Recent content by Tedjn

Yes, this works. Here is what I meant by not using s, however. See if this also makes sense as a simplification: We know sup A < sup B. Let ε = sup B - sup A, which is positive. By Lemma 1.3.7, there exists a b ∈ B such that b > sup B - ε = sup A. Thus, b is an upper bound for A.

First you must explicitly mention that Y = X\B ∈ [X]R, which I imagine you might already have done. Once this is done, consider this other Z ∈ [X]R disjoint from B. We know that X\Z ∪ Z\X ⊆ B. What does this say about Z relative to X?

Why do you think c must be complex? All the inner products you are using result in real numbers.

You are essentially correct but here are some thoughts. This element s you claim is any arbitrary upper bound for A. Being any arbitrary upper bound for A there is no guarantee that s is less than sup B (hence epsilon is positive). You would need to argue that such an s exists between sup A and...

Here is a hint: let's say you have some element g that maps to some nonzero integer m. What does gn map to? Is there something wrong that inevitably happens?

Thank you everyone. This has been very helpful. I feel ridiculous for forgetting that of course we can map curves in Rn back to curves in the manifold. I think I have convinced myself then why the vector space structure of Rn induces the vector space structure of TxM. Landau, I did mean...

Hi everyone, On the Wikipedia page for Tangent space there is a definition of the tangent space at a point x using equivalence classes of curves. It mentions that the tangent space TxM is in bijective correspondence with Rn. My first question is simply: is there an easy geometric way using...

A little embarrassing, but I have had very little exposure to anything involving manifolds and am trying to work through these notes over spring break. I will have many questions on even the simplest concepts. In this thread I hope to outline these as I encounter them, and if anyone can help I...

Here's a nice problem. This one is quite fun and something that does not require any higher mathematics. 2010 Putnam Problem B3 There are 2010 boxes labeled B1, B2, ..., B2010, and 2010n balls have been distributed among them, for some positive integer n. You may redistribute the balls by...

I am not certain to what the question is referring, but it could be the following: for every 45 students that travel by bus there are 5 students that travel by car. We can either choose the 45 bus students (method 1) or we can choose the 5 car students (method 2).

Ah, you are trying to integrate by parts again, but I should have been more specific -- it can be done by u-substitution. That is, you let u = something and du = something dx and substitute. If we think about this carefully, then it is true that this is argument is supported by the chain rule.

The integral \int \dfrac{\cos x}{\sin^2 x}\,dx can be done with a substitution. Do you see how?

Saying y(x) has no x-intercept is the same thing as saying that y(x) is never equal to 0. The quadratic y(x) is continuous; if you have studied or are studying calculus, you will know that from the intermediate value theorem that y(x) is therefore always positive or always negative. If you have...

How are you able to conclude that ON = OQ for example? In particular, I doubt that P'Q' hits the origin. This would only be true if (by similarity) PP' = QQ', and there doesn't seem to be any reason why that must happen.

If the series converges, which means the integral you did is finite, it means that log(log x) evaluated at infinity must be finite. This means that the function log(log x) must be bounded at infinity. When a function is always increasing, as in this case, you can picture two ways that the...