What do you mean by "The real decimals (base ten) are divided exactly in half at .5000..."? Are you considering just the unit interval? If so then there are only two naturals, 0 and 1, in the unit interval so you haven't got an argument. But if you mean that the real line can be divided "in half" at any point, that doesn't prove much since there are no natural numbers left of zero. Can you clarify your question?

I don't see where you're getting A = {1, 2, 3, 4, 5, ...}

Also what do you mean the reals are divided "exactly in half?" It's true that we can pick any arbitrary real x and biject the reals to the left of x and those to the right (with or without x in either subset). But "exactly in half" isn't well defined, since each subset has length infinity.

user 4894: the real decimals are all infinite decimals on the open interval (0, 1). They divide into (0, .5] and (.5, 1). Don't know what to call it but a division into halves. The associated naturals aren't 0 and 1 as endpoints of the interval, but the naturals associated with the form y.xxx... of real numbers. If the real numbers map onto the real decimals, then all naturals do too. I'm not sure why this isn't clear. Reals are just a product of the naturals and the real decimals, hence uniformly distributed over them.

Accumulation. There are infinite real decimals in the interval (0, .5]. Real decimals have the form y.xxx..., where y is a natural number. The reals map onto the real decimals, therefore so do the naturals. The infinite sequence of naturals is 1, 2, 3, 4, 5, ..., which must be associated with the reals mapping onto the interval (0, .5], otherwise you would have a finite number mapping onto (0, .5] and an infinite number mapping onto (.5, 1) with no logical basis for the distinction.

user 4894 #2. In the bijection between the reals and the real decimals, the real decimals function as a sort of divisional template. You are bijecting the sets as wholes (that is what "bijection" means). So if the real decimals divide in half at .5, the associated reals divide correlatively. Suppose the evens could be divided in half and N is in bijection with them, then a division would be produced on N and the first part of it would be {1, 2, 3, 4, ...} since it is infinite. E can't be so divided, but the real decimals can. Map a set onto a divided set and you induce a division in it.

@Nightspore Please preface handles you're addressing with '@' so we get notified. Ok so you divide (0,1) in half so let's say .1 is in the left half and .6 is in the right half. Now for each natural number such as 5, say, you have 5.1 and 5.5. So what happens next? Walk me through this step by step since your exposition's a bit murky.

@Nightspore "If the real numbers map onto the real decimals, then all naturals do too. I'm not sure why this isn't clear." It's not clear because it's false. The natural numbers do NOT map onto the real numbers. That's the famous Cantor diagonal argument. en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

@user 4894. If the reals are (by assumption) in bijection with the real decimals, then for every real decimal there is one and only one associated real. I have no idea, nor do I need to, how they are specifically related. All I need here is that the two sets are in bijection. For some decimal there is a real mapped to it. That's it. What I can say is that there are then two sets of reals created by this bijection, and the first, given the form y.xxxx..., contains the values of y collected in A. I am not obliged to get any more detailed than that: the division of naturals exists.

@user 4894. You are reading something into what I said that I never intended. The naturals do map onto the reals as respectively densely infinite aggregates constituting the value of the variable y in y.xxxx... So you have an infinite aggregate of 0's, of 1's, of 2's, and so on. Let's just name the infinite aggregate of 0's "0", of 1's "1", etc. All of these are mapped either to one or the other half of the division. We can count the respective aggregates: 0, 1, 2, ... Either an infinite number of such aggregates exists in both divisions or one is finite. But the latter has no rationale.

You are wrong because you are playing with the idea of a set being "half" of another set without giving a formal definition of that concept. The entire idea behind why axiomatic set theory was started because Cantor realized that imprecise language about how big sets are led to issues. That's why we have cardinalities. Take A={1,2,...} and B={2,4,...}. Intuitively, or informally, you might say "look, B is half of A" but that doesn't make sense, both sets have the same cardinality, half of A isn't in B and B isn't half as big as A.

"The real decimals (base ten) are divided exactly in half at .5000..." this is absolutely wrong. Both of those sets have the same cardinality, they both have the cardinality of the continuum, they weren't "divided in half". There are just as many natural numbers in both of those sets because they have the exact same cardinality as before, it doesn't matter what interval you take. That's the entire point of Cantor's theorem. Have you ever worked with infinite cardinals and ordinal arithmetic before?

Here's just something to think about and hopefully you understand why you're wrong. Can you create a bijection between {1, 2, 3, ...} and {2, 4, 6, ...} (I'll give you too big of a hint: f(x)=2x)? If you can, then your second paragraph about A and B makes no sense. All naturals are "accounted for" in {2, 4, 6, ...} because there's a bijection, and guess what my friend, the cardinality of a set with the cardinality of the naturals union a set with the cardinality of the naturals still has the cardinality of the naturals.

@Not_Here. Well one of us is wrong. I gave no cardinal interpretation to the word "half" and am quite aware of what you point out. That is not what I am saying. The division in the real decimals is purely sequential and as such immediately obvious: (0, .5] and (.5, 1). I don't know how to make that clearer. I never said anything about what "half of (0, 1)" means in terms of cardinality. In fact it is essential to my argument that the cardinalities are the same. Don't read what I didn't say.

@Nightspore What you're saying makes absolutely no sense, you are incorrect and it is very clear that you have at best a mediocre understanding of cardinality and infinite sets. The second half of your second paragraph where you talk about your "paradox" makes absolutely no sense and anyone who has had a basic education in cardinal and ordinal arithmetic can see why. I implore you to check out an introductory textbook if people's arguments here don't illuminate why you're incorrect enough for you to see it.

@Nightspore And you would understand why you're wrong if you were familiar with the subject.

@Nightspore "onto" is a technical term. Do you mean to say that the naturals map "onto" the reals? That means than we can map the naturals to the reals in such a way that every real gets hit. This is simply false, Cantor showed this can't be done in 1874.

@user 4894. Yes, you're right. I conflate two senses here that I shouldn't. The naturals as such do not map onto the real decimals. I was conflating that with the mapping of what i've called the respective infinite aggregates of 0's, 1's, 2's, etc. That was sloppy of me, but the basic point still stands. I have added a clarifying paragraph to the original question that addresses this. Rest assured I do know the difference. Thanks.

@Nightspore Ok. So now walk me through this, your exposition is very unclear. You have a real such as 5.1234... and that maps onto some real in (0,.5), say. Maybe it maps to 1/3. Now what? The phrase "respective infinite aggregates" is not clear to me, I do not know what you mean. Can you give an example?

@user 4894. The real numbers have the form y.xxxx... where y is a natural and .xxxx... is a real decimal. You start with 0 followed respectively by all the real decimals, then 1, 2, 3, etc. So you have an infinite aggregate first of 0's as the values of y, then 1's, and so on. All I am doing is making a simple point about the naturals being thus associated with the real decimals. Either the number of such aggregates in (0, .5] is finite or not. But it can't be finite. So all the naturals show up mapped onto (0, .5]. But then what naturals are in (.5, 1) as values of y? Paradox. That's it.

@Nightspore "Either the number of such aggregates in (0, .5] is finite or not. But it can't be finite. So all the naturals show up mapped onto (0, .5]. But then what naturals are in (.5, 1) as values of y? Paradox." dude this is exactly where you are wrong. Answer me this question, do yo understand that {1, 2, 3, ...} and {2, 4, 6, ...} have the same cardinality? Do you understand that {2, 4, 6, ...} can map onto exactly the same sets that {1, 2, 3, ...} can? Do you understand that {2, 4, 6, ...} union {1,3,5,...} has the same cardinality as either of those first sets???

It's not a paradox; you do not understand how infinite sets work. Just like what someone on math.se said where your question was already closed and dealt with, "infinity is weird", but it is not paradoxical, you do not have an actual contradiction, you haven't shown anything that changes anyone's views of infinity. All you've done is show that you don't understand bijections or infinite sets. Just as a thought experiment, given your original set up, why can't A ={2, 4, 6, ...} and B = {1,3,5,...}? What happens to your paradox then? Both sets have the same cardinality, there's no paradox.

@Not_Here. I swear to you I understand everything you are saying. I studied set theory in a doctoral program in philosophy. I was taught by a professional. I've done logical metatheory, etc. Please believe me, I get what you are saying. The problem I had on the math site is the one I'm having here. People are responding to basic points they think the argument misses, but doesn't, and not to what it actually says. It is extremely simple. I don't mind being wrong, but it's frustrating to be criticized for what I did NOT say. No offense,, but you're missing the point badly. Please try again.

@Nightspore nobody who has done any of the things you've claimed to do would make the statement that you can map the naturals onto the reals, nor any of the other plainly wrong things you've claimed to have proven in this post.

@Nightspore Your exposition is extremely difficult to understand, that's why people don't understand it. Try harder to be clear. I got as far as mapping, say, 5.1234... to some real in (0,.5) but I do not understand anything else about your exposition.

@user4894. Think of all the occurrences of "0." in the reals. An infinite number associated with the real decimals. Call it "P0" for "packet of 0's". Same for all the occurrences of "1." as values of y in the form y,xxxx..., call it "P1", and so on. Therefore we have, because of the bijection between the reals and the real decimals, the secondary association of P0, P1, P2, ... with the real decimals. That is just what a bijection between the reals as the product of the real decimals and the naturals means. Now just ask: how many such packets map to (0, .5]? Finite? Infinite? Can't be finite.

@Not_Here. Well I've done all I claimed to do, as well as teaching mathematical logic at a university level, and you're right: I wouldn't say what you say I'm saying. Problem is, what you say I'm saying isn't what I happen to be saying. I've lost all hope of getting through to you on that point. You are not getting what I am saying. Maybe it's time for you to move on to someone else. I simply do not know how to make what I'm saying any clearer to you.

@Nightspore Because what you're saying is nonsense, everyone including professionals have been trying to get that through to you. Three years ago you asked an extremely basic question about ordinals, there is no possible way that you went from not understanding that question to being qualified to teach mathematical logic at university level in three years, sorry. Maybe provide some sort of proof instead of just claiming things when the available evidence points the exact opposite way.

@Nightspore Oh I see. Ok, how many elements are in P0? It's the cardinality of the reals, right? Same for P1, P2, ... Now suppose we biject P0 to (0,.5) and then bijection the union of P1, P2, ... to [.5,1). Then only one packet got mapped to P0, right? Am I now understanding your construction? So what is the issue? ps -- The bijection of P0 to the left half of the unit interval can certainly be order preserving. Let me know if I'm understanding your construction.

@user4894. Yes, so far so good. As I posted in response to Daniel Mroz's answer, I can sustain any number of counterexamples, because the real problem concerns the possibility as such. What we seem to have is a situation in which ANY bijection leaves a finite number of packets in one half and an infinite number in another. Assume a bijection, the problem in general is how many packets can be moved to (0, .5] while preserving the mapping overall? The answer is either "all of them" or "a finite number". I reject the latter as a limit: the P-asymmetry (finite-infinite) has no rationale.

@Nightspore Well you can map the union of the even P-bundles to the left half of the unit interval and the union of the odds to the right half. But if you want the bijection to be order preserving, then it does seem like you have to map finitely many bundles to the left and the rest to the right. So ok, we understand that. What's next in your argument? What are A and B?

Regardless of the merits of your theory or your credentials or training, where is your question about philosophy? This seems baldly to be a non-formal proof provoking responses to both its structure and the truth of its conclusion rather than a question....

@virmaior. Fair enough. For me the question was implicit, since this is a work in progress, about the philosophical relation between an intension and its extension in the infinite case. My real interest in infinity has to do with the notion of inordinate greatness in the ontological argument. All this is subordinate to that, chiefly the problem of deriving categoricity from what seems to be pure hypothesis. A question is asked in hope of getting some help understanding a situation. I wanted a little feedback to help me refine this problem. Not all questions have to end with a "?".

Er, maybe you should read up on how this SE is supposed to work? (not all statements end in a period). It's not really well-suited to helping with philosophical works in progress. It's meant to help people who have questions about philosophy. See philosophy.meta.stackexchange.com/questions/474/… ...

@user4894. Picture the P-sequence as a uniform infinite array. "Uniform" because there is no convergent limit. Granting there may be no constructive way to do it, is there nevertheless an ideal possibility of dividing such a sequence in half, given that the real decimals are a template for doing so? What would it mean to say there is no way to divide it into symmetrically infinite halves, when it is in bijection with that template? We are then imposing something like a potentiality or constructive criterion on a pure ideal. A and B represent an ideal possibility: two equal (infinite) halves.

@Nightspore Please don't go off in a different direction, we're making progress. If A and B are the natural numbers left and right of the center point, you just convinced me that A consists of 1, 2, 3, ..., n; and B must be n+1, n+2, ... There is one exceptional case which is if one of the P-bundles straddles the center point. But then its leftmost element is the integer y.0 so that doesn't change anything.This all seems perfectly clear now and you are not explaining to me what you think the problem is. Please make an effort to be clear.

@Nightspore ps - As Daniel Mroz noted below, it's perfectly clear that A must be finite and B infinite. As you patiently walked me through your construction, you convinced me of this with perfect clarity. So I don't see why you're confused. If you have any order-preserving bijection between the nonnegative reals and the unit interval, it's clear that only finitely many natural numbers can be mapped to the left of any arbitrary point. There's nothing special about .5, the same analysis works for any point in the unit interval.

@Nightspore pps -- I see your core problem. Forget the reals. Say you have the natural numbers 1, 2, 3, ... and you want to partition them into two infinite sets. You can pick the evens and the odds, or the primes and the composites (arbitrarily throwing 1 into one set or the other), etc. But if you insist that the naturals stay in their original order, then one set must be 1, 2, 3, 4, ..., n, and the other set is everything n+1 and greater. This is the essence of the situation, the reals are a red herring.

@Nightspore ppps -- Re your irritation with virmaior, he is perfectly correct. SE is an awful venue for this type of conversation. You'd be far better off on any of the many philosophically/math/science-oriented online discussion forums. SE is for specific questions that have answers. The conversation you and I have had has been enjoyable for me as I have come to understand your idea. But this convo is 100% against the rules of this site. I don't like it but that's how this site is.

@Nightspore pppps -- I know what you want. You want a copy of the naturals followed by a copy of the naturals. That is the ordinal number ω + ω = ω2. That is NOT the order type of the naturals, which have order type ω. That is the perfect essence of your issue.

@user4894. Everything you say is correct, and that IS what bothers me. Both you and Mroz are putting the constructive case very well. No argument---except this thing about the uniformity of the sequence. The way I see it, applying (good, sound) constructive reasoning is logically equivalent to introducing something like a convergent limit. You CAN do it. It DOES make sense. But something essential is being brushed aside. A uniform sequence extends from one end of the unit line to the other, is in bijection with a divided template, but there is NO WAY of dividing it into equal infinite parts??

@Nightspore The SE software is already complaining about the extended discussion. I've certainly gotten to a full understanding and can't add anything. This has nothing to do with the real numbers. It's simply a manifestation of the fact that you can't subdivide ω into ω2 while preserving the order, because they are different ordinals. BY DEFINITION you can't equate ω and ω2 while preserving the order. You can only do it by changing the order as in listing evens before the odds as in 2, 4, 6, 8, ..., 1, 3, 5, 7, ... Nothing else I could possibly say.

@user4894. OK. Guess I'll have to pack up and find a site more amenable to my way of doing things. I was never all that good with rules and don't intend to start now. As for wanting omega plus omega, no, it's something else. But I suppose all that will have to wait. Odd that people think philosophical questions can just be asked and answered. Didn't they ever hear of a guy named Socrates?

@Nightspore Socrates would be banned from SE. (That's still a better fate than he got in real life!) Just how it is. SE was never intended for philosophical questions and philosophy is indeed a poor fit for the SE ecosystem. But still, you can't sensibly expect to partition ω into ω + ω and preserve the order. That's an unreasonable expectation since it's contrary to the definition of ordinals.

@Nightspore It's not an issue with the rules of the site, it's the issue that your question is nonsense and multiple people have explained thoroughly why your argument is incorrect. The fact that you can't accept that and are resorting to "I'm being censored, I thought this was a place to philosophy!" is highly indicative of this, it happens all the time here. Feel free to post any proof that you've taught mathematical logic at a university level though and maybe people will take your misunderstandings about ordinals more seriously.

@user4984. Well one last comment. I wanted to show that it is necessary, as an ideal possibility, for the division to be made in such a way that N, as intensionally defined, is completely in (0, .5]. That's the paradox. It has to do with the difference between ideality and constructivity in infinitary thought. My research into the concepts has led me to conclude that the introduction of constructivity changes the entire schematism. A lot of infinitary thinking is based upon what may be considered systematic equivocation between these modes. Thanks for the discussion.

@Nightspore Don't you see that the naturals are not "uniform?" They're asymmetric, finite on one end and infinite on the other. If you line up the naturals and slice them with a meat cleaver you'll always get finitely many on the left and infinitely many on the right.

@user4894. Yes I do see that. And in one sense, yes, you're right. I really do get it. But in ideal terms that isn't quite what is going on, or at least there is another legitimate way to look at it. There has to be. In fact, if you think about it, all of this, at a sufficiently fundamental level, HAS to end in paradox. I heard Hugh Woodin lecture on this not too long ago and even he's not sure of ANY of it. Actually paradox is the safest bet. You are moving from a constructive view of the sequence to an ideal one, seamlessly. That's the way it is often done, but it's dangerous too.

@Nightspore: You should take advice from experienced users more seriously. Yes, you may state your questions whatever way you like. But the understandability to others and the clarification of its philosophical content is in your responsibility. You will have to accept when others look at things differently.