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Is there a natural topology on the set of open sets?

Do you need topology or algebra to come to a conclusion about the cardinality of a set?

  • For example, suppose you want to show R (the reals) is uncountable. One famous proof is based on Cantor's diagonal reasoning. But to use it you must know every real number has a decimal representation, actually you have to work not on R but on the field (R, +, *). You must know there are relationships between real numbers. Another proof is based on a sequence of nested closed intervals. It doesn't require algebra, but you must know what the closed sets of R are, so you need a topology defined on R. This proof won't work if, instead of the topology induced by the Euclidean metric, you pick the discrete metric. So, my question is: Does the cardinality of a set depend on topology, or on algebraic structures, or on the relationships between its elements or on something other than the pure nature of such elements? I think so, because if the only information you have about a set A is that A is a set, then I don't see how anyone can affirm anything about its cardinality. Thank you.

  • Answer:

    There is something that needs to be pointed out here: all topological and algebraic notions can be enterely formalized within a formal Set Theory. This implies that, for determining the cardinality of a given set (if it can be determined: it's possible to show that the most common Set Theory, ZFC, is too weak to decide several "natural" questions about sets, but this is not relevant for the present matter, because no amount of Topology or Algebra will settle them either), Algebra and/or Topology (or any other common mathematical field) is not, strictly speaking, necessary, for those are just abbreviations for set-theoretical notions. Of course, no one will attempt to formalize any proof of, say, the uncountability of the reals entirely within ZFC (after all, medical costs are high); instead, we use abbreviations that we know will do the job. But let it be noted that all different proofs of this fact (Cantor diagonalization, Cantor's nested intervals, the impossibility of a surjection that maps N onto R, etc.) may be done entirely from the axioms of ZFC and any formal system of first-order logic (by the way, Cantor proved this before having a rigorous definition of the reals). So, the impression that we need other subjects to decide set-theoretical questions is mainly a result of: (1) Forgetting that these subjects are themselves set-theoretical notions, and not something that stands outside of it. (2) Forgetting that a set is not, strictly speaking, given to us fully formed, but it's a result of a cumulative process, in which sets are formed from simpler sets, starting with the empty set. When you speak of R with the discrete topology, you are no longer speaking about the set of reals, but of another set: a topological space that, as a set, is P(R) (R with the usual topology is also not R, but a subset of P(R)).

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Well, you certainly need something. Saying "the set of reals" doesn't make any sense unless we have some sense of what exactly the reals are. I think your average Joe (who cares about such things) would say that the reals are exactly the set of all infinite decimal representations--we would of course need to exclude digit sequences which end in all 9's. In this case, the popular diagonal proof suffices, so long as we note that different representations are different numbers (again, excluding infinite-9's). Otherwise, we don't really have a good sense of what the elements of R really are. In the axiomatic construction of the reals, the structure of the set is never defined. We're given that it satisfies the field requirements under +,* along with an ordering and that every non-empty subset that's bounded above has a least upper bound. That's enough to completely define the reals--well, supposedly. My current analysis text glosses over that part a bit, referring instead to the decimal representations. In any case, I suppose that having these axioms, one could come up with an uncoutability proof...I'm not sure how to go about such a thing without appealing to decimals, and I think that would be rather interesting to see. I haven't seen the nested closed interval proof, but of course it would require the topology of the reals. Of course I think we could define some other metric on the reals, changing the form of the closed sets...so would any valid metric give us the same basic proof? I'm tempted to say that the cardinality depends only on the elements themselves, not necessarily on the algebraic or topologic structure. Using such structures though make proofs about countability much easier to work through, especially if we aren't explicit about the elements that make up the set (e.g., the field definition of reals). Just because I found it while thinking about your question, here's a link to some different constructions of the real numbers: http://en.wikipedia.org/wiki/Construction_of_real_numbers

Ben

I think that you have answered your own question. In one case, you find the cardinality of a set by looking at closed intervals, a purely topological construction. On the other hand, there may be no topological stucture, but you may be given an algebraic structure. In that case Algebra will be the only tool that you have to determine cardinality. In the first case, since you only have a toplogy, you do not have any knowledge of algebraic structure, and thus there is no relationship between elements. In the second case, you are only given relationships between elements with no knowledge of what an open set is. So my answer is that the cardinality is given by what knowledge you have about the set. It could be algebraic or topological. EDIT: JCS has it right. Much better than my response. We get so into our own little fields that we forget that every field is basically trying to do the same thing. Each field (Topology, Algebra, etc.) is simply a different approach and interpretation of much more basic concepts. I (personally) would do well to remember this more often. Fantastic answer, JCS! You get my vote for best answer!

squeek

I would suggest thinking about the Dedekind construction of the reals from the countable set of rational numbers and the total order on the rationals. You might then be able to prove that such a construction on any countable totally ordered set is uncountable, thus showing that topology has nothing to do with the cardinality. Conjecture: If S is a totally ordered countable set with the property that if x≤y and x≠y, then there exists z such that x≤z≤y, where z≠x and z≠y, then the Dedekind construction based on this set and total order is uncountable.

ksoileau

As others have remarked, you clearly need to know something beyond the name of the set. I have an interesting example from a previous question of mine: "Consider the set of all possible sequences {a_n}, where the terms of each sequence a_1, a_2, a_3 ... are each non-negative integers. What is the cardinality of this set?" The best answer, by "Pascal", used facts about transfinite arithmetic. I managed to "concretize" this by engineering this into a specific mapping of sequences to real numbers. So neither topology nor algebraic structures were assumed, but it was an awful lot of hard work. I hope you can find it, from the reference below.

nealjking

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