those lambda_j that happen to be equal to lambda
this came up in a patent i was looking at a while ago, where the applicant and patent office got around it by typesetting a^[complicated] as a [complicated]
For the same proof I sent, I now understand the notation and the last 2 lines however I don't understand what axler means by: "This will imply that the behavior of R on the eigenvectors of T is uniquely determined"
@AkivaWeinberger I didn't even think about it before I wrote that.
I still haven't thought about it. I'm not really interested in trivia or unmotivated problems. Tell my why I should care, and I might think about it, but "What is the growth rate of [f]?", devoid of context, is not a question I find terribly interesting.
igame: i would read that as, "this will imply that the function R is known on the set of eigenvectors for T" and hence (because R is linear and the eigenvectors span) that R is known
i.e. that you know how to compute Rx for x an eigenvector of T
kinda weird to see pleonasm like that in a modern text
(any particular value of a function is 'uniquely determined,' that's what it means to have a function in the first place)
i guess he's also rolling in, the fact that this info is known for R assuming only that R is positive and R^2 = T proves the uniqueness of a "positive square root for T," but it's not really written very well
Although most college courses in the US do almost no proofs because we're teaching engineers.
My policy when teaching the standard science/engineering calculus courses was to do only proofs that give one insight into what's going on, not proofs that are clearly "mathematical masturbation."
Many proofs are written to obscure what's going on; that's not going to convince students that they should learn proofs.
But one should do some derivations/proofs in basic calculus courses. I just think that many college teachers take the path of least resistance and skip them.
