This Question attempts to get at the heart of a key distinction many have rambled about on here. Mods please see final section first.
Question
Below is an incorrect proof that the cardinality of the set of even numbers is greater than the cardinality of the set of natural numbers.
Question: What logic can be given to explain why this proof is incorrect, but cannot also be used to claim the standard diagonal $|\mathbb{R}|>|\mathbb{N}|$ proof is incorrect? Familiarity with that proof might be useful.
The “Proof”
Bijection from naturals to positive evens cannot be done. Proof:
Suppose a listing is proposed, naturals in order on the left mapped to some listing of even numbers, on the right:
$1 \mapsto E1$
$2 \mapsto E2$
$3 \mapsto E3$
$…$
Take every $E$ in the list and add $2$ to $n$th digit to the left of the of the decimal.
(Example if we have gone through the first three lines and our number is $284$, and $E4$ is $78$ then the new number is $2284$, because $E4$ has a $0$ four places to the left of the decimal. If $4 \mapsto 25476$ then $7284$, if $4 \mapsto 29476$ then $1284$)
If applied to the standard ordering of $1 \mapsto 2, ~ 2 \mapsto 4, ~3 \mapsto 6 ~…$ Then our new number is developing as: $… ~ 2222224$ continuing to the left as we go. Like the standard diagonalization proof; it yields a new number each time. Whatever line $n$ you go to, the new number will vary from that one in the $n$th place to the left of the decimal.
Applied to any listing that can be proposed, for any item we reach on the list, the new number will vary from that $n$th number in the $n$th digit to the left of the decimal. Therefore the new number varies from every number and is not on the list. Therefore the list is not complete.
The Question Again / Note
Again, the proof is incorrect. Explain how it is by using logic that cannot be analogously employed to contradict the normal proof for the cardinality of reals vs. naturals.
Thanks.
Possibly Related w/o Solution
I’m hoping the minefield of past Cantor ramblings won’t bias mods. A good answer to the above will go a very long way for us.
This questioner argues with “anti-Cantor Cranks” which I am not. In his whole argument, it seems like every statement defending the diagonal proof could apply to my above incorrect proof. What am I missing?
Refuting the Anti-Cantor Cranks
Also maybe slightly related: proving cantors diagonalization proof
Despite similar wording in title and question, this is vague and what is there is actually a totally different question: cantor diagonal argument for even numbers
Similar I guess but trite: Cantor's Diagonal Argument
This is more similar but solved by periodicity not applicable here: Why does Cantor's diagonalization not disprove the countability of rational numbers?
Best Answer
Even numbers (meaning, integers) have only finitely many nonzero digits when written in base 10.
Your procedure certainly produces a sequence of digits. But unless you know the digits produced are $0$ after a certain point, you will not obtain an even integer: you will just obtain an infinite sequence of digits which does not correspond to an even integer.
Under your procedure, in order for the sequence of digits to determine an even integer we need to make sure that there exists some natural number $N$ such that for every $n\geq N$, the $n$th digit of the $n$th number on the list is $8$. That way, when you construct your new sequence of digits, the resulting digit in position $n$ will be a $0$ after position $N$.
But consider the numbers $2$, $22$, $222$, $2222$, and so on. Since you start with a bijection between $\mathbb{N}$ and the even numbers, they must all be on the list. But since there are more than $N$ of them, for any given $N$, at least one of them appears after the $N$th term on your list.
That means that there is no $N$ such that for all $n\geq N$, the $n$th number on the list has $n$th digit equal to $8$: for every $N$, there is an $n\geq N$ such that the $n$th digit of the $n$th number is either $0$ or $2$. So the sequence of digits you produce contains $2$s and $4$s in arbitrarily "late" positions. They never become "all zero after this point."
This shows that the sequence of digits you are producing is not in fact an even integer. So the fact that it is not in your list of all even integers is not a surprise: it doesn't belong to the club, it doesn't have to be on the club's roll.
This doesn't occur in Cantor's argument because every sequence of digits corresponds to a real number between $0$ and $1$.
There is a misunderstanding here. The standard argument does not "produce a new number each time." The argument produces a sequence of digits, that we then put together to create a new number. It is that end result that we compare to the $n$th number on the list to conclude it is not equal to that.
A key thing that is not often mentioned (as Noah Schreiber notes) is that for any sequence $(d_1,d_2,d_3,\ldots)$ of digits (so $d_i$ must be one of $0,1,\ldots,9$), $$\sum_{i=1}^{\infty}\dfrac{d_i}{10^i} = 0.d_1d_2d_3\cdots$$ is definitely, absolutely, no question, a real number between $0$ and $1$. Any sequence yields a real number when you have the completed sequence.
Your process also yields a sequence of digits. But you are trying to put the digits $(d_1,d_2,d_3,\ldots)$ into a number by doing $$\sum_{i=1}^{\infty} d_i10^{i-1} = \cdots d_4d_3d_2d_1.$$
But this is a number if and only if the sum is finite: if and only if $d_i=0$ for all sufficiently large indices $i$. Otherwise, what you have is not a number. Numbers don't extend infinitely to the left of the decimal point.
That is a key difference between real numbers and integers: real numbers can extend infinitely to the right of the decimal point. Integers cannot extend infinitely to the left of the decimal point.
Your error is to think that the resulting object is a "number", let alone an "even number". It isn't. As I show above, what you produce cannot be a number at all.