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You are focussing on words rather than the meaning of words. I suggest you first try to understand how the facts you state imply that O(n) is not connected. * connectedness is a property of a topological space, not of a group or a set * O(n) is partitioned into the 1 and -1 fibers of the...

Well, it consists of all elements of the form r(x^2+1)+s(x+1) with r,s ring elements. So it cannot contain, e.g., the element x.

You cannot decide whether a given set is a field. You can decide whether a given ring is a field. After all, a field is by definition a special kind of ring, namely a commutative ring in which every nonzero element is invertible. In particular, a field has no zero divisors (i.e. a field is a...

A subnet is not so hard to define, although there are slightly different ways which give the same nice properties: a set is compact iff every net in it has a convergent subnet, a net converges to x iff every subnet converges to x, and the like. I was just watching the last hours of Day 6 of 24...

Yes, that's what I meant as well (larger in the sense of inclusion). I am sorry, that is of course what I meant (I wrote mistakinly 'discrete' instead of 'trivial'). The finer=larger=stronger the topology, the less convergent nets. I think I understand what you are saying. You seem to be...

A topology is uniquely determined by its convergent nets; in general sequences do not suffice. So yes, if you start with a collection sequences with a specified limits, and demand that these are all the convergent nets, then you get a topology whose closed sets are the sequentially closed...

No it doesn't. <x,0>=x1x2.

You just said it yourself. View U(n) as subspace of R^{2n^2}. You know the open sets of R^{2n^2}, hence of every subspace of it.

Ah, so by 'induced on [0,1]' you don't mean the subspace topology. Could you define the fine topology for me? Is it the initial topology on X w.r.t. all convex functions X->R?

I think he wants to show that this equivalence relation (or rather, 'being tangent') is chart-independent, i.e. that if two curves are tangent in some chart, then also in any other chart. This follows directly from the chain rule.

What do you mean by 'equivalent' topologies? I am not familiar with the fine topology, but if by equivalent topologies you simply mean 'the same topology' (i.e. the same open set), then it is of course a tautology.

No, the author was correct: he did not mean that Nx \cup Ny is a filter base, but a filter subbase. A filter subbase is a collection subsets, such that every finite intersection is non-empty. Then the collection of those sets which contain such a finite intersection forms a filter, it is the...

You have proven that 1/0 does not obey the usual rules of computing with fractions. This should not be too surprising, and doesn't have much to do with 'infinity'.

I agree with DH. Also see here.

Mod out by [G,G], its commutator subgroup. E.g. see here.

Three and a half years ago JF started the same thread: https://www.physicsforums.com/showthread.php?t=203040 [Broken]. Personally, I don't understand anything what that guy is saying. His mastery of the English language is also kind of off-putting.

There still seems to be miscommunication. You probably made a typo in the last sentence of your previous post, where you say you want to show that g is holomorphic. So my understanding is that you DO want to prove that, for every rho, I_rho is holomorphic. Yes? Please be clear about this. If...

The OP still seems to ask something else than the above: So you agree that I_\rho:\mathbb{C}\to \mathbb{C} is holomorphic for every \rho? But you want to show that in fact \mathbb{C}^2\to \mathbb{C} (\rho,s)\mapsto I_{\rho}(s) is holomorphic?

I was suggesting that the claim of your notes implies that the hypotheses of Morera's theorem are satisfied. Forget about your I_\rho for a moment; we are trying to prove the following: Suppose f:\mathbb{C}\to \mathbb{C} has the property that \forall R > 0, \exists M > 0 such that \forall s...

I may be mistaking, but doesn't this follow from [URL [Broken] Theorem[/url]? If f is bounded on every disc, then I guess its integral over every closed curve must vanish.

I think we can generalize this to any category: if * is a terminal object and A is any object, then the product of A and * is canonically isomorphic to A. We just show that A has the universal property of the categorical product of A and *, which we know to be unique up to unique isomorphism...

Proving my hint is the most work. Once you've done that, use the homeomorphism \mathbb{R}/2\pi \mathbb{Z}\to S^1 and the fact that a surjective continuous map sends a dense set to a dense set. To prove my hint, prove the generalized statement: \mathbb{Z}+r\mathbb{Z} is dense in...

It should be only two lines, so you might want to see if you can shorten your reasoning.

Hint: \mathbb{Z}+2\pi \mathbb{Z} is dense in \mathbb{R}.

I don't have much experience with these little and big O things, but it seems you are using a different definition than the one on wikipedia. It says: f(x)= O(g(x)) as x->\infty iff there exists a K>0 and x0 such that |f(x)|\leq K|x^2| for all x>x0. So you just have to find ONE K>0 such...

Both answers are correct. It doesn't matter at all how sharp your estimates are. That's the point of big-O-notation: you only care about some limiting behaviour. Preno's approach is equally valid, but I suspect that you haven't seen (or aren't allowed to use) this criterion.

Yes, I believe I explicitly said that: I will think about your other two questions later :p May I ask: what do you think about Rudin's book, besides it not being suitable for a beginner? Have you (been forced to) use(d) it in a course? Have you done many exercises? Do you use it as a reference?

It seems trivial. If (a_n) is a Cauchy sequence w.r.t. the graph norm, then both (a_n) and (Aa_n) is a Cauchy sequence w.r.t. the usual norm. Hence they converge to some a and b, respectively. As A is closed, we get that a is in D(A) and b=Aa.

I am sorry if my question annoyed you. In fact I am familiar with the mathematical terminology; I was under the impression some people in this thread use different terminology in the context of physics. By the way: Like I said, for me hermitian and self-adjoint are synonyms (although I...

For what it's worth: I find Rudin's book very good; the exposition is beautiful and I often look up stuff in it. So I use it mainly as a reference. I don't know how good it serves as a "beginner's book". I haven't done any of the exercises, but scanning over them I see there are some darn hard...

To me, Hermitian and self-adjoint are synonyms. Could you give your definitions? Perhaps one of them means symmetric in the sense that <Ax,y>=<x,Ay> for all x,y in the domain, which need not be the whole Hilbert space? And the other want the domain to be all of the Hilbert space? Or want...

Why on Earth would you erase the question? The point of a forum is that you can have discussions and read/refer to them back.

So have you found a function continuous at the origin only?

Yes. If R is any ring with two ideals I, J, and q:R\to R/I is the quotient map, there is an obvious map R/(I+J)\to (R/I)/q(J) namely r+I+J\mapsto q(r)+q(J) It is easily shown to be a (well-defined) ring isomorphism.

You probably mean "injects in" instead of "is contained in". Anyway, could you share you understanding of this fact? Because if you understand this, then the other way around is not so hard.

You can't really say that some subject is of a "higher level" than another. There are people who do research in linear algebra. Obviously this is not about the stuff you learn in a typical introduction to linear algebra. You might say something like "high school level", "bachelors level"...

I outlined a proof here, which shifts the hard work to some lemma (which does not require AOC itself).

Sorry, I hadn't read your other question. Yes, I agree with you! I opened my pdf file of Kreyszig and I don't understand why he's doing it either: the series (8) on page 390 already converges absolutely for |\lambda|>||T||. In fact, he's quoting Theorem 7.3.4, in which he's quoting Theorem...

No. He's using the following result (here c and \mu are scalars, A a bounded operator between Banach spaces): Lemma: If \sum_{k} (\mu A)^k converges in norm, then \sum_{k}(cA)^k converges absolutely for |c|<|\mu|. Proof: Let M=\sup_k\|(\mu A)^k\|. Note that M is finite: \|(\mu...

@totentanz: How can one try 50 textbooks?

To actually prove this, one of course needs to make the notion of 'natural isomorphism' precise, which has been done in category theory. See here for a proof of the fact there there is no natural isomorphism between the identity functor and the dual functor.

In short, I would describe them as 'conceptual inverses'. I found this summary helpful when while struggling with the concept myself. Good examples to keep in mind are the left adjoints to a forgetful functor, which are the functors that assign to a set the free object on that set: group...

Right. So the order on N is definitely not dense. And then there's the term dense-in-itself, meaning 'containing no isolated points' (which of course depends on the superset you're considering.)

Or simpler, the 1x1-matrix (a) has inverse (1/a), and these have norms a and 1/a, respectively :p In general, it's good advice to test statements in functional analysis in the easy case of finite dimensions first.

What kind of 'denseness' do you have in mind here? Some measure theoretic concept?

There isn't really a definition of a set. It's an abstract thing (quoting Jarle), on which most of mathematics is built. Just think of it as "a bunch of stuff" in the most general sense. The 'objects' of a set (or 'elements' or 'stuff it's made up of') can be anything. (Well, to some degree...

The pointwise limit of (real-valued) measurable functions is measurable. That is one of the most basic and important results in (elementary) measure theory. If that's not in the book you're reading, then I'm pretty sure that's not a book about measure theory :) E.g. see here.

The most concise proof is using the most elegant, coordinate free definition. Namely if L is an endomorphism of the n-dimensional vector space V, then the induced map \hat{A}:\bigwedge^nV\to \bigwedge^nV is a linear map between 1-dimensional spaces. The scalar by which it acts is the...

Well, it's true if you mean the right thing: the Cauchy-Riemann equations are equivalent to \frac{\partial f}{\partial\bar{z}} = 0 where \frac{\partial}{\partial\bar{z}}= \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right)

Let M be an oriented m-manifold and \alpha an m-form on M with compact support. Given an oriented atlas (U_i,k_i), you take a smooth partition of unity (f_i) subordinate to it, and define \int_M\alpha:=\sum_i\int _{k_i(U_i)}(k_i^{-1})^*(f_i.\alpha). It can be shown this is independent of...