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Is there such a thing as consecutive complex numbers? I thought that this is true according to Zermelo's theorem that states that every set can be well-ordered.

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  • $\begingroup$ No, there aren't. The complex numbers cannot be given a total order in the way the real numbers can. Zermelo says that those numbers can be well ordered, but I doub't anyone has the faintest idea how to achieve that...not even with the real numbers! $\endgroup$ – DonAntonio Oct 14 '16 at 16:00
  • $\begingroup$ "How to achieve that" requires some version of the Axiom of Choice. In particular there is no explicit construction of such a well-ordering. $\endgroup$ – Robert Israel Oct 14 '16 at 16:02
  • $\begingroup$ Every set can be well-ordered in lots of ways. That doesn't mean there is one way that is useful to call two "consecutive." Any finite set can be well-ordered, but it doesn't make sense to say that two pennies are consecutive. $\endgroup$ – Thomas Andrews Oct 14 '16 at 16:14
  • $\begingroup$ "but it doesn't make sense to say that two pennies are consecutive" um, why not? I take my pennies out and randomly call them Bob, George, Fred and John and randomly declare Bob < George < Fred < John. Doesn't it make perfect and complete sense to say George and Fred are consecutive? $\endgroup$ – fleablood Oct 15 '16 at 1:07
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The coins in my pocket are a finite set, so they can be well-ordered in a constructive way, but it doesn't make sense to talk about "consecutive coins in my pocket" until someone gives a useful definition.

So if Sheila takes the coins out of my pocket and places them left-to-right in a line (well-ordering them) then we can talk about consecutive coins.

The problem with well-ordering the complex numbers, or even the real numbers, is that:

  1. We can't do it. We can say "out there in the ether, there is a well-ordering of this set." But we can't find a computable definition that we can use. Basically, Sheila can't take these and give me a single well-ordering that we can talk about. We can say "given that we have a well-ordering, here's what we can do with it..." but it is, in some way, not concrete. (We actually know that it isn't possible to do this, because there are models of set theory without the axiom of choice where such a well-ordering does not exist. The axiom of choice is an appeal to existence without the ability to specifically "find" it in some constructive way.)

  2. There isn't a useful well-ordering of the complex numbers that elicits any new knowledge about the complex numbers, because the desire is for an ordering on an algebraic structure to interact usefully with that structure. There also is not a useful well-ordering of the real numbers - that is, a specific well-ordering that gives us any insight into the real numbers beyond an understanding of the set's cardinality.

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  • $\begingroup$ Nice explanation. $\endgroup$ – User Oct 15 '16 at 7:08
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Every set can be well-ordered (assuming the well-ordering principle), but any such ordering will probably not be compatible with any of the meaningful algebraic or topological structure of the complex numbers, so it's not really useful for studying them (ignoring for now the fact that the well-ordering principle tells us nothing about how to construct one). Look up the definition of an ordered field-- it is a theorem that the complex numbers are not an ordered field, i. e. there is no ordering on them that satisfies the definition. There is also no order on the complex numbers such that the order topology is homeomorphic to the standard Euclidean topology.

Note that I'm not just talking about well-orders here, but orders in general.

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  • $\begingroup$ The point about topology is clear: removing a point that is not the first or last disconnects a set in the order topology, while the complex numbers with one point removed is still connected in the usual topology. $\endgroup$ – Robert Israel Oct 14 '16 at 16:04
  • $\begingroup$ Right, thanks. I edited my answer. $\endgroup$ – Vik78 Oct 14 '16 at 16:07
  • $\begingroup$ Not does the theorem not say how to construct such a well-ordering, but we know that there is no constructive well-ordering. $\endgroup$ – Thomas Andrews Oct 14 '16 at 16:20
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"Consecutive" implies there has been some order imposed on a given set. And for that, yes complex numbers can be ordered, albeit the question is what kind of order would that be and what do we want it to be? Two important properties for complex numbers is its topological property and its algebraic properties. Those together is what defines complex numbers and what we care about, so the question is more

Does there exist an ordering for $\Bbb C$ that is compatible with the topology and algebraic structure of $\Bbb C$?

For that it's a resounding, "No". As a field the requirement to order it, while respecting the field structure, is that the sum of squares cannot equal to $0$ or $-1$ and in complex numbers we have $1^2+i^2=0$ and as such they cannot be ordered.

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Theoretically it is possible to give well ordering to the set of complex numbers. But is difficult to exhibit it. Usually, each order(total order) on a non empty set X induces a topology on X and we can have the corresponding product topology on XxX and discuss the continuity of various binary operations on X. This enable us to define topological group, ring, field etc. For example, in the set of complex numbers we can give a total order namely dictionary order in which to compare two complex numbers we first compare their real parts and only if they are equal then we compare their imaginary parts. This order induces a topology (order topology) on the complex plane but the usual operations like addition, subtraction, multiplication, division of complex numbers are not consistent with this topology (in the sense that the operations are not continuous). By this reason, this order is rarely discussed except in the case of some counter examples in connection with some topological properties.

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