## Infinite Sets and their Cardinalities

For the discussion of math. Duh.

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maninblack
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### Infinite Sets and their Cardinalities

First off I'm going to apologize for murdering the common notation. I'm not sure how to insert the characters that I have in mind. That said while studying set theory in college I came across some things that struck me as intuitive that did not hold true when considering the Cardinalities of infinite sets.

A commonly accepted and widely used theory regarding cardinalities states Given two sets A and B, A is a subset of B implies that |A| <= |B|.

Intuitively it would seem to follow that.... Given two sets A and B, A is a proper subset of B implies that |A| < |B| (that is the cardinality of A is STRICTLY less than the cardinality of B). This is not always true. It is true for finite (and thus trivial) cases, and is true in some infinite cases (where A is the naturals and B is the reals for example).

My thought, and what I would like you all to consider is, what happens if we believe this (my above implication) as an axiom. I understand this would change how we consider infinite cardinalities, and how we prove that two sets have the same number of elements, but it my make the whole mess of infinite cardinalities more understandable....it may not.
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Fafnir43
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### Re: Infinite Sets and their Cardinalities

What happens when you need to compare the sizes of two sets which aren't subsets or supersets of each other, and differ by an infinite number of elements (like R \ [0,2] and R \ [1,3])? I don't see a way of showing those two sets have equal order under your axiom, or even comparing them at all, without bringing bijections back into it...
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maninblack
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### Re: Infinite Sets and their Cardinalities

They/We/Someone would have to define a way to describe infinite cardinalities. I have thought about that, and it is a significant problem. I, however, would like to solve it without the use of bijections. (As the use of bijections leads to counter intuitive results.)

My initial thought on that matter would be to define some unit to be the cardinality of [0,1] (for the moment I will call this unit r). My thought then would be to also define a convention stating that any range of the reals [0, k] would have the cardinality of k*r....or the more general range [m,n] would have the cardinality (n-m)*k. This would allow for the cardinalities of your two ranges to both be 2r. This was my initial thought, but the intuitive sense goes right out the window when you start talking about infinite units and their coefficients.

Some convention would need to be defined, and the one I described would seem to 'work' (in the loosest sense of the word).
addams wrote:It was like a game of Rugby and I was the Puck. Yeah. I was taking a beating and I was in the wrong game.
roband wrote:If this is a serious question, the answer is no.
If it's a philosophical one, I have no idea.

ThomasS
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### Re: Infinite Sets and their Cardinalities

This route would also give counterintuitive results. Probably enough so to not make an actual theory.

I mean, cardinality is useful because it lets us say that {1,2,3,...} has something in common with {1, 1/2, 1/3,...}. But if we can map n to 1/n and have something in common, why not go to 1/(n+1) or to (n+1).

antonfire
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### Re: Infinite Sets and their Cardinalities

In short, you can't apply that to arbitrary sets (only to sets of real numbers) because there are too many ways to map a set to a subset of the reals, which will give different answers. If you only apply it to (some) subsets of the reals, then what maninblack is describing generalizes to a "measure".
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Robin S
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### Re: Infinite Sets and their Cardinalities

Measures aren't restricted to be on the reals.
This is a placeholder until I think of something more creative to put here.

maninblack
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### Re: Infinite Sets and their Cardinalities

antonfire wrote:In short, you can't apply that to arbitrary sets (only to sets of real numbers) because there are too many ways to map a set to a subset of the reals, ...

Therein, however, lies part of the problem with this "measure." One using my axiom would have to define another way to describe what are currently considered countably infinite sets.

Consider how the integers relate to the naturals. If one would define n to be the cardinality of the naturals it would seem reasonable to attempt to describe the cardinality of the integers in terms of this n. Would it be as simple as the intuitive 2n? Or would we need to also consider zero. Remembering, of course, that I have asked that the use of functions (mapping, bijections...and so on) be thrown out because they can cause counterintuitive results.
addams wrote:It was like a game of Rugby and I was the Puck. Yeah. I was taking a beating and I was in the wrong game.
roband wrote:If this is a serious question, the answer is no.
If it's a philosophical one, I have no idea.

antonfire
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### Re: Infinite Sets and their Cardinalities

If you throw out functions, why do you assume that the negative integers have the same "cardinality" as the positive integers?

Robin S wrote:Measures aren't restricted to be on the reals.
Never said they were.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Robin S
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### Re: Infinite Sets and their Cardinalities

Ah, I misinterpreted what you were saying.

maninblack, the problem with discounting counterintuitive results is that sometimes it renders what's left meaningless, and even when this is not the case it can cause more counterintuitive results to appear.

For example, even if you don't use functions in your definition of size, you still have to consider what happens when you apply a function to members of a set. Take the function f(x) = x + 1, defined on the natural numbers. By your reasoning, this should have n distinct inputs, but n-1 distinct outputs, despite every input being associated with exactly one output and vice versa. Put another way, if you list all of the ordered pairs (x,f(x)) then there are n such ordered pairs, but if you only list f(x) then there are only n-1 values. Where did the last value go?

For another example, consider the function f (x) = ((-1)xx - (x mod 2))/2, defined on the non-negative integers. By your reasoning, there should be n+1 distinct inputs, but 2n+1 distinct outputs.

As soon as bijections stop having to preserve cardinalities of sets, you come up against some real problems.
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maninblack
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### Re: Infinite Sets and their Cardinalities

antonfire wrote:If you throw out functions, why do you assume that the negative integers have the same "cardinality" as the positive integers?

Robin S wrote:Measures aren't restricted to be on the reals.
Never said they were.

Comfort......"seems like they should be"....."it's just intuitive".....(said one clinging to a belief that is obviously not going to hold any water)
antonfire....touche....

And to Mr. Robin... I believe the word is SNAP.

So one following my axiom would have to completely redefine functions starting from the domain/codomain......So basically in an attempt to alleviate a little discomfort, I have asked mathematics to be completely rebuilt......which..... is so not going to happen.

Interesting how the seemingly convenient can have disastrous consequences.

My real issue comes from asymptotic functions, and I suppose with that the concept of infinity. I don't know what I'm trying to say here other than *pouts* my axiom obviously won't work.
Last edited by maninblack on Thu May 08, 2008 3:17 pm UTC, edited 2 times in total.
addams wrote:It was like a game of Rugby and I was the Puck. Yeah. I was taking a beating and I was in the wrong game.
roband wrote:If this is a serious question, the answer is no.
If it's a philosophical one, I have no idea.

Robin S
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### Re: Infinite Sets and their Cardinalities

Yes. And for the record, I'm male.
This is a placeholder until I think of something more creative to put here.

yeyui
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### Re: Infinite Sets and their Cardinalities

If you define [imath]|A| < |B|[/imath] to mean [imath]A \subset B[/imath], then what you have done is define a partial order (but not a total order) on the class of sets. Partially ordered sets (posets) are a rich and still active topic. for an introduction see Stanley's Enumerative Combinatorics. However, this is not going to give you a new theory of cardinality if you want cardinality to be a total order on sets.

maninblack
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### Re: Infinite Sets and their Cardinalities

yeyui....What you have stated is the converse of what I was considering. But I guess I have trouble understanding where, to which sets, it applies. If you are talking about infinite sets (which I assume you are) then the only sets that statement applies to would be A being a countably infinite set and B being uncountably infinite.

Now if I'm understanding what you are saying correctly you intend to imply that the natural numbers are a sub set ALL ranges of the reals......which is obviously not true.

I hope I've missed something and someone can point out to me what I have missed.
addams wrote:It was like a game of Rugby and I was the Puck. Yeah. I was taking a beating and I was in the wrong game.
roband wrote:If this is a serious question, the answer is no.
If it's a philosophical one, I have no idea.

BeetlesBane
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### Re: Infinite Sets and their Cardinalities

maninblack,
You may want to view a BBC documentary called "Dangerous Knowledge", it runs around 2hrs. Clips are available on the web; I watched it as a single clip and have seen it listed split into both 3 and 4 parts. It' covers the consequences some have paid for working with counterintuitive aspects of non-finite cardinality.

yeyui
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### Re: Infinite Sets and their Cardinalities

What I was said was intended to be (almost) the same as in the initial post.

In initial post it was proposed to take [imath]A \subset B \Rightarrow |A| < |B|[/imath] as an axiom. Well, to do that you must change the definition of one of those the symbols or else the axiom will cause a contradiction. If you resolve the contradiction by redefining [imath]|A|<|B|[/imath] (" A is smaller than B ") to simply mean than A is contained in B, you have a defined a partial order. It is important to note (which I did not do earlier) that this necessarily changes the meaning of [imath]|A|[/imath].

Now that I am making a second post about this, I will point out a few other things. The above really is defining a partial order on sets, not on "cardinalities" of sets. This doesn't actually define a partial order on cardinalities for the simple reason that we have not defined such a concept. If we accept the proposed axiom, the above definition, then the usual definition of cardinality will lead to a contradiction.

BTW: I do not suggest trying to define a notion of cardinality that works with your intuition. Your intuition is based on your experience with finite sets and infinite sets simply behave differently. My guess is that if you insist that [imath]A \subset B \Rightarrow |A| < |B|[/imath], you will have difficultly defining what [imath]|A|=|B|[/imath] means and that, regardless of your definition, you will still be forced to accept things that are "clearly wrong" (read: counterintuitive).

Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

yeyui: At the very least, a definition of the "size" of a set should have the property that finite sets have the same size iff they contain the same natural number of elements. In particular, there should be different sets with the same size. All sets have a different (most often incomparable) size with your definition.

I don't know of any definition of size other than cardinality, but I doubt a reasonable one exists. The one property I'd want is that renaming the members of a set doesn't change its size. Well, that's precisely what a bijection does!

Macbi
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### Re: Infinite Sets and their Cardinalities

I think there might be more worth to this idea than some of you think. Clearly |A|<|B| defined only in terms of A being a subset of B will only produce a partial order, but there does at least exist a definition of a total order where A being a subset of B implies |A|<|B| (Just the normal definition with this tagged on would work, though it would be inelegant).
However in his second post maninblack seems to want to do this by ascribing a number to describe the size of each set, this doesn't seem likely to work.
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Owehn
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### Re: Infinite Sets and their Cardinalities

I haven't actually tried, but it seems like a straightforward application of (industrial strength) Zorn's lemma demonstrates the existence of a total extension to every partial order, so there should be some linear order on Set that agrees with inclusion as a partial ordering.
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ThomasS
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### Re: Infinite Sets and their Cardinalities

Owehn wrote:I haven't actually tried, but it seems like a straightforward application of (industrial strength) Zorn's lemma demonstrates the existence of a total extension to every partial order, so there should be some linear order on Set that agrees with inclusion as a partial ordering.

I think this "works" in the sense that you can extend the partial order on sets which is induced by [imath]\subseteq[/imath] to a linear order. However, if you just did that, you wouldn't know, even in principle, how to tell if one set was bigger than another. e.g. You would know that either [imath]\{1,2\} < \{3,4\}[/imath] or [imath]\{3,4\} < \{1,2\}[/imath] but you wouldn't know which.

Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

Macbi wrote:there does at least exist a definition of a total order where A being a subset of B implies |A|<|B| (Just the normal definition with this tagged on would work, though it would be inelegant).
Could you be more precise? Do you mean this?
$\newcommand{\card}{\text{card }} |A|<|B|:=\cases{\card A<\card B&\text{or}\\A\subset B}$
In that case, which sets have equal size? If you change the strict inequalities above to non-strict, the extra case does nothing, and you have the regular definition of the cardinality.

edit: typo.
Last edited by Torn Apart By Dingos on Sat May 10, 2008 3:43 pm UTC, edited 1 time in total.

Macbi
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### Re: Infinite Sets and their Cardinalities

Torn Apart By Dingos wrote:
Macbi wrote:there does at least exist a definition of a total order where A being a subset of B implies |A|<|B| (Just the normal definition with this tagged on would work, though it would be inelegant).
Could you be more precise? Do you mean this?
$\newcommand{\card}{\text{card }} |A|<|B|:=\cases{\card A<\card B&\text{or}\\A\subseteq B}$
In that case, which sets have equal size? If you change the strict inequalities above to non-strict, the extra case does nothing, and you have the regular definition of the cardinality.

I meant:

$\newcommand{\card}{\text{card }} |A|<|B|:=\cases{\card A<\card B&\text{or}\\A\subset B}$
Since in the first post it was said:
maninblack wrote:Intuitively it would seem to follow that.... Given two sets A and B, A is a proper subset of B implies that |A| < |B|
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Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

Yeah, I mistyped. That's what I meant. Then which sets have equal size?

Macbi
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### Re: Infinite Sets and their Cardinalities

Not sure what you mean by that question.
Most sets that had equal size before still have equal size, and all of the sets of equal size still have bijections between them.
Unless you just want a definition, in which case:
|A|=|B| iff ( not(|A|<|B|) AND not(|B|<|A|) )
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Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

Okay, that works. I thought there was a contradiction there someplace but I must've been wrong.

antonfire
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### Re: Infinite Sets and their Cardinalities

Then "equal size" is not transitive. The negative integers have "equal size" with the positive integers and the positive even integers, but the positive even integers are "smaller" than the positive integers.

There's no contradiction, it's just that that notion of "equal size" sucks ass.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Macbi
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### Re: Infinite Sets and their Cardinalities

Oh. Damn. Transitivity of equallity is rather important.
Okay, scrap that. Is there some total order with the properties we want?
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Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

Wait, that is a contradiction. A total order needs to be transitive.

Macbi
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### Re: Infinite Sets and their Cardinalities

I think that the order still is, only equality isn't.
evens=<positives=<negatives
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Token
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### Re: Infinite Sets and their Cardinalities

Unfortunately, the order isn't transitive, because we have positives<=negatives<=evens, but evens < positives.

Even if this weren't the case, you'd have to deal with the fact that, I believe, the equivalence classes under this order would be a proper class, not a set, so the order would be badly defined anyway.
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skeptical scientist
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### Re: Infinite Sets and their Cardinalities

Well, you can just define a partial order [imath]a<b \leftrightarrow a \subset b[/imath]. Then you get a perfectly reasonable notion of size with [imath]a \subset b \rightarrow a<b[/imath]. The trouble is, most things are incomparable.
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the tree
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### Re: Infinite Sets and their Cardinalities

Okay I'm vaugely confused, but isn't it the convention that with with infinite sets then [imath]|A|\leq|B| \Leftrightarrow \exists \mbox{injection}:A \rightarrow B[/imath]? That's consistent with [imath]|A|\leq|B| \Leftarrow A \subset B[/imath] and thanks to Schroeder-Bernstein it creates a natural ordering on sets with finite and infinite cardinalities.

BeetlesBane
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### Re: Infinite Sets and their Cardinalities

the tree wrote:Okay I'm vaugely confused, but isn't it the convention that with with infinite sets then [imath]|A|\leq|B| \Leftrightarrow \exists \mbox{injection}:A \rightarrow B[/imath]? That's consistent with [imath]|A|\leq|B| \Leftarrow A \subset B[/imath] and thanks to Schroeder-Bernstein it creates a natural ordering on sets with finite and infinite cardinalities.

If you review the OP, the initiating question posited strictly less than being dependent on proper subset.

maninblack
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### Re: Infinite Sets and their Cardinalities

skeptical scientist wrote:Well, you can just define a partial order [imath]a<b \leftrightarrow a \subset b[/imath]. Then you get a perfectly reasonable notion of size with [imath]a \subset b \rightarrow a<b[/imath]. The trouble is, most things are incomparable.

Too Roight... with the "axiom" stated in the opening post you do get a partial order with a perfectly reasonable notion of size/cardinalities. The trouble as the skeptic describes it is what I was trying to avoid by attempting to assign a cardinal number to infinite sets. I feel that it has been shown (through the thread) that such assignments will not work, and our order is not very useful where set containment is not relevant.

I'm quite surprised that this discussion has continued, but I am interested to see where it will go from here.
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Macbi
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### Re: Infinite Sets and their Cardinalities

I just think it might be possible to extend that partial order to a total order while keeping the nice properties of it, and with all sets of the same size sharing a bijection.

What if we say that (lets see if I've got the hang of this LaTeX thingy) |A|<|B|[imath]\Leftrightarrow[/imath]([imath]\exists[/imath]p,q[imath]\epsilon\mathbb{Z}|\forall[/imath]x[imath]\epsilon[/imath]A, (x(-1)p+q) [imath]\epsilon[/imath]B) (Or even: lets see if I've got the hang of this set notation thingy)

That is, A is smaller than B if there is rigid transformation on the real line that makes it a strict subset of B.

This still doesn't make it a total order, but it does expand it, and is surely a property we would want.
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Torn Apart By Dingos
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### Re: Infinite Sets and their Cardinalities

I found a counterexample to that, Macbi. Take an infinite sequence x_n such that 0<x_n<1. Let A contain
x_1,
x_1+1,x_2+1,
x_1+2,x_2+2,x_3+2,
and so on.
Then A+1 is a strict subset of A. So |A|=|A+1|<|A|.

Interesting. I thought at first I'd have to use non-measurable sets (since this reminded me of Banach-Tarski, which was why I originally got doubts).

ThomasS
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### Re: Infinite Sets and their Cardinalities

Invariance under translation is one of the nice properties of the standard measure of the real line. In some sense it is what makes it different than other measures on the real line. The full implications of what these types of constraints do to measures theory is one aspect of "geometric measure theory", which is a fairly deep field.

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### Re: Infinite Sets and their Cardinalities

Torn Apart By Dingos wrote:I found a counterexample to that, Macbi. Take an infinite sequence x_n such that 0<x_n<1. Let A contain
x_1,
x_1+1,x_2+1,
x_1+2,x_2+2,x_3+2,
and so on.
Then A+1 is a strict subset of A. So |A|=|A+1|<|A|.

Okay, if we can't have that property then I concede that this is a bad idea after all.
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### Re: Infinite Sets and their Cardinalities

It just occured to me that A=(0,\infty) works as well. My counterexamples are often needlessly complicated. I recall another time when I wanted to come up with an equation in one variable with more than one solution (a physics book had said that a (non-linear) system of n variables should have n solutions, which I found suspect), and, since it reminded me of the fact that there is a bijection from R to R^2, my first example was something like f(x)=1, where f(x) removes every second binary digit from x (so f(0.123456...)=0.246...). Later, having woken up some more, I realised I could've just used x^2=1 (hell, even x=x).

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### Re: Infinite Sets and their Cardinalities

yeyui wrote:What I was said was intended to be (almost) the same as in the initial post.

In initial post it was proposed to take [imath]A \subset B \Rightarrow |A| < |B|[/imath] as an axiom. Well, to do that you must change the definition of one of those the symbols or else the axiom will cause a contradiction. [snip] If we accept the proposed axiom, the above definition, then the usual definition of cardinality will lead to a contradiction. [snip]

That contradiction is from the very definition of infinite cardinality, right? A set has finite cardinality if all proper subsets have different cardinality, and a set has infinite cardinality if there is a proper subset with the same cardinality. So insisting there are no proper subsets with the same cardinality as the improper subset would essentially be denying the existence of infinite cardinalities.

I propose, tongue in cheek, the following definition for infinity: That for which your intuition fails. But then, most of physics from last century would be infinity...
LOWA

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### Re: Infinite Sets and their Cardinalities

Nimz wrote:That contradiction is from the very definition of infinite cardinality, right? A set has finite cardinality if all proper subsets have different cardinality, and a set has infinite cardinality if there is a proper subset with the same cardinality. So insisting there are no proper subsets with the same cardinality as the improper subset would essentially be denying the existence of infinite cardinalities.

That's only equivalent to being finite (in the intuitive sense) with dependent choice. Something like [imath]X[/imath] is finite [imath]\Leftrightarrow \exists[/imath] an injection [imath]f: X \rightarrow \mathbb{N}[/imath] with [imath]f(X)[/imath] bounded is better, I think.
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