Recent content by shoescreen

Hi, I have been reading about metric spaces and came across an elementary property that I am having difficulty proving. A quick search on these forums and google has also failed. Given a metric space with distance function d, and an increasing, concave function f:\mathbb{R} \rightarrow...

If by F(x) you mean an anti-derivative of f(x), you are correct.

Actually, applying geometric ideas to metric spaces is a very active area of research. Although you don't get a very good analogue of angle per sey, you are able to equip a fairly robust notion of geometry. The notion of (ricci) curvature, for example, is crucial for studying geometries in...

Hello, I am learning about smooth manifolds through Lee's text. One thing that I have been pondering is describing manifolds such as |x| which are extremely well behaved but not smooth in an ordinary setting. It is simple to put a smooth structure on this manifold, however that is...

Hello all, I've recently used a property that seems perfectly valid, yet upon further scrutiny I could not come up with a way to prove it. Here is what I would like some help on. Given two sets X and Y and functions f and g mapping X into Y, with the property that f is injective and g is...

I've been reading a complex analysis book which had an example showing \sum^\infty_{n=1}1/n \cdot z^n is convergent in the open unit ball. I'm now looking at the case when |z| = 1. Clearly z = 1 is the divergent harmonic series, but i know this series is in fact convergent for all other |z| =...

so if I use |a_n - A| < f(A,B) |b_n - A| < g(A,B) I can split up each inequality and ultimately end up with something like a_n > -f(A,B) + A b_n < G(A,B) + B so a_n - b_n > -f(A,B) - g(A,B) + A - B so if i want to show a_n - b_n > A - B, the function -f + -g must be positive. But this is...

Well I was asked to prove the this by contradiction before and i was able to do it with (a-b)/2 and i still don't see how to do it directly :( EDIT: hah i think i got it! EDIT: just kidding...

i should have mentioned earlier, the problem statement explicitly states do not prove by contradiction.

Homework Statement Given the sequence {a_n} converges to A and {b_n} converges to B, and a_n <_ b_n for all n>_ n*, prove A <_ B Homework Equations x+ epsilon < y for every positive real epsilon, implies x <_ y A - B = (A - a_n) + (b_n - B) + (a_n - b_n) The Attempt at a Solution...

I take back what i said before, the inequality is in the correct direction. It's frustrating, I had all of this before yet trashed it all because of that inequality sign. Ugh, guess i wasn't in math mode. Thanks for the help!

hmmm, but the inequality is in the wrong direction. If want to compare |a_n - A|\|sqrt(a_n) + sqrt(A)| with a smaller sequence, I need something bigger in the denominator, and since root(A) <_ root (a_n) + root(A), I can't guarantee anything about the original sequence.

Homework Statement given the sequence {a_n} converges to A (non zero), show sqrt(a_n) = sqrt(A) Homework Equations The Attempt at a Solution I've tried to expand |sqrt(a_n) - sqrt(A)| as |a_n - A|\|sqrt(a_n) + sqrt(A)| since that gives me the numerator to work with, but I can't...

Hello all, I have been thinking of a way to prove divergence of a sequence that should work, but can't move past one road block. Here's the idea, given a sequence a_n, say that i know for any consecutive numbers in the sequence, |a_(n+1) - a_(n)| > d, where d is a constant. Now this...