This is a follow-up to the question: [1] Paradox: Creating an uncountable set of natural numbers.
There is also a relation to this q: [2] Limits of sequences of sets. Are the limit points always included?
In this version I'm attempting to use a similar technique to create a set of real numbers incrementally, as a sequence of sets. The question is (a) is the technique valid, and (b) how to correctly compute the cardinality of the resulting set.
Goal: create the set of real numbers in the open interval $(0,1)$ by iteration: successively adding more digits after the "decimal" point and updating the set. Binary representation of the numbers will be used.
Step 0.
Start with empty set(s): $A_0=X_0=\{\}$.
Step 1.
Enumerate the first number 0.1; this is all numbers (in the interval) with 1 bit (which literally means binary digit) after the point. Collect these numbers in a set $A_1=\{0.1\}$, and likewise $X_1=\{0.1\}$.
Step 2.
Enumerate all numbers with 2 bits after the point. Collect the new numbers from this step in set $X_2=\{0.01, 0.11\}$. Collect all the numbers so far in $A_2=\{0.1, 0.01, 0.11\}$.
Step $k$.
In step $k$, for $k>0$, continue the previous enumeration, adding all the numbers with $k$ bits after the point and a "1" in the final position worth $2^{-k}$. Collect all the new numbers in the set $X_k$ and all the numbers in total so far in $A_k$.
Note that the cardinality (number of members) of $A_k$ is $2^k-1$ while that of $X_k$ is $2^{k-1}$.
Note that
(a) The sets $A_k$ can be viewed as a sequence of sets $(A_k)$ with increasing index $k=0,1,2,\dots$. Likewise for $(X_k)$.
(b) Embedded in $(X_k)$ are many sequences of (rational) numbers $(x_k)$ with increasing index $k=1,2,\dots$. One example is the seq. $0.01, 0.011, 0.0111, \dots$ which converges to $0.011111\dots=0.1$ (not a very interesting example).
(c) The successive cardinalities of $(A_k)$ can be viewed as a sequence of natural numbers $(s_k)$ with increasing index $k=0,1,2,\dots$; where $s_k=2^k-1$.
(d) It takes an infinite number of steps to reach infinite number of digits after the point, required for constructing irrational (and cyclic rational) numbers. So $k$ must approach infinity. Not only approach - it needs to reach infinity (such as $\omega$ or $\aleph_0$).
So now one would like to do $$ \lim A_k $$ In order to get a set $A_\infty$, or rather $A_\omega$ or $A_{\aleph_0}$, containing all the numbers.
But now you may argue: I have basically described an enumeration. Enumerations are inherently countable and not powerful enough to construct something uncountable, such as the real interval $(0,1)$. Having iterated for all [finite] $k$, all I will get is all the rational numbers between 0 and 1 such that their binary repr. is non-cyclic.
Here is my counter-argument.
First, let's not use the "set-theoretic limit" based on a discrete metric. Let's use the most general definition of "limit of a sequence of sets", and specify that we are using the standard Euclidean metric for convergence of numeric sequences.
Second, It is slightly easier to reason about $(X_k)$ than $(A_k)$.
Third, I will actually do $$ \limsup X_k $$ because $\limsup$ is the most liberal in accepting limits of all convergent subseqences of $(X_k)$, as members of the resulting set.
Fourth, observe that $ \limsup X_k $, according to its definition, does include any limit value that corresponds to any convergent subseq of numbers found across all $X_k$. Surely this is going to include an amount of irrational numbers.
Can I prove $$ \limsup A_k=(0,1), $$ as in containing all the real numbers in $(0,1)$? No. But I think I have a strong argument that $ \limsup$ of $X_k$ and $A_k$ at least contains more than just rational nums.
Regarding $\limsup A_k$, observe that every new member contributed to $A_k$ for some $K$ will remain for all $k>K$. Every such member is a constant number sequence that will survive in $\limsup A_k$ (as well as in $\liminf$).
That's a very long text again.
Repeating the questions.
Do you believe that this construction can yield all the reals in $(0,1)$? If not, why, and is there a correct way to get there using an exponential growth process? If not, is there another way?
How does one evaluate the cardinality of the final step, $\limsup A_k$? I suppose it is not by simply taking $\lim s_k$ as defined above (seq of successive cardinalities of $A_k$).
[Addendum 2024-01-15, sharpening the argument]
Can I prove $$ \limsup A_k=(0,1), $$ as in containing all the real numbers in $(0,1)$? No.
Later I came up with a formal argument in favor of this, that to me appears to work as a proof.
Consider any real number $x$ in $(0,1)$. In order to make it more interesting, consider in particular when $x$ is irrational, though this is not required for the argument.
For any $\epsilon>0$ there exists a $K$ such that for every $k>K$, there exists a number $x_k \in A_k$ such that $|x_k-x| < \epsilon$. This paraphrases the definition of a convergent number sequence. This also paraphrases the existence of a convergent number sequence within the sequence of sets $(A_k)$, as stated in the general definition of "$\liminf A_k$". Due to obeying said definitions, $x \in \liminf A_k$.
To see more clearly that such a $K$ always exists, let $\epsilon_K=2^{-K}$, and observe that this is the finest resolution in $A_K$, and therefore for any $x$ we will find an $x_K \in A_K$ such that $|x_K-x| \le \epsilon_K$. For $A_{K+1}$ and beyond, we can ensure that strictly less than will hold.
When the above is understood, it is easily understood that $x$ will likewise qualify as a member of $\limsup A_k$, so we do have $$ \limsup A_k = \liminf A_k = \lim A_k = [0,1] \text{ in } \mathbb{R}. $$ Note that by necessity the interval became closed: due to the workings of $\lim A_k$, limit points $0$ and $1$ must also become members.
I will furthermore claim that the same is true for $\lim X_k$. (The argumentation needs slight modification but to the same result.)
Example: Binary number $0.101001000100001\dots$ corresponds to convergent sequence $\{$ $x_1=0.1$ in $A_1$, $x_2=x_1=0.10$ in $A_2$ as well as $A_1$, $x_3=0.101$ in $A_3$, $x_4=x_3=0.1010$ in $A_4$ and $A_3$, $\dots\}$. I am "hand-picking" the numbers to fit the sequence I want, but this is within my interpretation of the rules of the game.
For $\lim X_k$, the same example can be turned into this sequence instead, where each steps always adds one significant digit to the string of digits: $\{0.1, 0.11, 0.101, 0.1011, 0.10101, 0.101001, 0.1010011, \dots\}$.
Why am I writing all this? Well, initially I wanted advice in understanding how this works, but then I came up with this addendum, turning into an argument against people saying that it is not possible to construct the reals using an iterative process like the one I described. Of course, I would like to be told if I made amateur mistakes in the reasoning. I challenge you, the reader, to prove me wrong!