A question on completeness condition from Ahlfors' book

Question

I'm sorry I can't think of any better title for this question. Anyways my question is : In Lars Ahlfors-Complex Analysis , the author said that Real Field satisfies completeness condition . He said , for an increasing and bounded sequence : α₁<α₂<α₃....<α_n... and where an assumed real number B is bigger than αn for all n , there will exist a number A=limit of α_n when n tends to infinity with a special property : given any e > 0 there will exist a natural number n₀ such that A - e < α_n < A for all n >n_o. I don't understand this . To be specific , my questions are -

1. Why did he leave the sequence without terminating ? Couldn't he write the equation like this :α₁<α₂<α₃....<α_n...<B ? Or is B an arbitrary number from the real field which is not part of the sequence ?
2. And why do I need n to be near infinity ? I could take a quite small sequence where n is countable ..right? Then shouldn't I write : A - e < α_n ≤ A ?
   
   Thanks for your time.
   
   Edit: Off topic question : x²+1=0 has no roots in Real Field because of Order Relation as far as I understood . But in Complex Field there exists two roots for the equation . So does the complex field have Order Relation?

Answer 1

As already mentioned in the comments, a sequence $\alpha_n$ is called bounded (from above) if there is a constant $C$ such that $\alpha_n\le C$ for every $n$.

To your question 1: The notation is a matter of style. However, $B$ cannot be an element of $\{\alpha_1,\alpha_2,\ldots\}$, because the sequence is strictly increasing and thus every $\alpha_n$ is necessarily strictly smaller than $B$ (for if $B\le\alpha_{n_0}$ for some $n_0$, then $B<\alpha_n$ for every $n>n_0$, contradicting the definition of convergence).

Your question 2 makes no sense: Of course, $n$ is taken from a countable set, namely from the set of natural numbers (this is tacitly understood by the term “sequence”). But the other part of the question has nothing to do with countability: It is not possible to have $B-\varepsilon<\alpha_1\le B$ for every $\varepsilon>0$ (unless $\alpha_1=B$), and similarly, for any other fixed $n$, it is not possible to have $B-\varepsilon<\alpha_n\le B$ for every $\varepsilon>0$. On the other hand, if $B$ was chosen “correctly” (recall that completeness of $\mathbb R$ means that such a choice is possible), then for every $\varepsilon>0$ there does exist some $n$ such that $B-\varepsilon<\alpha_n\le B$. Necessarily, the smaller you choose $\varepsilon>0$, the larger $n$ has to be, because $\alpha_1<\alpha_2<\ldots$.

Note also, that completeness of $\mathbb R$ does not only mean that the limit $B$ exists for some bounded increasing sequence, but actually that every such sequence has a limit. To get a feeling that this is a non-trivial property, consider the particular sequence $\alpha_1=1$, $\alpha_2=1+\frac1{2^2}$, $\alpha_3=1+\frac1{2^2}+\frac1{3^2}$, $\ldots$, $\alpha_n=1+\frac1{2^2}+\cdots+\frac1{n^2}$ and the similar sequence $\beta_n=1+\frac1{2^3}+\cdots+\frac1{n^3}$. It can be shown that both sequences are bounded from above. Hence, the completeness of $\mathbb R$ implies that they converge to some limit. For the first sequence, Euler has shown that the limit is $\frac{\pi^2}6$ (Basel problem) (BTW, by implicitly using complex analysis), but about the limit $B$ of the second sequence still almost nothing is known (a relatively new famous result about this particular limit is that it is an irrational number, and one can get some numerical estimates, but that's it, essentially). Anyway, although we cannot calculate the limit yet (and maybe we will never be able to), we know at least that the limit does exist, because $\mathbb R$ is complete.

That being said, I would recommend that you read a text book on real analysis where all such things are well explained before you start with a text book on (advanced) complex analysis. I am sure you will need such basics from real analysis in much more places on this book. Moreover, the book of Ahlfors is somewhat aged meanwhile. This does not mean that there is something wrong with it, but it is likely that some notations are hardly used anymore which might confuse you when you eventually read a more modern treatment of complex analysis. (Though knowing older notations can sometimes also be an advantage, but you should be aware of it.)