John Lane Bell, is his paper "Abstract and Variable Sets in Category Theory" (go to Bell's Homepage to download it), defines an abstract set as follows:

"An abstract set is then an image of pure discreteness, an embodiment of raw plurality; in short, it is an assemblage of featureless but nevertheless distinct 'dots' or 'motes' [footnote 3: "Perhaps also as 'marks' or 'strokes' in Hilbert's sense.]. The sole intrinsic attribute of an abstract set is the number of its elements." (pg. 10)

With this in mind let us consider a countably infinite set of such strokes

{|,|,|,...} = S

and attempt to define the power set P(S) of S. If one allows the existence of the empty set { } one can define P(S) as follows:

{ { }, {|}, {|,|}, {|,|,|},...}

which is fine as long as the subsets are finite. However, things seem to break down when one tries to define the countably infinite subsets of S because by the definition of S, any countably infinite subset S' of S is simply S, which seems to make the cardinal number|P(S)| of P(S) Aleph-null.

Although Bell states two paragraphs down that "an abstract set cannot be regarded as the extension of an attribute...", it seems that one needs the notion of an 'attribute' to be able to distinguish countably infinite subsets S' of S from S (and from each other) in order to make |P(S)| greater than Aleph-null. But this seems to equate the cardinality of |P(S)| with the number of attributes which allows one to distinguish the countably infinite subsets S' of S from one another (such attributes may, for lack of a better term, may be designated as 'extensional attributes'). Herein lies (at least for me) the puzzle, if in fact it IS a puzzle.

  • I rather suspect (having kept an occasional eye on this stackexchange and been, shall we say, unimpressed!) that you'll do better asking any even semi-serious logic questions on the often excellent math.stackexchange. Certainly I'd have thought that that's true for something at this kind of level. Nov 8, 2012 at 8:07
  • The root of the problem exists in a simpler form: there are two interpretations of {{|},{|,|}}, depending on whether we see the first set as being a subset of the second. I have no idea as to how to answer this question with respect to Bell's theory, but the theory of ur-elements in set theory provides an answer. I agree with @Peter that math.sx is probably more useful than here for you. Nov 8, 2012 at 9:00
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    @PeterSmith: philo.sx welcomes any honest attempt to make this environment more… impressive. On a personal note, I might add that I have been following your blog for quite some time, and I for one would be delighted to have you on board! (Though I am myself a rather intermittent contributor…)
    – DBK
    Nov 8, 2012 at 14:44
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    @PeterSmith: Although philo.se is not targeted at an academic/professional audience, I think a critical mass of users with an academic background/engaged professionally in philosophy would help to raise the profile of both Q & A.
    – DBK
    Nov 8, 2012 at 14:45
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    If the sole intrinsic attribute of an abstract set is the number of its elements, then the sole intrinsic attribute of even the set of natural numbers N is its number of elements. Consequently, either N and N-1 don't differ in terms of their intrinsic attribute (which makes no sense since you've defined them differently), or there exists a theory such that N and N-1 have a different number of elements. As I recall, Lakoff and Johnson in "Where Mathematics Comes" tried to come up with a theory where N and N-1 have a different number of elements, I don't recall how it works though. Nov 9, 2012 at 16:09

3 Answers 3


You have a number of incongruities in your question, I think:

  1. If S is made up of infinite strokes, as you presented your question, its cardinality is 0
  2. So, the set S is countably infinite
  3. This also means that the set ℘(S) is not countable since |℘(S)| = 20 which, assuming continuum hypothesis, is equivalent to 1
  4. Ergo, all possible subsets of S (otherwise known as ℘(S)) are not countable; keep in mind that 20 > ℵ0 which is the same thing as 1 > ℵ0

Also, when looking at a power set ℘(S) of a set S, there is only one element in ℘(S) that's simply S.

Consider L = {a, b, c}
Then ℘(L) = {Ø, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

L is only equivalent to the last discrete element in ℘(L), so I guess one could axiomize α ∈ ℘(α) since the same holds for countably infinite sets (and I think uncountably infinite sets, as well but I could be wrong).

I don't think there needs to be any distinguishing between L and (in our case) the last element of ℘(L) as they represent the same "thing" - in one case L is a set made up of discrete "things" and in the second, L itself is a "thing" inside a larger set. I don't think this breaks Bell's discreteness (or Set Theory, for that matter).

If your question was about breaking set identity (which Bell argues is simply the number of elements - also known as cardinality) when using the power set, worry not! Directly from Wikipedia:

Cantor's diagonal argument shows that the power set of a set (whether infinite or not) always has strictly higher cardinality than the set itself (informally the power set must be larger than the original set).

So there is never any ambiguity. Or maybe I missed the entire point of the question.

@ThomasBenjamin, to answer your comment:

  • I have the set of all natural numbers N
  • I have the set of all even natural numbers, lets call this Ne
  • I have the set of all odd natural numbers, lets call this No

Set theory says that |N| = |Ne| = |No| - that is, the cardinality of the set of natural evens is equivalent to that of the set of natural odds and to that of the set of all naturals. This can be confusing because Ne and No are subsets of N.

However, that doesn't really seem to be a problem for Bell. In a sense, you're right to worry about |N| = |Ne| = |No|. How can we say that the even numbers are the same thing as the odd numbers?

Well, it looks like Bell argues that all sets of cardinality 0 or greater are abstract sets (as opposed to concrete sets like L = {a, b, c}). So you could say L possesses the attribute a, but you could not say Ne possesses the attribute 2. So then how can we know that 2 is in Ne? Well, Bell, at the bottom of pg. 10, shows us how to create a relation between two abstract sets. In our case, it would maybe be f: n → 2n.

So, going back to your question (and comment), there is no way to distinguish between two sets if they have the same cardinality unless we map them. So until we specify that the set of all even naturals maps to n → 2n and the set of all odd naturals maps to n → 2n + 1, the two are identical.

  • Within countable models of set theory, one can speak of "collapsing cardinals", even (as I understand it) all cardinals to Aleph-null. If one is to believe what is called "The Naturalistic Account of Forcing", the countable models of set theory (say of ZFC) are considered 'toy models' representing what can happen in the 'real universe' V of ZFC. In theory, then, one should be able to have a universe V' of ZFC where all infinite cardinals are collapsed to Aleph-null. How would this take place? This might be a way to resolve this puzzle, if it is a puzzle Nov 9, 2012 at 16:47
  • should read "all infinite cardinals in the countable model to Aleph-null" Nov 9, 2012 at 16:55
  • How, in the example I give would you be able to be able to have 2^Aleph-null > Aleph-null countably infinite subsets of S? On the face of it, all countably infinite subsets of S are identical to S. Nov 9, 2012 at 17:22
  • @ThomasBenjamin - you're right. I amended my answer a bit, but essentially you point out the reason why abstract sets need a mapping function to mean anything. Nov 9, 2012 at 18:27
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    @ David: Let's assume such a mapping exists for S, that is, to 'construct' a countably infinite subset S' of S (call it, say, cisub--it obviously must be 1-1 and onto) and use the order the countably infinite subset was constructed in as the distinguishing attribute. Define S as having been constructed at stage 0, S' at stage 1, then use cisub on S' to Construct a new countably infinite subset S'' at stage 2, cisub on S'' to construct S''' at stage 3, etc. Now since the stages 'i' seem at first glance to be ordinal numbers, one can seemingly extend this construction into the transfinite. Nov 10, 2012 at 6:54

I'm not sure what "intrinsic" means, but if the sole intrinsic property of sets is the number of elements, you can't have power sets in the normal sense.

2^{|,|} = {x|x is one of ({}, {|}, {|}, {|,|})} = { {}, {|}, {|,|} }

If you work through this, you may end up defining a power set operator that is identical to the successor operator. And that seems to me to be what you are doing (I have not checked to see if Bell makes the same mistake; I hope not).

Calling successor "power set" strikes me as unhelpful. But, anyway, yes, if you only have successor, you get stuck at ℵ0.

Incidentally, this is also why the natural numbers are usually defined as {}, {{}}, {{},{{}}}, ...--so you have a way to distinguish the elements.

Alternatively, a countably infinite subset of a countably infinite set has the same cardinality as the original (by definition), but it is not the same set. For example, N = {0, 1, 2, ...} but Z = {..., -2, -1, 0, 1, 2, ...}, and they are not the same because -1∈Z but -1∉N. So if there is more to a set than its mere cardinality, then we're okay.

  • Inasmuch as this answer somewhat echoes mine, it's completely correct. Furthermore a proper reading of Bell shows that even though Z = N (in an abstract sense), the mapping of Z is different than the mapping of N; namely, the mapping of Z is f: { n → 2|n| if n < 0, n → 2n - 1 if n > 0} (I simplify the bijection here). Maybe Z is not that great of an example because it's not as intuitive as even or odd numbers :P Nov 9, 2012 at 23:56
  • @Rex: I would definitely agree with you, if the sole intrinsic property of an abstract set is the number of elements, you cannot have power sets in the normal sense. You do seem to need the notion of attribute in order to distinguish countable sets from one another. I wonder If, in category theory, one could define an abstract notion of 'extensional attribute' solely in terms of arrows? Nov 10, 2012 at 6:36
  • @DavidTitarenco - You and I made different points, didn't we? You kept the traditional definition of set, although the question appears not to, and then argued that |2^S| > |S|. I accepted the characterization of set as equivalent to cardinality of that set, and showed that the idea of power set no longer works. I'm not sure how that's an echo. (Also, I really don't understand how "no negative numbers" is less intuitive than "no odd numbers".)
    – Rex Kerr
    Nov 10, 2012 at 9:39
  • @ Rex: Note that Bell's definition of "attribute" given on pg. 11 of his paper, i.e. "maps X --> 2" where "the members of 2 [are] playing the role of truth values: {} 'false' and 1 'true' (the set 2={ {}, 1} plays the role of " 'subobject classifier' or 'truth-value object' in [the category] Set") seems to break down as well on S for 'uncountable subsets' of S (that is, for identical (equivalent?) but numerically distinct subsets--see Neil's answer and comments) but holds if arrows are considered as primary in Category Theory, which they are.... Nov 13, 2012 at 11:22
  • oops--that is countably infinite subsets of S--sorry! Nov 18, 2012 at 11:49

Perhaps the most important thing about a set S in itself is its cardinality, as the labels of the elements are not important to the set-in-itself (compare to the notion of a thing-in-itself, which is to metaphysics as a set-in-itself is to the philosophy of mathematics). As soon as you entertain the power-set, you are no longer considering the set in itself: although the elements of the set may as well be anonymous (albeit distinguishable) if you are only considering the set in itself, the distinguishability of the elements is significant to the distinguishability of the subsets, in which case the cardinality no longer characterizes the subsets.

If every subset of order k is indistinguishable from every other subet of order k, perhaps because we have adopted a dogma that all sets of order k are "essentially the same", then indeed one cannot obtain a power-set ℘(S) which has a cardinality larger than S for infinite sets S. However, the question one should ask is why all sets of a given cardinality should be regarded as being for all intents and purposes as equivalent. I certainly do not regard a set of two ten dollar bills as being equivalent to a set of two twenty dollar bills, nor to a set consisting of one ten dollar bill and one twenty dollar bill. The reasons for this may be uninteresting to set theorists, but this is only to say that the subject matter of set theory cannot be neatly protected from application to non-set-theoretic problem domains. In this respect, I have a motivation to care, if I have a set of bills consisting of two twenty dollar bills and two ten dollar bills, which subset of two elements I consider. Thus the structure of the set, which may fade away if I consider the elements to be interchangeable tokens due to some detachment I have from the contents of the set, reasserts itself through the significance I attach to the elements as a consequence of the significance of the elements to a wider context.

It is difficult to find a better example than the integers, if one wishes to consider an infinite set. Perhaps the difference in the significance of a googol compared to a googol-plus-one to me is pitiful, but I recognize that there is in principle a difference, and to the extent that I should care at all about the number googol, I should care about the difference between googol and googol-plus-one. Thus I should care about the difference between any two positive integers in principle; and so the proliferation of subsets of the integers of any given cardinality is of interest. If I grant this, it is difficult to avoid the fact of the uncountability of ℘(ℕ).

The fact that we interpret ℘(ℕ) as a set with an extension larger than any possible notation system is an interesting one; one might suppose perhaps not that there are different degrees of infinity, but that there are systems of concepts which exceed our ability to reason non-constructively, by virtue of the fact that we can always construct a subset of the naturals about which any enumeration of proofs says nothing. However, distinguishability of labels is in essence the foundation on which mathematics is established, and to suppose that all sets of objects of a given cardinality are equivalent, despite that their elements may be given distinct labels, is to suppose that we can abolish all of the structure which mathematics is meant to describe. It is perhaps a self-consistent viewpoint, but I would maintain that it is an unproductive one, to say nothing about whether it is a necessary one.

  • @Neil: A few questions, then a comment. Regarding "The fact that we interpret P(N) as a set with an extension larger than any possible notation system...", are you referring to systems of notation like Kleene's 'O'? Why does P(N) have an extension larger than any possible notation system? How would you show this in a system like ZFC? Are you saying that there are sets of natural numbers that are not definable in formal systems of set theory? Regarding your example of sets of paper currency, would you say that two sets of, say, ten dollar bills (one set in the left hand pocket, Nov 11, 2012 at 6:34
  • the other in the right hand pocket--let's make these two sets sets containing exactly one element each) are identical or not? Regarding the set S under consideration, let's consider the following thought experiment: cross off every second element in S to form the subset S' of S (because one crossed off every second element one might call S' a 'proper subset' of S). Now cross off every second element of S' to form a third 'proper subset' S'' of both S and S'. By the definition of S, S=S'=S'' though S, S', and S'' are all numerically distinct. Because of this, one can place the elements Nov 11, 2012 at 6:56
  • (because one can extend this construction of 'proper subsets' of S into the transfinite) of P(S) seemingly in 1-1 correspondence with ORD, the class of all ordinals. One in theory could do the same with the elements of N (with possibly the same results?) but wouldn't this show that CH is false? Please comment. Thanks Nov 11, 2012 at 7:07
  • @ThomasBenjamin: any notational system has cardinality at most ℵ0, as any element of that system is a finite sequence of symbols drawn from a finite alphabet, or can in any case be specified by such a string of symbols. This simply isn't enough to compete versus the extension of all elements of ℘(ℕ). Nothing more than Cantor's diagonal argument is required. (This is similar to the fact that the set of computable reals has measure zero, which is for the exact same reason.) Even if you allowed yourself an infinite alphabet you would have the same problem. Nov 11, 2012 at 14:17
  • @ThomasBenjamin: for the ten-dollar bills, it is a question of the world-model to which you wish to put these ten-dollar bills to use. If you are a collector of currencies, then perhaps you only care about which ten dollar bill is in better condition (that is, better represents the ideal set forth by the mint), in which case having a second example of a ten dollar bill is less interesting to you if it is of lower quality. If you only care to spend the money, the quality of the bill is uninteresting, and the two bills are interchangeable while remaining distinct: not identical, but equivalent. Nov 11, 2012 at 14:25

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