Recent content by g1990

Homework Statement Let X be an ordered set where every closed interval is compact. Prove that X has the least upper bound property. Homework Equations X having the least upper bound property means that every nonempty subset that is bounded from above has a least upper bound, in other...

okay- thanks!

well, I don't know how I can prove that there is no bijection. Again, I know I can find a bijection from Z+ to {1,2}xZ+, but listing the other set in order would entail TWO infinities. It would be (1,1),(1,2),(1,3)... (2,1),(2,2),(2,3)... is that enough to say that Z+ x {1,2} cannot be bijected...

oh yeah- Z+ is the set of positive integers

Homework Statement Both {1,2}x Z+ and Z+ x {1,2} are well-ordered in the dictionary order. Are they of the same order type? Why or why not? Homework Equations To be of the same order type, we must be able to construct a bijection that preserves order, that is, x<y => f(x)<f(y)...

Homework Statement 1. f(z) is a function that is analytic on all of the complex plane, and mod(f)<=mod(z). Prove that f=cz. 2. f(z) is analytic on all of the complex plane, and mod(f)<= sqrt(mod(z)). Prove that f is constant Homework Equations Liouvilles thm: the only bounded entire...

nvm- got it! f'(z)=du/dx+idv/dx. Taking these and the real and imaginary parts, we can get the CR eqns for f' using the CR eqns for f and the fact that the mixed partials are equal.

I have to solve this problem using only calculus. I can't use any theorems like using Taylor series. Maybe I can say that it is harmonic?

Homework Statement Let g(z) be an analytic function. I have to show that g'(z) is also analytic, using only the CR eqns. I am given that the 2nd partials are continuous Homework Equations let f(x,y)=u(x,y)+iv(x,y) CR: du/dx=dv/dy and du/dy=-dv/dx continuous 2nd partials...

Okay, so if I let N>2|w|, then my argument will work?

Oh, maybe I need that N>|w|?

Lets see, if |f|>N, then |g|=1/|f-w|. By the triangle inequality, |f-w|>||f|-|w||>||N|-|w|| I could let N=S|w|, so that then |f-w|>||N|-|w||=||S||w|-|w||=|S-1| Then, |g(z)|=1/|f-w|<1/(S-1). I don't get why having |w| in the bound of g(z) is a problem.

Alright, so pick an N. Then, when |z|>M for some value M, |f(z)|>N. Thus, when |z|>M, |g(z)|<1/|N-w|. When |z|<=N, then |g(z)| attains a maximum (on the boundary) because g is continuous and the disc |z|<=N is compact. Call this maximum P. Then, g(z) is bounded by max(P, 1/|N-w|). But, g is...

Lets see... the modulus of f is definitely bounded from below by 0. If I could get it to be bounded from above that would be a contradiction b/c of the maximum modulus principle. I'm not sure how to do that though. You could say that if z if large, f is above a certain value so that g is below a...

lets see, if f(z) never equals w, then g(z) is analytic everywhere on C. Then, as z approaches infinity, g(z) approaches 0 because f(z) approaches infinity. I'm not sure why this is a contradiction.