Does this prove the union of countably many countable sets is countable?

Question

I am wondering if the following argument proves that the union of countably many sets is countable. Number each set in the union by a number of zeroes, assigning an arbitrary set $0$, the next set we pick $00$, the next $000$ etc. And then assign each element in each of the sets a number of $1$'s in the same manner- and then have an injection from this union and the set of all binary strings, by our function conjoining the set whence an element came and its position in that set to get a binary string (if an element is in multiple sets we arbitrarily pick one set and its position therein to map it to). Then this union is countable due to the set of finite binary strings being countable.

I am unsure because proofs of this statement I have seen, such as this, use an argument which is more complicated, leading me to believe there is a fallacy in this seemingly simpler argument.

https://math.stackexchange.com/q/2560968 Given you are taking the base to be countable sets, the sequence may have the length aleph zero or infinity. Which makes the binary sequence uncountable. — Gurmukh Singh, Sep 25 '23 at 01:06
@GurmukhSingh The set of all finite binary strings is countable. — spaceisdarkgreen, Sep 25 '23 at 01:10
But countable sets can have infinite length, forcing you to create a binary string of infinite length. There this argument would fall off — Gurmukh Singh, Sep 25 '23 at 01:13
If you replaced countable union of countable sets with a finite union of finite sets this would definitely hold true — Gurmukh Singh, Sep 25 '23 at 01:14
@GurmukhSingh When do they consider an infinite length sequence? Every element of the union corresponds to a finite sequence of 0's and a finite sequence of 1's. — spaceisdarkgreen, Sep 25 '23 at 01:22

spaceisdarkgreen · Accepted Answer · 2023-09-25T01:28:01.540

It seems fine to me. The usual argument is pretty much the same: we assign to each set in the union some $i\in \mathbb N,$ and to each element in each set we assign some $j\in \mathbb N,$ so each element of the union maps injectively to a pair $(i,j)\in \mathbb N\times\mathbb N.$

In either case we need to use the fact that either the set of finite binary strings, or $\mathbb N \times \mathbb N$ is countable, and also need to know that a subset of a countable set is countable. Adding proofs of these facts would add some more length to the proof.

Also, the way we've described the construction is a bit hand-wavy, so we might want to be a little bit more explicit at the expense of the proof appearing more complicated. For instance, we write the sets $(A_i : i\in \mathbb N)$ and then for each $i$, enumerate $A_i = \{a^i_{j}: j\in \mathbb N\}.$ (Note even here there are some questions about whether the enumerations are injective that we'd need to iron out if we were doing this in full detail, but ultimately it works out fine.)

There are a few more details swept under the rug. For instance, you identified there was a hiccup if two of the sets had the same element, so if we want to be careful, we should say that we map it to the $a_{ij}$ for the least $i$ such that it is a member of $A_i$. There is also the detail that in assigning to each $A_i$ an enumeration, we are using the axiom of choice (but this usage is unavoidable).

But as I said, the idea is fine... I'm really just trying to give some justification for why we might see some seemingly more complicated proofs out there.

Does this prove the union of countably many countable sets is countable?

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