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Given an alphabet $\Sigma$ containing countably infinitely many letters, Cantor's diagonal argument can be used to show that the set of unordered strings of infinite length without repeating letters which draw from this alphabet is uncountably infinite, since these "strings" are essentially just sets of letters, or subsets of $\Sigma$, making the set in question equivalent to the power set $\mathcal{P}(\Sigma)$ of the alphabet.

Diagonalization arguments also show that the number of ordered strings of infinite length which can contain repeating letters from a finite alphabet is uncountably infinite.

When trying to figure out, out of curiosity, the cardinality of the set $S$ of ordered strings which don't contain repeating letters from the same alphabet $\Sigma$ as before, I was assuming my answer would be $\aleph_1$, and tried to draw bijections to $\mathcal{P}(\Sigma)$ and the set of ordered strings without repeating letters. However, I couldn't find any.

I know that since the strings in $S$ are ordered versions of the strings in $\mathcal{P}(\Sigma)$, $S$ contains all the permutations of each string in $\mathcal{P}(\Sigma)$, so does this indicate that $|S|=|\mathcal{P(P}(\Sigma))|=\aleph_2$? And does the set of ordered strings with repeating letters but this time, drawing from the infinite alphabet $\Sigma$, have the same cardinality as $S$, a greater cardinality, or the same cardinality as the set of ordered strings with repeated letters that come from a finite alphabet?

I feel like I'm out of my depth, and would appreciate any help, thanks!

Edit: Note that this question is different from the question size of infinite strings and infinite alphabets because that question only considered strings where letters can repeat, and it considered: infinite strings from a finite alphabet, and finite strings from an infinite alphabet, not infinite strings from an infinite alphabet.

Hanul Jeon
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user17301834
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  • @TankutBeygu This question seems different because it's about the set of nonrepeating infinite strings. – user14111 Apr 04 '25 at 23:53
  • "Unordered strings" are called sets (or multisets if repetitions are allowed); "ordered strings'" are called strings or sequences. – user14111 Apr 04 '25 at 23:57
  • Let $\Sigma={a_1,a_2,a_3,\dots}$. Let $T$ be the subset of $S$ consisting of all strings where the first two letters are $a_1$ and $a_2$ (in either order), the next two are $a_3$ and $a_4$ (in either order, and so on. Note that there is an obvious one-to-one correspondence between $T$ and the set of all infinite (ordered) strings over the alphabet ${0,1}$. – user14111 Apr 05 '25 at 00:02
  • @TankutBeygu, I have ammended to question to show how it is different from the question "size of infinite strings and infinite alphabets". – user17301834 Apr 05 '25 at 07:55
  • @user14111, thank you for your clarification on "unordered strings", I am aware these are basically just sets, and have editted the question to make that more clear. I referered to them as unordered strings to more easily relate the set of strings without repeating letters as the set of permutations of each "unordered string". If this is confusing though, let me know and I will edit the question to use the correct terminology. – user17301834 Apr 05 '25 at 08:00
  • @user14111, I see how your method can be used to draw a bijection between $T$ and the set of infinite binary strings, which is uncountably infinite. I also see that I think you're suggesting that applying the same operation repeatedly can be used to shuffle the alphabet and reach any string in $S$. However, there are strings which can only be reached by repeating this operation infnitely, and one problem I also ran into when trying to find bijections from $S$ is that strings don't necessarily need to contain every letter in the alphabet. (Please correct me if I'm wrong) – user17301834 Apr 05 '25 at 08:10
  • Because of this, I'm unsure how to get from the subset $T$ to the entire set $S$. – user17301834 Apr 05 '25 at 08:11
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    I think it will help you to identify "the set of ordered infinite strings with no repeating letters" with "the set $I$ of injective functions $\Bbb N \to \Bbb N$" and "the set of ordered infinite strings" with "the set $F$ of functions $\Bbb N \to \Bbb N$" (both for solving this problem yourself, and for searching). The way to go here is to come up with injections $\mathcal P(\Bbb N) \hookrightarrow I \hookrightarrow F \hookrightarrow \mathcal P(\Bbb N)$. This shows all these sets have the same cardinality (Schroeder-Bernstein). user14111's comment was assuming you knew this type of argument. – Izaak van Dongen Apr 05 '25 at 08:24
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    By the way, the standard meaning of $\aleph_i$ is not related to iterated powersets. If you want to use Hebrew letters, the standard notation is "beth" $\beth_i$ - but it's also totally fine to just write $|\mathcal P(\Bbb N)|$ or $2^{\aleph_0}$. Also, you have a little error in your intuition - if choosing a string in $S$ corresponds to choosing an element of $\mathcal P(\Sigma)$ and choosing a permutation, that indicates that $|S| = |\mathcal P(\Sigma) \times \mathcal P(\Sigma)|$, not $|\mathcal P(\mathcal P(\Sigma))|$ – Izaak van Dongen Apr 05 '25 at 08:24
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    Thanks @IzaakvanDongen, your comments have really helped to clear up my confusion on user14111's comment, and on the line of thought related to permutations of the power set. I was struggling when trying to research this topic online (since I haven't learnt it in school), and I think these comments will help set me on the right track. – user17301834 Apr 05 '25 at 08:37
  • I see that multiple people have suggested that this question is a duplicate of the question "size of infinite strings and infinite alphabets". My question has been answered, but I have already explained how it is different from that question. Please read both questions before marking this as a duplicate. – user17301834 Apr 05 '25 at 11:06
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    Voting to reopen: neither of the suggested duplicate-originals contemplates a restriction to infinite strings of unique elements. The OP's main question was whether or not that restriction makes a difference, so it can never be a duplicate of something that doesn't even mention such a restriction. – Troposphere Apr 09 '25 at 08:15

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Since your alphabet $\Sigma$ is countably infinite, we can assume without loss of generality that $\Sigma$ is actually $\mathbb N$.

There is a simple bijection between the set of all sequences of natural numbers $(a_1, a_2, a_3, \ldots)$ and the subset of sequences $(b_1, b_2, b_3, \ldots)$ that don't repeat any number.

Namely, given $(a_1, a_2, a_3, \ldots)$, let $b_n$ be the $a_n$th smallest element of $\mathbb N\setminus\{b_1,b_2,\ldots,b_{n-1}\}$.

To invert this, let $a_n$ be the cardinality of $\{1,2,\ldots,b_n\}\setminus\{b_1,b_2,\ldots,b_{n-1}\}$.

So there are equally many sequences with and without the no-repeats restriction.

(Patching up any off-by-one errors in this argument, according to whether $0\in\mathbb N$ for you or not, is left as an exercise for the reader.)

Troposphere
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