Recent content by ystael

This is a mostly correct argument: you have the right basic idea, but a couple of points come across as nonsense or misstated. You say "let p be a zero divisor and q be any other polynomial, then pq = 0". This is false. What it means for p to be a zero divisor is that there exists a...

Start by computing \tilde{u}(\vec{e}_j) for j = 1,2,3. (These are the components of \tilde{u} relative to the basis \{\vec{e}_1, \vec{e}_2, \vec{e}_3\}, or more properly speaking, relative to the corresponding dual basis.) Then expand the new basis vectors \widehat{\vec{e}_j} in terms of the...

You have complete freedom to choose both the set and the topology you give it. So one strategy you could try is to make your underlying set very large -- say, a large number of copies of some familiar space. Alternatively you could try starting with a familiar set like \mathbb{R}, but giving...

For part (a), you don't need to do a calculation to prove that A and B are normal in G; you could simply observe that G is abelian so all its subgroups are normal. Also, your proof that AB = G has the right idea but is phrased wrong. You need to put it in the form: given g = x + iy \in G, I...

For the first part, the problem is likely that you didn't show why the set of points where (f_n) converges should be a Borel set. It's not completely obvious.

Suppose \omega is a volume form on M (that is, a nowhere zero m-form). Think about how you can use \omega to convert a vector field X which is everywhere linearly independent from TN, into either a volume form for N or a smooth frame field for N. (By "frame field" I mean a family of m - 1...

I assume you mean that \{0, 1\} is to be given the discrete topology and then \{0, 1\}^\mathbb{N} gets the product topology based on that. Remember that the basic open sets of the product topology on a product \textstyle\prod_\lambda X_\lambda are the the sets \textstyle\prod_\lambda...

Notation in this subject varies widely. What is \tau_1(M)? Do you mean to say that X is a smooth vector field on M? (that's what I guess from the problem) Also, what are you using for a definition of 'orientable'? There are many ways to approach the definition.

Your definition of "convex set" is wrong; there is no function involved. A set S (in some real vector space V) is convex if, whenever x, y \in S and 0 \leq a \leq 1, then also ax + (1 - a)y \in S. Once you correct that, if you find yourself working too hard, you're doing something wrong...

Your explanation is sufficient: a function is (complex-)analytic at a point only when it is (complex-)differentiable on an open neighborhood of that point, and since your function is (complex-)differentiable only on a line, it is not analytic at any point. As for the implications of the...

Given any subset H \subset G, how would you attempt to prove that it is a subgroup of G? What properties of H would you attempt to verify?

Incidentally, there is a better way to make binomial coefficients than (^p_k), which will always produce "small" binomial coefficients and won't always place the arguments gracefully with respect to the parentheses. You can write \binom{p}{k} to get \binom{p}{k}.

1. How do you use differentiation to determine whether a function defined on a subset of \mathbb{R} is monotonic? Can you find a function whose values at the natural numbers give the terms of your sequence? 2. "Eventually monotonic" just means "monotonic after some large index N". So take...

A sequence in X is a function n \mapsto x_n from \mathbb{N} to X; it has infinitely many terms regardless of the cardinality of X. The range or image of the sequence is the set \{x_n \mid n\in\mathbb{N}\} of values taken by the sequence; it is a subset of X. If X is a finite set, then the...

Think about whether the sequence takes only finitely many or infinitely many distinct values. Also, remember that a sequence can converge to more than one point in sufficiently bad topological spaces.