Is 1 = 0.999... ? Really?

Your notation is a little unclear to me, so I’m going to make some assumptions, correct me if I’m wrong.

I assume you mean the element 1 in set A to be different from the element 1 in set N.
I assume you mean P to be the union of the two sets.
I’ll use (x_A) to mean x from set A and (x_N) to mean x from set N.

so P = ({a,1_A,1_N,2_A,2_N,3_A,3_N,…}), and (f(x) , P \to N):

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

This function uniquely maps every element in P to a corresponding element in N, and fully covers N.

No.

P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

That is the same set, just reordered. Just as both ({…,-3,-2,-1,0,-1,-2,-3,…}) and ({0,1,-1,2,-2,3,-3,…}) are the set of integers.

(f(x)) is a bijection, no matter how you order P.

It is not the “same set”. It is an ordered set. You cannot reorder it.

N = ({1_N, 2_N, 3_N …})
P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

What is your bijection function?

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

That (f(x)) is a bijection does not depend on the order.

1   2   3   4   5   6   7   ...
^   ^   ^   ^   ^   ^   ^
|   |   |   |   |   |   | 
|   |   |   |   |   |   |   
v   v   v   v   v   v   v   
A   1   2   3   4   5   6   ...   

The above is a visualization of the bijection between the sets (\mathbb N) and (\mathbb N \cup {A}).

Imagine that these are wooden blocks like a child plays with. Alphabet blocks. On each wooden cube we write one of the top row numbers on the block. So we have a set of wooden blocks labelled (1, 2, 3, \dots), one natural number for each block.

Now we come along and repaint the label on each block with the corresponding label from the bottom row. The block labeled (1) we repaint (A). The block labelled (2) we repaint as (1), and so forth.

Are not adding or subtracting any blocks or changing them in any way other than to label them differently.

There is a line of houses on one side of a street with addresses 101, 102, 103, 104, 105. One day the city tells them that their street is being converted to even numbers only, with the odd numbers across the street, and they’ll have to change their address.

Each resident goes down to the hardware store, buys some new decorative metal numerals, and nails the numerals to the front of their house. Now they read 202, 204, 206, 208, 210.

But the houses don’t change. Each house is identical before and after the readdressing, with the sole exception of the decorative numerals representing its address.

A bijection is nothing more than a relabeling or renaming of underlying objects that do not change.

So of course there’s a 1-1 correspondence between a given set before and after repainting. It’s the exact same set, completely unchanged except for a different label.

That function pairs only half of the set P with N (as if there was any other choice).

You have half of set A paired (with N).
And you have half of N paired (with N).

N = ({1_N, 2_N, 3_N…})
P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

…and it’s obvious.

You have a choice to pair the first half of sets A and N to N or to pair all of N to almost all of A.
But you cannot pair all of both A and N to N.

You are claiming that the same set N can be paired to itself as well as another infinite set. It is more than obvious that a set once paired with itself cannot also pair with more than itself.

(N = {1_N, 2_N, 3_N…})
(A = {a, N})

(P = {N, A})

Your function;
(f(x) = {N, a, N} \to N)

To pair P with N means to pair N with itself twice plus one more pairing for “a”.

Nothing can be paired with more than itself.

If that’s true, it should be trivial to identify a single, specific element in P that isn’t mapped to N. Can you?

That would be an erroneous hypothesis of yours. When dealing with infinite strings, one cannot point to what is happening “at the end” or beyond. That is what you use to hide the failure of your function to pair all members. And that is the failure of bijection as a principle.

Bijection cannot be used with infinite sets for that very reason. The word “infinite” is not well defined and thus cannot be precisely used in mathematics. Two sets can be infinite and yet of very different size or degrees of infinite (such as Aleph0 and Aleph1). Although once an infinite set is well defined, as in the hyperreals, normal reasoning applies again because math is merely logic applied to quantities.

You cannot know that you have paired an infinite set to anything other than itself. But you can know that once paired with itself, there is nothing left of it to pair with anything else. Do you disagree with that?

In Cantor’s day and prior, they didn’t have the hyperreals understanding.

There is no ‘at the end’; by their construction the natural numbers have no upper bound. The bijection I provided relies only on properties of the natural numbers, namely that every natural number has a successor (from which it’s straightforward to show that doubling a natural number also produces a natural number).

I don’t disagree. But the union of a set with cardinality (\aleph_0) and a set of cardinality 1 does not produce a set of cardinality (\aleph_1) (it produces a set with cardinality (\aleph_0)).

Yes, I disagree. A countably infinite set can be paired with a countably infinite subset of itself, so that e.g. the set of natural numbers can be paired with the set of even natural numbers (the bijection being (g(x) = 2x)).

Even using your ordering of P, all elements are mapped to elements of N. So
(f(1_N) = 2(1)+1 = 3 \
f(2_N) = 2(2)+1 = 5\
…\
f(a) = 1 \
f(1_A) = 2(1) = 2 \
f(2_A) = 2(2) = 4 \
…)
James, your objections here lack rigor. You’ve been presented with a proof that shows that the set of natural numbers has a bijection to the set of natural numbers plus (a), and you don’t like that conclusion. And you keep saying, in various ways, that you just really don’t like that conclusions. But that’s not a rigorous argument. The proof works, you’ve made no rigorous attempt to show otherwise, and you can’t identify an element that isn’t mapped by the bijection.

If you want to reject the conclusion, just point to a premise that you don’t accept (e.g., the construction of the natural numbers), or to a step in the proof that you don’t like (e.g. (x) is a natural number (\therefore 2x ) is a natural number), or any element for which it fails (e.g. (x=?) (can’t suggest one because one doesn’t exist)). Those would be rigorous responses.

Continuing to assert your preferred conclusion is not.

The other day I showed you the definition. A set is called infinite if it can be put into bijective correspondence with a proper subset of itself. Since I’m interested in math history I mentioned that this definition is due to Richard Dedekind. You responded by not only insulting me, which is par for the course; but also by throwing out some insults at Dedekind. That really makes no sense. It seemed a little off frankly.

But you can’t say infinite sets aren’t well defined in math because they clearly are.

Yes, infinite sets can have various cardinalities. Which does not alter the fact that infinite sets are perfectly well defined.

You’ve said that many times. “Logic applied to quantities.” But you have your own private definition of logic, and don’t even appear to accept the existential and universal quantifiers of predicate logic. But “logic applied to quantities?” It’s like “Better living through chemistry.” Sounds like it means something but it’s just a slogan.

We’ve established beyond all doubt at this point that the hyperreals don’t help any aspect of your argument.

And the hyperreals are a technical construction in modern set theory. You can’t reject modern set theory and predicate logic yet claim to believe in the hyperreals.

Pure BS and you know it.

Although I did figured out a way to “point to” an element that your function doesn’t pair with anything.

You use x as your pointer, a natural number, I presume. And you reference 2x and 2x+1. What happens when x goes to infinity? You have 2 * infinity.

So for 2x, x from 1 to infinity, your pointer is at;
2,4,6,8,10,… 2*infinity and those are paired with A

Then you have 2x+1, x from 1 to infinity;
3,5,7,9,11,… 2*infinity+1 and those paired with N

[b]What kind of number is “2x+1” such that it can go to 2*infinity+1?

And exactly what is “infinity+1”?[/b]

What is that a motto of your’s or something. Why do you join those who always accuse of their own guilt? You guys formed a club or something?

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).

And again, what is “infinity+1”?

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?

I didn’t say that it has to “map infinity”. I said it has to go 2*infinity in order to map A, and also to map the subset N, both within P.

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.