3

This may be a silly question; it was occasioned by exercise 17 of Doets Basic Model Theory book, in which he asks us to prove that, for an arbitrary language $L$, there are at most $|L| + \aleph_0$ $L$-terms and $L$-formulas. The cases when$ |L|$ is finite or countable are clear enough, but I was wondering about the uncountable case. The proof I saw used the fact that, if $|L|$ is uncountable, then at least one of the sets of constants, function symbols, or relation symbols must also be uncountable. Doesn't this assume the Axiom of Choice, though? It seems to me that, without Choice, it's possible to have the cardinality of a countable union of countable sets be uncountable, which would make the above assumption false (or, at least, not so straightforward). Is this correct?

Nagase
  • 5,467
  • You must have misread that proof; I can't imagine how the statement "if $|L|$ is uncountable, at least one of the sets of constants, function symbols, or relation symbols must also be uncountable" could possibly come up in a proof that there are at most $|L|+\aleph_0$ $L$-formulas. – Eric Wofsey Jul 23 '15 at 02:32
  • @EricWofsey - the proof I was thinking of is in Hinman's Fundamentals of Mathematical Logic; he's proving that, for any infinite cardinality $\kappa$, if $|L|$ is $\kappa$, then so is both the set of atomic formulas and the set of formulas more generally. He says: "Suppose now that $L$ has $\kappa$ many non-logical symbols for some uncountable cardinal $\kappa$. Then at least of the sets $Rs_L$, $Fs_L$, or $Cs_L$ must alone have cardinality $\kappa$." (p. 94) – Nagase Jul 23 '15 at 02:37
  • That is proving the opposite of what your question is asking about, namely that there are at least $|L|+\aleph_0$ terms and formulas. In any case, the axiom of choice is not needed to show that a union of three countable sets is countable. – Eric Wofsey Jul 23 '15 at 02:40
  • @EricWofsey - He actually proves that the sets have exactly cardinality $\kappa$, which proves my claim, right? And thanks for the point about AC; I thought "countable" could be also finite, but, from your comment, I gather it means "countably infinite" in this context? – Nagase Jul 23 '15 at 02:44
  • He only uses the sentence you quoted in proving the "at least" part though. I don't understand your second question. Without choice, it can be shown that a finite union of countable sets is countable, but it cannot be shown that a countably infinite union of countable sets is countable. Here "countable" means "countably infinite or finite". – Eric Wofsey Jul 23 '15 at 02:46

1 Answers1

2

As I said in my comments, the axiom of choice is not used in the way you are saying it is. Nevertheless, choice is still used in the proof, namely in asserting that forming finite strings of symbols from $L$ does not yield a larger set than $L$ itself if $L$ is infinite. For arbitrary infinite sets $L$, this is a quite nontrivial result that cannot be proven without heavy use of the axiom of choice.

In fact, the theorem you are asking about is actually equivalent to the axiom of choice. For instance, let $A$ be any Dedekind-infinite set, and consider the language $L$ with a unary relation for each element of $A$. By considering formulas of the form $a(x)\wedge b(x)$ for $a,b\in A$, it is clear that there are at least $|A|^2$ $L$-formulas. If there are at most $|L|+\aleph_0=|A|+\aleph_0=|A|$ $L$-formulas, this means $|A|^2\leq |A|$. But by a theorem of Tarski, $|A|^2\leq |A|$ for all Dedekind-infinite $A$ implies the axiom of choice.

Eric Wofsey
  • 330,363
  • This is excellent! If possible, could you develop a bit more the point on your first paragraph, or perhaps point me to a source which discusses the details of that result, including where choice principles come up? – Nagase Jul 23 '15 at 02:54
  • 1
    @Nagase: The key theorem is that if $\kappa$ is an infinite cardinal, then $\kappa^2=\kappa$. This can be proven by transfinite induction on cardinals; see the first half of this answer, for instance. Once you have that theorem, it follows by induction that $\kappa^n=\kappa$ for all finite $n\geq 1$. There are $\kappa^n=\kappa$ strings of $n$ symbols from an alphabet of size $\kappa$, and so there are $\aleph_0\cdot\kappa\leq \kappa^2=\kappa$ total strings of any finite length. – Eric Wofsey Jul 23 '15 at 03:07
  • Got it. That was very helpful, thanks again! – Nagase Jul 23 '15 at 03:17
  • I have a further question about the countable case. In that case, you don't need choice to show that the set of all strings over $L$ is countable, because you can use directly something like Gödel's $\beta$ function or Cantor's pairing function to show that there's an injection from the set of all finite strings into $\mathbb{N}$, right? – Nagase Jul 24 '15 at 01:37
  • 1
    @Nagase: Right. More generally, there are many particular infinite cardinalities of $L$ for which choice is not needed (such as any well-orderable cardinality, and even some not-necessarily well-orderable cardinalities such as $2^{\aleph_0}$). – Eric Wofsey Jul 24 '15 at 01:46
  • Sorry, a final question: in the case of arbitrary infinite $\kappa$, do you also need choice to state that $\aleph_0 \times \kappa = \kappa$? It seems so because otherwise this multiplication could be undefined (though you presumably don't need choice to say that $\aleph_0 \times \kappa \leq \kappa \times \kappa$? Or do you also need choice to prove that multiplication of infinite cardinals is monotonic?). – Nagase Jul 24 '15 at 01:58
  • 1
    @Nagase: Choice is not needed to define the product $\aleph_0\cdot\kappa$ or show that multiplication is monotonic, but it is needed to show that $\aleph_0\leq\kappa$ (and hence that $\aleph_0\cdot\kappa\leq \kappa^2$). – Eric Wofsey Jul 24 '15 at 02:16