Unless you’re changing the definition of your sets, infinity is not a member of P, so we never have 2*infinity. Every element of P is finite, even though P is infinite.
Not so. For you to match EACH element in A while skipping every other number, your 2x must go from 2 to 2* infinity;
2x ( \to ) 2,4,6,8,10,… 2*infinity and those are paired with A (an infinite subset of P).
Infinity isn’t a natural number. (x \notin P: x = \infty). So, granted, (\infty) isn’t mapped by the bijection, but (\infty) doesn’t have to be mapped for it to be a bijection between P and N.
Are you trying to take the limit of the function to show that it’s not bijective?
P has to have an infinite number of elements, which it does. But infinity isn’t one of those elements. The mapped elements never reach infinity.
Take another bijective function which is clearly a bijection, (f:\mathbb{R} \to \mathbb{R} ) where (f(x)=\frac{1}{2^{x+1}}). I hope you’ll agree that’s a bijective function from the reals to the reals, but by your logic where you need to plug in infinity, it’s nonsensical. Fortunately, infinity isn’t in (\mathbb{R} ), so the function is still bijective.
Similarly here: infinity isn’t in the domain of the bijective function I provided. It fails to map infinity, but infinity is not in the set P. Every element in P is a finite, natural number, and is mapped by the function to N.
I never said that infinity is an element. Why do you keep saying that?
And yet your function term “2x+1” must reach beyond infinity.
In your math model, your function terms “infinite + 1” and “2 * infinity” are undefined, similar to (\frac{x}{0}). So your function certainly can’t reach 2*infinity+1, yet must pass infinity in order to map A and N. You are using undefined terms to map the subsets of P, as explained;
That was your stated bijection function.
Since both subsets A and N are infinite, what is your intended range for x?
Or back to the simpler case where P = {A,1,2,3…), again your index pointer must extend to infinite+1 in order to pair every element of P to N. Yet the term “infinite+1” is undefined (meaningless). So your function is actually undefined.
I certainly disagree. As I just pointed out, “x+1” for “x = 1 to infinity” is undefined. If you can add 1 to it, it wasn’t infinite. So what would be the range for x?
P goes beyond simple infinity. And again, I never said anything about “infinity” being an element. If I had two subsets in set K and both subsets had only 2 elements, the set K would have 4 elements even though neither subset has an element “4”. In the case of set P, neither subset A nor N has an element “infinity”, yet the set P has twice that plus 1.
You keep acknowledging that infinity is not an element of P, and yet you simultaneously insist that a function whose domain is the elements of P must be defined for (f(\infty)). Don’t you see the tension there? Infinity isn’t in P, P is the domain of (f(x)), so it isn’t a problem that (f(\infty)) is undefined: (f(x)) need not be defined for any element that isn’t in its domain.
When you say that the function must “pass” infinity, it makes me think that you are under the impression that a function between two sets must operate on the elements sequentially, i.e. in order to map an element, it first must map every preceding element. It doesn’t. (f(1_N)) is the same whether P is ordered ({a,1_A,1_N,2_A,2_N,…}) or ({1_A,2_A,…a,1_N,2_N,…}). (f(x)) does not depend on where (x) falls in P . Which should be obvious, because an element’s ordering in P is not a variable in (f(x)).
Is it your position that there’s no such thing as a bijection between infinite sets? If not, can you give a non-trivial example (i.e. not the identify function)?
Infinity is not a number and certainly not an element. It is a property. Your term “2x+1” must have a similar property in order to match the subsets that have that property. I have said nothing about “infinity” being an element. Stop claiming that I have.
No. I wasn’t talking about the order. The term “2x+1” must have the property of being greater than 2*infinity.
So again, what is your intended range for x?
If it is less than infinite, it will not map the subsets.
If it is infinite, your function is undefined.
There is not a bijection between sets of differing degrees of infinite. That should be pretty obvious to anyone.
Right, the similar property that both sets are infinite. For (x \in P: x \in N \implies 2x+1 \in N ). This is satisfied. There are an infinite number of elements in P, and an infinite number of elements in N, and for every element in P, (f(x)) maps it to N.
Disagree. There’s no element in N that’s “greater than 2infinity", so no element from P needs to be mapped to an element "greater than 2infinity”.
Let me try a different way: is 2x+1 a function that is defined for all natural numbers?
Do you mean for (f(x))? If so, N. If not, I don’t know what you’re asking for.
But you also said that (f:\mathbb{R} \to \mathbb{R} ) where (f(x)=\frac{1}{2^{x+1}}) is not a bijection, even though it’s a bijection from (\mathbb{R} ) to itself, so it’s clearly not a function between “sets of differing degrees of infinite”. Your only objection to it was that ( \mathbb{R} ) is infinite, and x+1 where x ‘goes to’ infinity is undefined.
So I ask again, what’s a non-trivial example of a bijection between infinite sets?
I think you’re improperly combining the concepts of limits and infinite sets by trying to inject a notion of ‘going to infinity’. Defining an infinite set does not entail any ‘going to’. For example, the natural numbers are defined by (roughly) taking a first element (either 1 or 0), taking as a given that each element has a successor, and taking as a given that each successor is also a natural number. That constructs the set of natural numbers, without relying on any notion of ‘going to’. N doesn’t ‘go to’ infinity, it is infinite and contains the natural numbers.
And it looks like I missed this from you previous response:
For finite numbers like 2, cardinality fits our intuitions. Our disagreement is whether those same intuitions work for infinite sets. My position is that infinity is counter-intuitive, and a set P that contains two countably infinite sets, is itself a countably infinite set of the same cardinality. Your position is that P is a bigger infinity. That’s our disagreement, and repeating your position is not an argument in favor of it.
When you insist that (f(x)) can’t be a bijection because it’s undefined for infinity, and (f(x)) is a function whose domain is the set P, i.e. it only needs to be defined for the elements of P, you are implying, whether you intend to or not, that infinity is an element of P. We seem to agree, it is not. Then we should also agree that it is irrelevant to the question at hand that P is undefined for infinity. Infinity is not in the domain of (f(x)), and it doesn’t need to be in order to be a bijection between P and N.
P has MORE THAN an infinity of elements. It has TWO infinite sets within, A and N.
N only has a simple infinity of elements.
You are stuck.
BS. Of course you know what I am asking. And it isn’t “f{x}”.
You specified a function using a range of x. I am asking what that range is. Without you specifying it, it is undefined, thus your function is undefined either way.
You are stuck.
And not that it matters but;
A “non-trivial example” would be the odd numbered naturals combined with the even numbered naturals being paired with the naturals.
I really don’t know what you’re asking for. Let’s define some terms, so we aren’t talking past one another. These are the definitions I’ve been using, I’m open to others but these seem to match the standard definitions:
The domain of a function (f(x)) is the set of all (x) for which the function is defined. So for example the domain of the function (\frac{1}{x-3}) is the real line excluding 3 (because it’s undefined at 3).
The range of a function is the set of all (f(x)), i.e. the set of outputs when given the domain as inputs. So for example the range of the absolute value function (|x|) is the positive real numbers.
The function is the operation on an element in the domain that maps it to an element in the range. Note that the function acts on the elements of a set, not on sets themselves nor on properties of sets.
So, for my (f(x)), the domain is P, and the range is N. An element by itself does not have a range, thus my confusion in response to your question, “what is your intended range for x?”, and why I assumed you mean the range for (f(x)). As I understand it, functions have ranges, elements don’t. If you’re using a different definition of ‘range’, you’ll need to define it.
Again, point to an element of P that is not mapped to N by (f(x)), i.e. an element of P that, when plugged in for (x) into the function I defined does not result in an element of N. Please keep in mind that infinity is not an element of P.
For clarity, it’d be helpful if you could specify the function in terms of (x), instead of just specifying the domain and range of your function. There can be many functions between two sets, so it’s generally not enough to specify the domain and range.
Here, it seems like this is the identity function. For every element (x \in T, f(x)=x). For example, the element 1 in T equals the element 1 in N. 2=2. 3=3. That is the function I intended to exclude by specifying a “non-trivial” bijection between infinite sets. If that’s not the function you intended, could you clarify?
To be fair, it now seems you interpreted function to be an operation on sets rather than an operation on elements of those sets. That might explain a lot of our disagreement here. As I’ve been using the term, a bijective function is an operation on each element of one set, and produces each element of another set.
Your function, f{x}, has a variable, x. That variable represents an index pointer to an element. x is not an element. Rather the variable x might be pointing to element 1 in set P, element 100, or element (14*{10^{1000}}).
So the variable x in your function HAS A RANGE from 1 to some greater number or infinity. Tell us what that upper bound for x is. It is your variable. You certainly know its intended range.
Claiming that you don’t know about your own function is a little silly.
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})