Don’t you think it’s strange that, while you insist that there are infinity elements in P that aren’t mapped to N, you can’t name one of them? You can’t describe any property of them. You can’t define the set of those elements.
Let’s take a step back and talk about the elements of P, because I think part of your confusion is in the purposely unusual way you’ve described that set, and in your seeming belief that the way P is ordered affects whether or not it has the same number of elements as N. Do you think that P has more elements when it’s ordered ({1_A,2_A,3_A,…,a,1_N,2_N,3_n,…}) than when it’s ordered ({a,1_N,1_A,2_N,2_A,3_N,3_A,…})? How about if it were ordered ({…,3_A,2_A,1_A,a,1_N,2_N,3_N,…})? I hope you can see that the identity function is a bijection between these three sets (i.e. orderings of the set P), so they must have the same number of elements. Do you agree?
Oh I can, just as soon as you actually define the range of your x.
Because you have now stated that the range of your x has been infinite, It is clear that the second half of both infinite subsets A and N are not mapped by your function. You use x to begin mapping both A and N, but then your terms “2x” and “2x+1” become undefined while half of each subset is still unmapped.
The problem has been merely that you include an undefined term “2 * infinite + 1”.
The purpose of the order was to help expose the failure of your function. Other than that, it doesn’t really matter. But since you have used an undefined term in order to handle the order of P, it doesn’t really matter what else is going on.
You haven’t identified an actual bijection function yet.
Using your idiosyncratic meaning of ‘range’, I think the answer is P. (x) can be any element from P.
Where does the “second half” of the natural numbers start? Better yet, name one element in the “second half” and we can plug it in and see what happens.
Again, since infinity isn’t in P, it isn’t a problem that (f(\infty)) is undefined.
Glad we cleared that up.
Do you agree that the question of creating a bijection between P and N is the same as the question of creating a bijection between (\mathbb{Z}) and (\mathbb{N}), i.e. the integers and the natural numbers? The bijection from P to (\mathbb{Z}) is:
(f(x)=\begin{cases}0&x=a\-x& x \in A \land x \neq a\x& x \in N\end{cases})
Can you just define what you mean by range? How is it different from domain? Why are you plugging the Jamesian range into (f(x)), when (f(x)) is expressly defined as an operation on the elements of P and nothing else? And why don’t you know it, why do you need me to tell you what it is, and why can’t you name an element of P that doesn’t get mapped until you know it?
Quite so. And yet, it’s pretty straightforward to define an (f(x)) that maps one uniquely to the other. Infinity is counter-intuitive, after all.
I have several times. Your effort to escape through pedantics isn’t going to work with me. You should know that.
No. It isn’t “counter-intuitive”. You, as well as very many in the past few hundreds years, merely accepted an absurdity as “Our Real Truth”. You can’t identify a bijection between the integers and the naturals for the same reason that you can’t for P and N.
One of the variety of explanations that I provided:
And now, I think we have had enough of this.
You didn’t and can’t provide a bijection between the provided P sets and the set of natural numbers. You can’t because the term “infinity” or “infinite” is not well defined such that the issue of greater degrees of infinite, “infinite + x”, would have any meaning. Your older simple-minded math model doesn’t allow for greater degrees of infinite other than what Cantor provided with Aleph1. He caught hell to his grave from people like you merely for that one example.
Later, I’ll look back to find out why any of this has been relevant to the topic…
For me part of the problem is that dividing by zero itself creates a problem - it is undefined. Dividing by anything greater than zero could not possibly give infinity - infinity is undefined.
I would say that it is grossly or vulgarly defined, not completely undefined.
InfA and H are sufficiently defined to allow normal math operators to make sense. But “standard math” holds to the word and concept of “infinite” to mean merely “endless”, without concern for what degree of endlessness is involved. That distinction didn’t gain authority in mathematics until around 1947 with Hewitt and still isn’t being taught much to many even today. And because of that, many simple minded paradoxes are accepted as nonintuitive truths which are in fact merely poor semantic conflations (e.g. “They are both infinite therefore they are equal length”).
Haha, yeah, I’ve forgotten how we got down this rabbit hole as well. But I think we got to the same place during our last back-and-forth in this topic, so it seems like it matters. And I find it interesting, and trying to convince you has improved my understanding, as it always does.
So I don’t really care why it’s relevant, we’re here and you’re wrong and I intend to fix that.
This is a pretty unclear definition. As I asked before, can you distinguish the concept of the “range of x” from the concept of the domain of (f(x))? Why couldn’t you provide it? What does it tell you about the elements of P that you don’t otherwise know? Do you have some theorem about how it relates to… anything? What’s the use of this concept?
I think I see how you’re getting confused.
You have invented a concept of the “range” of a variable in a function (and it’s distinct from both the common meaning of a “range” of a function and the domain of a function; I’m going to use (range_J) to distinguish it from the common meaning of “range” in set theory)).
(range_J) is something like the largest and smallest numbers that get plugged into (f(x)).
Since in this case x has no upper bound, you say the (range_J) is (1 \to \infty ).
(here’s where the confusion comes in) Since you’ve decided that ( \infty) is the ‘largest number that gets plugged into x’, you assume (\infty) is in the domain of (f(x)).
You note, correctly, that the function is undefined for (\infty).
You conclude that the function can’t be a bijection because it’s undefined for certain important values.
The mistake in here is that, whatever the (range_J) is, however important it may be (big clue here to its lack of importance is that it’s being invented here as opposed to in a academic article or similar serious forum), it isn’t the domain, and P having no upper bound does not entail that (\infty) is in the set P nor that (f(x)) needs to be defined for (\infty). It isn’t, and that doesn’t affect whether or not (f(x)) is a bijection.
Still waiting. Keep in mind that (\infty) isn’t an element of P.
You never provided a defined bijection function and I explained why you couldn’t. So asking me to point to something very specific that it didn’t do is nonsense. What does graphiticality not refer to? Your terms were undefined and undefinable in your math model. Your function was nonsense due to the lack of specificity concerning the word “infinite”.
I have explained this many times now. I said before that you never stop bantering, regardless of how wrong you have been shown to be. But for me, you have been sufficiently shown the flaw in your thinking whether you admit it or not.
So unless you come up with a new confusion, this has been enough.
So you propose to map an infinite set with a finite variable?
Your function is undefined because you refuse to define its terms, specifically the range of x, although you let it slip that x is “unbound”, aka “infinite”. But if x is infinite, “2x” and “2x+1” are undefined in your antiquated math model.
P has TWO infinite sets. You must have at very least one infinite variable to even try to map it. But when you do that, your function becomes incomprehensible, aka “undefined”.
What I said was that your x must be endless. And you agreed by saying that it is “unbound”, aka “infinite”. Thus if x is infinite, the term “2x+1” is undefined. And that means that your function is undefined.
