Recent content by Scousergirl

Homework Statement Prove that there is no holomorphic function f in the open unit disk such that f(1/n)=((-1)^n)/(n^2) for n=2,3,4... Homework Equations The identity theorem: Let f and g be holomorphic functions in the connected open subset of C, G. If f(z)=g(z) for all z in a subset...

for any arbitrary open cover of [a,b] there is a finite subcover for covering compact but I the question isn't asking to use the definition of compact just covering compact.

Right so I assume U is my arbitrary open cover. Covering compact implies compact but not vice versa right?

The question asks to prove directly that the closed interval is covering compact - U= an open covering of the closed set [a,b] I started by taking C=the set of elements in the interval that finitely many members of U cover. Now I need to somehow use the least upper bound theorem to show...

The normal topology...this is what i came up with...not sure if its right: Claim: they are not homeomorphic Proof: Assume they are. Then There exists a continuous function f from Q to N. Therefore all of the sequences in Q are mapped to a sequence in N preserving limits. But since sequences...

Is Q homeomorphic to N? I understand that there exists a bijection from Q to N but I cannot figure out how this function is continuous and it's inverse is also continuous.

How can you deduce that nad bounded sequence in R has a convergent subsequence?

Prove that the monotone sequence condition is equivalent to the least upper bound theory. I can't seem to get around how to prove that the two are equivalent. (it seems trivial).

Homework Statement Prove: Let (Xn) be a sequence in R (reals). Then (Xn) has a monotone subsequence. Homework Equations Def: Monotone: A sequence is monotone if it increases or decreases. The Attempt at a Solution I know it has something to do with peak points...that is there...

Thank you so much this helps a lot. I guess that the proof would be easiest assuming there is no such s and then show that b-epsilon is thus greater than all s thus giving a contradiction to b being the least upper bound. This problem seems almost trivial now i know how to tackle it. :)

hmmm...I think maybe I am confusing multiple concepts here. The question that I am having difficulty with is how to prove that such an s exists for all subsets of the real numbers. (b-epsilon)<= s <= b. I tried tackling it by contradiction: Assume there does not exist an s in S such that...

Ok, I think I understand the least upper bound part of the question, I just don't know how to go about proving that ANY subset of the real numbers (S) that is non empty and bounded above by b must contain an element s such that (b-epsilon)<= s <= b. 1* is a dedekind cut (the real number 1...

I understand the definitions of cauchy sequences and the cauchy criterion, but I don't see how I can relate them to this problem. I think what I am confused with is what kind of S we are dealing with here. FOr example, doesn't S={1*, b} satisfy these conditions but i don't see how i can use the...

I'm having a little difficulty understanding Epsilon in the definition of convergence. From what the book says it is any small real number greater than zero (as small as you can imagine?). Also, since I don't quite grasp what this epsilon is and how it helps define convergence, I am having...

I'm having a little difficulty understanding Epsilon in the definition of convergence. From what the book says it is any small real number greater than zero (as small as you can imagine?). Also, since I don't quite grasp what this epsilon is and how it helps define convergence, I am having...