6

While reading the Wikipedia article on infinite sets I found the following quote:

A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number.

If the axiom of choice holds, then a set is infinite if and only if it includes a countable infinite subset.

This raised an interesting question; without AC, is it possible to have a set where, given any natural number, you can find a subset with cardinality great than that number, but not necessary one equal to it?

For example a set where all definable subsets have even cardinality?

Edit to clarify the question: are there sets that satisfy the intuitive meaning of infinat but not the definition as quoted above?

FWIW, id be interested in both the case where "subsets larger than n exist" requires those subsets be finite and where they don't. (Though I suspect the second case is uninteresting as most of the interesting properties would be trivially true.)

BCS
  • 633

2 Answers2

6

No, this is impossible.

If $A$ has a subset $B$ such that $B$ has size greater than $n$, then there is, by definition, an injective function $f\colon\{0,\dots,n-1\}\to B$, which is also an injective function to $A$. Taking the image of $f$ is a subset of $A$ with exactly $n$ elements.

Asaf Karagila
  • 393,674
  • 1
    That construction assumes that $B$ can be ordered. Without AC, is that always true? For example, could a set exist where the only definable binary operation on its elements is equally? – BCS May 29 '23 at 18:57
  • 2
    No, the assumption is that $B$ is finite. Namely, there is some $n$ such that there is a bijection between $B$ and ${0,\dots,n-1}$. – Asaf Karagila May 29 '23 at 18:59
  • Te set of undefinable numbers has no countable subsets without axiom of choice. – Anixx May 29 '23 at 19:02
  • 1
    @Anixx: Numbers, what do you mean? Definable, in which structure in what universe? – Asaf Karagila May 29 '23 at 19:05
  • Good to know you’re still kicking – SBF May 29 '23 at 19:09
  • @AsafKaragila an infinite set without non-empty finite subsets: all real numbers minus all definable numbers. – Anixx May 29 '23 at 19:13
  • 2
    @Anixx: Definable in what structure? In what language? Using which logic? – Asaf Karagila May 29 '23 at 19:14
  • 1
    @S.L. Th set of all undefinable numbers is obviously not finite but also not infinite under the given definition. Of course, this raises the question, definable in what language. I think, this is not that important. – Anixx May 29 '23 at 19:26
  • For instance, a set of numbers, whose decimal expansion is a random sequence. This set obviously has no finite non-empty subsets. – Anixx May 29 '23 at 19:28
  • 3
    @Anixx: The question of language is *insanely* important. Adding all the numbers as constants makes all numbers definable. In the grand universe of sets, or even in an initial segment thereof, it is quite possible that all real numbers are definable (see more in "pointwise definable models of set theory"). In some infinitary logics we have infinite disjunction and conjunctions and we can write the complete type of a real number. There are many ways to make all reals definable. – Asaf Karagila May 29 '23 at 19:29
  • 1
    But would not that kind of logic require or imply axiom of choice? – Anixx May 29 '23 at 19:31
  • 1
    @Anixx Depending on the situation, it may or may not imply that the reals can be well-ordered. But not in the example when we add all the reals as constants to the language. – Asaf Karagila May 29 '23 at 19:37
  • "B has size greater than n, then there is, by definition, an injective function f:{0,…,n−1}→B" - by what definition? Obviously, there is no function from a finite set to the set of numbers whose decimal expansion is a random sequence. – Anixx May 29 '23 at 19:37
  • 4
    @Anixx The existence of the injection is given by the very definition of "greater cardinality". You keep mentioning "random sequence", I don't think it means what you think it means. – Asaf Karagila May 29 '23 at 19:41
  • "The existence of the injection is given by the very definition of "greater cardinality" " - okay, so, the problem is in this definition. Basically, it turns out that the size of random reals is not greater than a finite set. But it has a number of subsets which is greater than a given finite set (for instance, all sets of randoms between n and n+1 is greater than any finite set), – Anixx May 29 '23 at 19:49
  • 3
    @Anixx: If you want to reject the standard definitions and come up with your own, that's fine. But that isn't the question here. So if that's all neatly wrapped up, I think this conversation is over. – Asaf Karagila May 29 '23 at 19:52
  • @AsafKaragila is AC required for every finite set to have a mapping to the natural numbers less than or equal to its cardinality? Or can there be sets where any ordering is nesseaaraly arbitrary (which IIRC requires AC)? – BCS May 29 '23 at 19:53
  • 1
    @Anixx Every non-empty set has non-empty finite subsets. Proof: Suppose $X$ is non-empty. By definition of "non-empty", there exists $x\in X$. Since $x\in X$, ${x}\subseteq X$, and ${x}$ is non-empty and finite. Thus there exists a non-empty finite subset of $X$. – Alex Kruckman May 29 '23 at 19:54
  • @AlexKruckman doesn't that require AC in the general case? Or am I misunderstanding AC? – BCS May 29 '23 at 19:57
  • No, you don't need AC to pick a single element from a single non-empty set. For further discussion of this, see here, for example. – Alex Kruckman May 29 '23 at 20:01
  • 1
    @AlexKruckman so what is the proper term for the assumption that non-constructable subset don't exist? Or to put it another way; "you can't pick a single element from a non empty set unless that element has a distinguishing property". – BCS May 29 '23 at 20:34
  • @BCS Such a principle is simply incompatible with set theory formalized in first-order logic. If the definition of "$x$ is non-empty" is $\exists y, (y\in x)$, then the rule of Existential Instantiation in first-order logic allows you, given a non-empty set $x$, to use $y$ such that $y\in x$ in proofs about $x$. – Alex Kruckman May 29 '23 at 21:03
  • @BCS If you don't like this, you either need to use a non-standard definition of "non-empty" or use a non-standard logic. Neither of these choices will make it easy to communicate with other mathematicians (see Asaf's last comment to Anixx above). I should also mention that due to Tarski's Undefinability of Truth Theorem, it is impossible to turn "$x$ has a distinguishing property" into a formal definition in any set theory to which Tarski's Theorem applies. – Alex Kruckman May 29 '23 at 21:10
  • @AlexKruckman can a finite set with indistinguishable elements exist? For instance, we have a set of $10$ elements but the elements are absolutely indistinguishable. Can we chose one element from such set? What about infinite set with such property? – Anixx Jun 11 '23 at 10:30
  • @Anixx What does "absolutely indistinguishable" mean? I challenge you to give a precise definition. – Alex Kruckman Jun 11 '23 at 11:36
  • @AlexKruckman consider a set of two elements, whose all properties coincide. Can such set be regarded as defined? Consider another case: the two elements $x$ and $y$ of $S$ are indistinguishable but there is an operation $$ defined on set $S$ such that $xx=-1$, $yy=-1$ but $xy=1$? For instance, can there be a set of imaginary numbers ${i,-i}$ if $i$ and $-i$ are not defined by ordered pairs of reals but by these properties: $x^2=-1$, $y^2=-1$, $x+y=0$? – Anixx Jun 11 '23 at 11:51
  • @Anixx All this is too vague to make any sense of. As I commented to BCS above, you cannot formalize "whose all properties coincide" due to Tarski's theorem. I have no idea how to interpret the question "can such set be regarded as defined?" I also don't understand what $i$ and $-i$ are supposed to be in your counterfactual universe where they're "defined by properties" instead of as an actual construction. – Alex Kruckman Jun 11 '23 at 11:59
  • @AlexKruckman related question: https://math.stackexchange.com/questions/4716621/are-random-reals-possible-without-axiom-of-choice?noredirect=1#comment9989912_4716621 – Anixx Jun 11 '23 at 12:05
5

No, and to be asking the question you must have misunderstood the passage you quote.

A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number.

There is no mention of AC here, and so this part is true even without AC.

Without AC the second part of your quote, about having a countably infinite subset, does not hold. This means you can (without AC) have an infinite set such that all its infinite subsets are uncountable, but it must still have finite subsets of every possible order.

Especially Lime
  • 40,828
  • You are assuming the given definition is correct. I'm questioning that assumption. The requirement that subsets of every cardinality exist seems overly restrictive. If the set of cardinalities of subsets is infinite, is the original set infinite (even if there are cardinalities for which no subset exist)? -- I suspect the punchline is that for any set theory; if it permits the existence of any given non-empty finite subset then it also permits the existence of a subset of cardinality one less. By induction, infinite sets, by any definition, must satisfy the original quote. – BCS Jun 02 '23 at 04:24
  • 1
    @BCS, yes, that is true, or even more straightforwardly if a set is non-empty it contains an element. Remove that element, and continue. If your initial set was infinite you can do this $k$ times and form a set whose elements are the things you've removed. You don't need AC, because you're only making finitely many choices (and you don't need AC to imply that any of your choices is from a non-empty set). – Especially Lime Jun 02 '23 at 07:44
  • 1
    You skipped over my point, kinda. It seems to me that a set theory could exist where the operation of "pick a single element from a non empty set" is not automatically possible, for example you might have sets where the "pick" operation would always result in two elements. My conjecture is that while such a set theory might work, it would be impossible to describe in it a finite set that has that issue. – BCS Jun 03 '23 at 03:23
  • @BCS As I commented above, the "pick operation" is not part of set theory, but rather part of the underlying first-order logic. So to come up with a set theory like the one you describe, you would need to use a non-standard logic (where the existential quantifier works differently). – Alex Kruckman Jun 11 '23 at 12:03
  • 1
    @AlexKruckman Am I in error in my understanding that in standard FOL, ∃ is a predicate and not an operation? That is that ∃ non-constructively asserts about the existence or non-existence of something (with specified properties) without necessarily identifying that something? -- Also, IIRC without a very careful formulation of set-theory, FOL on sets leads directly to Russell's Paradox suggesting things are more complicated than just (∃x x∈S) ⇒ {x} – BCS Jun 11 '23 at 14:47
  • @AlexKruckman I wonder if a set can be considered well-defined if its definition does not provide means of destinguishing its elements from each other, only a criterion if something belongs to the set or not. Can we end up with a set of two identical elements? – Anixx Jun 11 '23 at 15:47
  • @BCS $\exists$ is neither an operation nor a predicate. It's a quantifier, which has different status than both. I refer you to my comment to Asaf's answer above that begins "Every non-empty set has non-empty finite subsets." and contains a proof (in first-order logic, using only the axioms of pairing and extensionality!). To both you and Anixx: this is my last comment here. I strongly recommend that you read a book on axiomatic set theory (Hrabacek and Jech is a good choice) rather than continuing to speculate wildly about how it works. – Alex Kruckman Jun 11 '23 at 16:28
  • 1
    @Anixx Maybe a set where the only property defined on its elements is a equality? Two elements form the set only allow asking if they are the same? -- Actually what might be more interesting is an ordered countably-infinite-in-both-directions set with no distinguished elements. Say the set of positions on a zeroed out turning machine tape. Despite it being meaningful to to ask about the order of two elements, without already having an element in hand, any finite subset of is fundamentally arbitrary, indistinguishable from infinite other distinct subsets. – BCS Jun 11 '23 at 21:36
  • @BCS I am thinking about two types of sets with indistinguishable elements. In the example of sets with two elements, there are two types of indistinguishability, "bosonic": replacement of any single element in any function does not change the function's result and "fermionic": mutually swapping the elements in any function does not change the function's result but replacing only one changes the result (e.g. in complex numbers $i^2=-1$ and $(-i)^2=-1$ but $i(-i)=1$). – Anixx Jun 12 '23 at 00:00
  • @BCS relevant: https://mathoverflow.net/questions/423796/why-we-can-analytically-define-%ce%b5-in-dual-numbers-so-to-distinguish-%ce%b5-from – Anixx Jun 12 '23 at 00:17
  • @BCS Also relevant: https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb – Anixx Jun 12 '23 at 00:25