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Background: I was reading the following comment from the meta site:

"One can easily show that transcendental numbers exist, but to show that a single particular number is transcendental is a very different question."

The usual "easy" proof that transcendental numbers exist goes like this:

  1. The set of all algebraic numbers is countable, because there are countably many polynomial equations with integer coefficients and every such equation has finitely many roots.
  2. The set of all real numbers is uncountable.
  3. Therefore some real numbers are transcendental.

However, let $P_n$ be the set of all degree-$n$ polynomials with integer coefficients. Then $P_n$ can be enumerated in a way similar to the enumeration of rationals. The set of all algebraic numbers, $\bigcup_{n\in\mathbb N}\bigcup_{p\in P_n}\bigcup\{\omega:p(\omega)=0\}$, is therefore a countable union of countable union of finite sets. I have read that "it is consistent with ZF that $\mathbb R$ is a countable union of countable sets". So, without the help of some sort of choice, a countable union of countable sets is not always countable. I know only very little set theory, but this makes me question the validity of step 1 above.

Here are my questions: without assuming any kind of axiom of choice, is the proof above valid? If not, is Liouville's constructive proof (using $\sum_{k=0}^\infty 10^{-k!}$) already the easiest existential proof?

Ramen Nii-chan
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  • The theorem that says that the countable union of countable sets is a countable set already uses an axiom of choice, so the model you linked does not interfere with $\mathbb{R}$ being uncountable. – cnikbesku Jan 08 '25 at 16:58
  • You can prove that the algebraic numbers are countable because polynomials with integer coefficients are finite tuples of integers, which are countable. We can then use a Cantor argument to show that the reals are uncountable. You seem to think that expressing the reals as a countable union of countable sets is necessary; it is not. – John Douma Jan 08 '25 at 17:04
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    Liouville's is certainly more constructive/explicit than a cardinality argument. :) – paul garrett Jan 08 '25 at 17:13
  • @paulgarrett Pet peeve: the cardinality argument if written correctly is perfectly constructive and explicit. The differences are (i) the complexity/naturality of the reals the arguments produce, and (ii) the applicability of the arguments. Liouville's results in a simpler description of a real number, but also is less broadly applicable (e.g. the cardinality argument also works for numbers built from algebraic numbers and $x\mapsto e^x$, or etc.). – Noah Schweber Jan 08 '25 at 17:19
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    @paulgarrett (Technically my previous comment isn't quite right, since depending on the way we represent reals there may be issues, e.g. as in Bauer/Hanson. But the general idea that diagonalization is highly constructive is, I would argue, still true.) – Noah Schweber Jan 08 '25 at 17:23

1 Answers1

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There is no need for choice. Although ZF can't prove that a countable union of finite sets is countable, it can prove that a countable union of finite sets of real numbers is countable. This is just because the real numbers have a linear ordering (provably in ZF), so any finite set of real numbers is already ordered for you and you don't need to choose the ordering of each set when you do the proof.

(And there is no issue with showing that the number of polynomials with integer coefficients is countable either, as here we are just doing a count of finite sets of integers, which are well ordered already.)

spaceisdarkgreen
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    @RamenNii-chan Doesn't matter. The point is you have sets like $X_n = {a_n^0,\ldots a_n^{k_n}}$ where $a_n^0 < a_n^1 <\ldots < a_n^{k_n},$ and then you can put them into a sequence $a_0^0,\ldots, a_0^{k_0},a_1^0,\ldots, a_1^{k_1},\ldots$ and this is a surjection from the natural numbers onto $\bigcup_n X_n.$ (Alternatively you can just repeat each sequence infinitely many times and use Cantor's zigzag or something like that.) The part that requires choice in general is choosing the internal order of the finite sets, but with the reals (or any linearly orderable set) that is done. – spaceisdarkgreen Jan 08 '25 at 17:07
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    @RamenNii-chan That doesn't matter. For example, let's say our countably many sets are $$A_1={a^1_1<...<a^1_{n_1}}, A_2={a^2_1<...<a^2_{n_2}},...$$ The resulting surjection $\mathbb{N}\rightarrow\bigcup A_i$ is given by $$1\mapsto a^1_1, 2\mapsto a^1_2,..., n_1\mapsto a^1_{n_1}, n_1+1\mapsto a^2_1,...$$ – Noah Schweber Jan 08 '25 at 17:07
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    @RamenNii-chan Note that this doesn't show a countable union of countable sets of reals is countable, since the ordering (under the standard order of the reals) of a countable set of reals doesn't necessarily go like $a_0,a_1,a_2,\ldots$, it could be ordered like the rationals, or something more complicated, or whatever. Whereas the order of a finite set always looks the same. – spaceisdarkgreen Jan 08 '25 at 17:12
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    @RamenNii-chan I'm just talking about arbitrary finite sets of reals, not necessarily sets of roots of polynomials, and $k_n$ is the size of set $X_n$ (well, one more than the size since I started at zero... wish I hand't). So if it were sets of zeros of polynomials $k_n$ would be (one more than) the number of zeros of the given polynomial. – spaceisdarkgreen Jan 08 '25 at 17:16
  • Sorry, may I ask a follow-up question? Come to think of it, why is the set of all polynomials with integer coefficients countable? There are countably many possible degrees and for each fixed degree $n$, there are countably many degree-$n$ polynomials with integer coefficients. This is a genuine countable union of countable sets. Is countable choice needed to prove its countability? – Ramen Nii-chan Jan 08 '25 at 17:27
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    @RamenNii-chan You can represent a given polynomial as the sequence of its coefficients, which is a finite sequence of integers. The integers are countable, so can be put into the same order as the natural numbers e.g. $0,-1,1,-2,2,\ldots.$ Then the dictionary order (based on the above order) of the finite sequences will be an ordering of the same type as the natural numbers, so will provide a bijection between the finite sequences and the natural numbers. – spaceisdarkgreen Jan 08 '25 at 17:34
  • Thank you. I fully understand now. – Ramen Nii-chan Jan 08 '25 at 17:36
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    "Then the dictionary order of the finite sequences will be an ordering of the same type as the natural numbers". Not quite -- in fact the dictionary order of all the finite sequences is dense: no element has an immediate successor. The usual trick here is to start by grouping the polynomials by, say, their degree plus the sum of the absolute values of all the coefficients. Within each of those groups there are only finitely many polynomials, and the dictionary order is good enough for fixing the order internally in each group. – Troposphere Jan 09 '25 at 21:36
  • @Troposphere Yes, that’s right, thank you. – spaceisdarkgreen Jan 09 '25 at 21:48
  • (I was a bit too quick when I said no element has a direct successor, though -- they all have, but they don't all have immediate predecessors. And the order is not even well: Consider ... < 1112 < 112 < 12 < 2). – Troposphere Jan 09 '25 at 21:49