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”.
P is an infinite set, but every element of it is finite. So (f(x)) does not and need not map (\infty).
What you just said is that “(f(x)) does not and need not map endless”. Your sentence doesn’t even make sense.
Fine by me, you’re the one suggesting that it does need to map “endless”.
So (f(x)) being undefined at “endless” is also nonsense. So you objection to (f(x)) is nonsense, and (f(x)) is a bijection from P to N.
Anything for a word game, huh.
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.
Carleas, your recent argument has been that your x has no upper bound and thus no matter what value is plugged into it, there is always and forever another value that is higher, “x+1”. And that is what infinite means - that there is no upper bound, no end. The only problem comes in when you add to or multiply that infinite variable.
Yet when it comes to the idea that 1 = 0.999…, somehow the infinite string of 9’s gets treated as if an end was eventually reached, that it finally got to “infinity”.
My argument all along has been that the infinite string of 9’s doesn’t have an end, upper bound, or terminal destination, and thus cannot ever reach 1.0. It cannot get to infinity. Your argument favoring the use of your x in your function, is exactly the argument that I have been stating concerning the infinite string of 9’s. - There is no end to reach, thus the difference between the string of 9’s and the value 1.0 can never become 0. They can never be equal.
This is just a misunderstanding of how a variable works. Try to take what you’re arguing here and apply it to any other domain where we use functions. Are you really going to argue that every function on the real numbers is undefined?
There’s a distinction between the properties of a set and the properties of its elements. P is infinite, each element is finite. (x) is a variable that stands in for an element of P, not the set itself. For all (x \in P), (x) is finite.
I am not buying that you are that stupid nor that you think that I am, so let’s move on…
By the argument that you just gave and have given several times now, the function that yields 0.999… also maintains a finite variable that never reaches its infinite limit, because “infinite” mean that there is no limit to be reached.
And since that is so, the string of 9s never, ever achieves its limit and therefore can never equal 1.
There are no variables involved with 0.999… unless you use a series expansion to represent it and that’s not necessary.
1/3 + 1/3 +1/3 =1
0.333… + 0.333… + 0.333… =1
All fixed values!
No.
Acknowledge that for all (x \in P), (x) is finite.
P is 2 times infinite plus 1 long.
(P = {N, A})
(N = {1,2,3…} – ) infinite
(A = {a,1,2,3…} – ) infinite + 1
P could have been;
(P = {N_1, N_1, N_3,…})
Then P would be infinite² long.
And S could be;
(S_0 = {P_1,P_2,P_3…} – ) infinite³ long
(S_1 = {S_{01},S_{02},S_{03}…} )
(S_2 = {S_{11},S_{12},S_{13}…} )
(S_3 = {S_{21},S_{22},S_{23}…} )
.
.
And then
(T_0 = {S_1, S_2, S_3…} – ) infinite(^{infinite}) long
And each and every x could still be a finite number.
The problem isn’t x.
The problem is the range of x being doubled, squared, cubed, or added to in any way.
Your “2x+1” term is undefined because the RANGE of x is infinite and you don’t have a “2 * infinite + 1” in your math model. So you cannot fulfill the range of x before the term becomes meaningless. And that is why it fails to map P.
What would your x term be for set (T_0) ?
I know, I know … you don’t know what “range” means, right? Or is it that you don’t know what "2 * " means?
0.333… + 0.333… + 0.333… = 0.999…
As stated before,
1/3 leads to 0.333…
It does not equate to 0.333…
Because 0.333… is not a value or quantity.
This is a different definition of P. If this is the definition, then P is a set with two elements, so it’s cardinality is 2.
I assume this is a mistake, that you really mean that P is the union of N and A. But you’re playing fast and loose with your set construction, and it’s important because it is possible to create an infinity of a different cardinality (this is one of our few points of agreement, we just disagree on when that happens).
I know (range_J) is a concept you made up, from which nothing follows.
If it’s not a value or quantity, then what is it?
Different than what?
And I don’t care what “cardinality” you place on it. We are discussing the bijection between P and N.
And no, I don’t mean the union. You already tried that. P has two infinite subsets.
More BS. We aren’t even talking about cardinalities. And you would have to be very seriously stupid to not know what “range” means, especially since it has been explained several times.
It is an endless string of digits that results from trying to represent a fraction in decimal form. The endless string never reaches a satisfaction point where it would finally form an equivalence to 1/3. It is a misrepresentation of the fraction 1/3.
You’re confusing a map and the terrain. The “endless string” is the map. The map represents a fixed value.
It represents the exact value 1/3 in the real number system.
In a hyperreal number system , at best, it has a value 1/3 minus a fixed infinitesimal.
Different from ({1_A,2_A,3_A,…,a,1_N,2_N,3_N,…}). Your original set had an infinite number if elements. By your more recent definition, it has 2 elements.
A set that is an element of another set is not a subset. A subset is a collection of elements of a set. For example, if (Q = {1,2,3}), and (R = {Q,4}), (2) is not an element of (R), and (Q) is not a subset of (R) (it’s an element).
By contrast, if (W) is the union of two sets, (W = {1,2,3} \cup {4}), then (2) would be an element of (W), and (Q) would be a subset of (W).
These differences matter, James.
It has, poorly. But enough to show that it’s just a concept you pulled out of your ass about which there are no theorems that apply to the current conversation. Nothing follows from (range_J).
BS semantic games … again.
Face it. You are wrong.
There is no bijection between P and N.
Assuming you mean your original definition of P (since I agree there’s no bijection between a set of 2 elements and an infinite set), there is, the one I provided, the one for which you can’t point to an element that isn’t mapped.
Let’s try a different way.
Is it true that, if x is a natural number, 2x is a natural number?
Is it true that, if x is a natural number, 2x+1 is a natural number?