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When we say a sequence is monotonic, does that imply the sequence is Real Number Sequence? And other propositions about monotonic, all real-valued?

When I see some mathematical analysis books, sometimes they talk about some properties/facts like convergence of sequences in a metric space, for example, real field and complex field, sometimes the results are the same.

However, when I see some propositions about "monotonic" are limited within real numbers.

Maybe they write for simplicity of the theorem?

For example:

Definition: A sequence $s_n$ of real numbers is said to be monotonicaly increasing if $s_n\leq s_{n+1}$.

Complex sequences may also have something monotonicity?

Basic fact I know is that we cannot compare two complex numbers, maybe we should use partial relation? norm of complex number(this is a real number)? dictionary order?

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User19912312
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    Well, it meast at least that we're working with a sequence in some space with some order in it...]\ – DonAntonio Jul 25 '13 at 08:46
  • It seems to me this question is related to a programming task, is this correct? – David Jul 25 '13 at 08:52
  • @DonAntonio How about complex field. complex sequences – User19912312 Jul 25 '13 at 09:03
  • @spuorg-imes: You can place a (non-strict total) order on the complex numbers. For example, take $N: a+ib\mapsto a^2-b^2$. Then $x\leq y\Leftrightarrow N(x)\leq N(y)$. Also, by the Well-Ordering Principle, you can place an order on any set. So long as you are not bothered by the Axiom of Choice. (Although such an arbitrary order won't necessarily be preserved by the field operations.) – user1729 Jul 25 '13 at 10:02
  • @user1729 is this ordering of the complex numbers useful? I find it strange as it does not even extend the ordering of the reals. – Ittay Weiss Jul 25 '13 at 10:58
  • @IttayWeiss Sorry, a typo slipped in. It should be $a+ib\mapsto a^2+b^2$. It is useful because $N(x)N(y)=N(xy)$. I do not actually know if it has any explicit uses in the general context of $\mathbb{C}$, but if you are working over the Gaussian Integers (so $a, b\in\mathbb{Z}$) then this lets you factorise elements. It also works if you replace $i$ with $\sqrt{n}$ for $n$ an arbitrary integer, and it becomes $a+\sqrt{n}b\mapsto a^2+nb^2$. Again, if $a, b\in\mathbb{Z}$ you get a ring and this can be used to factorise elements. – user1729 Jul 25 '13 at 11:29
  • @user1729 ok, but now it's just the absolute value of $x$ (squared) and still the ordering you define does not extend the ordering of the reals. I fail to see the connection of normed rings to the question. – Ittay Weiss Jul 25 '13 at 11:36
  • @IttayWeiss In the comments the OP asked, in response to DonAntonio's comment about a space with order "How about complex field. complex sequences". I interpreted this as "can you put an order on the complex numbers?". So what I said puts an order on the complex numbers. – user1729 Jul 25 '13 at 11:45
  • @user1729 but is this a useful ordering? is there any useful ordering on the complex numbers to speak of? – Ittay Weiss Jul 25 '13 at 11:49
  • It is useful (for some value of useful) because it preserves multiplication. I do not know if the complex numbers can be ordered in such a way which (a) respects some facet of the field structure and (b) preserves the usual ordering of the real numbers. Sorry. (Although I believe that the complex numbers do not have a strict total ordering, that is, there must exist $a\neq b$ such that $a\leq b$ and $b\leq a$). I wasn't trying to be "deep" with my comment, I was just trying to answer the OPs comment off the top of my head...which is why it was a comment, not an answer... :p – user1729 Jul 25 '13 at 11:54
  • @IttayWeiss (Oh, I should say, you can place a total order $\mathbb{C}$ in a way compatible with the usual ordering of $\mathbb{R}$, but this is not compatible with the field operations. You can do this using the lexicographical ordering, where $a+ib\preceq c+id$ if either $a\leq c$ or ($a=c$ and $b\leq d$). EDIT: This ordering preserves addition, but not multiplication.) – user1729 Jul 25 '13 at 12:02
  • @user1729 Hi, I think you can organize your comments an one answer, that will be also useful. – User19912312 Jul 25 '13 at 12:16

3 Answers3

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To speak of monotonicity one needs to have a notion of order. As long as the set of objects you are considering your sequences to come from is ordered in a reasonable way, you can speak of monotonicity. It is common to consider partially ordered sets, or simply posets. A poset is a pair $(S,\le )$ where $S$ is a set (which can be any set at all) and $\le $ is a transitive, reflexive, and anti-symmetric relation on $S$. In the context of posets, so for sequences of elements from a poset, monotonicity makes perfect sense.

The real numbers are ordered by the usual meaning of $x\le y$. However, the complex numbers are not ordered in any natural useful way, so we don't speak of monotone sequences of complex numbers. An example of a poset which is useful in the context of analysis is the poset of, for instance, all functions $f:\mathbb R \to \mathbb R$. This poset is ordered by $f\le g$ precisely when $f(x)\le g(x)$ for all $x\in \mathbb R$. Then you can speak of monotone sequences of functions.

Ittay Weiss
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At the request of the OP, I have gathered my comments regarding the ordering of the complex numbers onto this answer. This answer basically gives two different orderings of the complex numbers, and wonders what the best is we can do to order $\mathbb{C}$. These orderings mean that we can talk about monotonic sequences in $\mathbb{C}$.

Ordering 1: Norms. A norm is a function which assigns elements in a ring a real number, $N: R\rightarrow\mathbb{R}$, such that $N(a\cdot b)=N(a)\cdot N(b)$. This can be made into an ordering by saying $a \preceq b\Leftrightarrow N(a)\leq N(b)$. For the complex numbers, we can define $N(x+iy):=x^2+y^2$. Note that this is not a strict total order as, for example, $N(1)=N(-1)$, and indeed that this does not preserve the usual ordering of the real numbers. Note also that, in general, $N(a+b)\neq N(a)+N(b)$.

Ordering 2: lexicographical ordering. Define $x_1+iy_1 \preceq x_2+iy_2$ if and only if either $x_1\leq x_2$ or ($x_1=x_2$ and $y_1\leq y_2$). Note that this is a strict total order which preserves addition and the usual order of the real numbers. However, it does not preserve multiplication.

We then have a question: Can we give $\mathbb{C}$ a total ordering which is compatible with the operations of the field? Well, no, it turns out we cannot. This means that either addition or multiplication must fail. Clearly Ordering 2 is the best we can do for addition: It preserves addition as well as the underlying ordering of the real numbers. Ordering 1 is not quite so good for multiplication. We therefore have the following question,

Question: Does there exist a total ordering of the field of complex numbers which preserves multiplication and the underlying ordering of the real numbers?

user1729
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A monotonic sequence $(x_n)$ is a sequence for which $\exists n_0$ such that $\forall n \ge n_0$ we have that either $x_n \ge x_{n+1}$ or $x_n \le x_{n+1}$.

Generally, it should be apparent from the context whether the sequence is real-valued or not. A real-valued sequence will usually be denoted $(x_n)_{n=1}^\infty$, whereas sequence made up of natural numbers will usually be denoted $(n_k)_{k=1}^\infty$.

But this is only a convention; it does not guarantee anything. You should always consider the context of the problem.

David
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    I feel that this doesn't quite answer the question - when can it be not real-valued? As natural numbers are also real numbers, you seem to be implying that they are always real-valued... – user1729 Jul 25 '13 at 09:59
  • Consider a sequence $(a_k){k=1}^\infty$ of natural numbers and a sequence $(x_n){n=1}^\infty$ such that $\forall n: x_n = n^{a_n}$. Now exponentiation can be understood as simply multiplying $n$ as many times as $a_n$ equals to. If the sequence $(a_n)_{n=1}^\infty$ was real-valued instead, then the problem would demand much deeper understanding of what $a^b$ means. So it does make sense to distinguish between natural and real numbers, despite the fact that natural numbers are real also. That is why context of the problem is important. – David Jul 25 '13 at 10:33
  • @user1729 I see what you mean now. The question had been extended, so it is now clear. Thanks for pointing it out. – David Jul 25 '13 at 10:35
  • Ah, I hadn't realised it had been edited either! You answer certainly makes more sense in the context of the original, un-edited question. – user1729 Jul 25 '13 at 10:38