Cauchy Sequences - Complex Analysis

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The discussion focuses on understanding proofs related to continuity and Cauchy sequences in complex analysis. A function is continuous if it satisfies the limit condition involving sequences converging in the complex plane, requiring an epsilon-delta approach for proof. The concept of a Cauchy sequence is clarified as a sequence where terms become arbitrarily close together, indicating convergence in complete spaces. It is noted that every Cauchy sequence in the complex plane converges, and the integers, being a discrete set, also have limits within themselves. A solid grasp of these definitions and theorems is essential for tackling the problems presented.
MurraySt
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Hope someone could give me some help with a couple of problems.

First:

Proof of -

A function f:G -->Complex Plane is continuous on G iff for every sequence C(going from 1 to infinity) of complex numbers in G that has a limit in G we have

limit as n --> infinity f(C) = f(limit as n --> infinity C)

Should I use the definition of the complex limit? Limit proofs always scare me.


Next I need to provide a proof of "Every Cauchy sequence in the complex plane is convergent"

I'm a little shaky on the definition of a Cauchy sequence if someone could provide one in more layman's terms that would be greatly appreciated.


And lastly, were asked to "show that every Cauchy sequence of integers has a limit in the set of integers (Z)

Again I feel with a better understanding of the definition of a Cauchy sequence I would have a shot at getting it.
 
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Hi MurraySt! :smile:

MurraySt said:
Hope someone could give me some help with a couple of problems.

First:

Proof of -

A function f:G -->Complex Plane is continuous on G iff for every sequence C(going from 1 to infinity) of complex numbers in G that has a limit in G we have

limit as n --> infinity f(C) = f(limit as n --> infinity C)

Should I use the definition of the complex limit? Limit proofs always scare me.

I fear that you'll have to do some epsilon-delta things here. You'll need to prove

\forall \varepsilon>0:\exists n_0:\forall n\geq n_0:~|f(x)-f(x_n)|<\varepsilon

Apply the definition of continuity here and the definition of x_n\rightarrow x.

Next I need to provide a proof of "Every Cauchy sequence in the complex plane is convergent"

I'm a little shaky on the definition of a Cauchy sequence if someone could provide one in more layman's terms that would be greatly appreciated.

Do you know the corresponding theorem for the real numbers? I.e. do you know that R is complete? That should help you here.

Intuitively, a Cauchy sequence is a sequence such that the terms lie closer and closer together. It are sequences that should converge in a space that's nice enough. And nice enough means complete here.

And lastly, were asked to "show that every Cauchy sequence of integers has a limit in the set of integers (Z)

Again I feel with a better understanding of the definition of a Cauchy sequence I would have a shot at getting it.

Prove in general that a closed set of a complete space is complete.
 
micromass gave a good intuitive definition, so I'll give a slightly more technical one. A sequence is Cauchy if, given any arbitrarily small positive number epsilon, we can find a point in the sequence after which all terms of the sequence are closer together than epsilon. Formally, for any \epsilon >0 there is some N such that for any m,n \geq N, |s_m - s_n| < \epsilon.

micromass said:
Prove in general that a closed set of a complete space is complete.

This seems like a little overkill for this problem (which is not to say it wouldn't work). The integers are a discrete set, so if we have, say |a - b| < 1/2 for integers a and b, what else can you say about a and b?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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