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I'm trying to understand this math.SE answer by Captain Lama. The question is about how to prove that always either $|X| \leq |Y|$ or $|Y|\leq|X|$ for arbitrary sets.

Captain Lama firstly remarks that—due to the axiom of choice—the problem reduces to showing that if there is no surjective map $X\twoheadrightarrow Y$, then there is however an injection $X\rightarrowtail Y$. Now he defines a partial order $(A, f)\leq(B, g)\iff(A\subseteq B\land g|_A = f)$ on the set $E:=\{(A, f) : A\subseteq X, f : A \to Y\text{ injective}\}$. I guess he does this to apply Zorn's lemma, because as a next step he says:

Clearly this order is inductive (just take the union in a chain to get a supremum), so there is a maximal element $(A,f)$.

I don't know what "inductive" means in this context but I think he means that each chain has an upper bound (why does he speaks about "supremum"?). So my question is: How to prove that each chain $T\subseteq E$ has an upper bound? Does one need totality ($\forall x, y\in T: x\leq y\lor y\leq x$) in the proof? Captain Lama says that one should take the union. So the domain of $(A, f)$ is going to be $A = \bigcup_{t\in T}\pi_1(t)$, I guess, but what can we take as function $f?$

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"Inductive" here (I believe) means that every chain has a supremum (EDIT: but see Daniel Fischer's comment), that is a least upper bound; of course this is stronger than what we need to apply Zorn, but it's worth noting.

As to what the supremum is, exactly, you have to take the union of both coordinates of elements of the chain - that is, $$\left(\bigcup_{t\in T}\pi_1(t), \bigcup_{t\in T}\pi_2(t)\right)$$ where the union of a family $\{f_i: i\in I\}$ of functions is the set $\{(x, y): \exists i(f_i(x)=y)\}$. Note that this is only a function if the $f_i$s "agree" with each other; but they do in this case, since $T$ is a chain (so in particular, yes, one does need totality in the proof to guarantee that the union of the functions is a function).

It will follow immediately that this is an upper bound of $T$, and it's not hard to show that it is in fact the least upper bound of $T$.

Asaf Karagila
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Noah Schweber
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  • Thanks for your fast answer. Before accepting it, let me think about it a bit. I'm wondering: why do you use the notation ${f_i : i\in I}$ for a family? Concerning to the notation I've learned, a family is denoted $(f_i)_{i\in I}$ or $(f_i : i\in I)$, whereas your notation ${f_i : i\in I}$ with curly brackets denotes a set (note that if we have a family, we can construct the image of it, but a family contains more information than the image set). –  Feb 23 '17 at 16:58
  • @user419308 Often the notation "${f_i: i\in I}$" is used to denote "$(f_i)_{i\in I}$". But in this case, I only need the set, not the map from indices to functions - the point is that any two functions in the set need to not disagree with each other. – Noah Schweber Feb 23 '17 at 16:59
  • Mhm? If you have for each $i\in I$ a function $f_i$ this indicates that you have a function $i\mapsto f_i$ rather than a set. Or what do you mean? –  Feb 23 '17 at 17:00
  • Ahhh, thanks for clarification. I haven't heard of this kind of abuse of notation. –  Feb 23 '17 at 17:01
  • Bourbaki also defines an inductive set to be a partially ordered set that satisfies the hypothesis of Zorn's lemma when nonempty. In German, an "induktiv geordnete Menge" is precisely one that (when nonempty) satisfies the hypotheses of Zorn's lemma, so every chain has an upper bound, not necessarily a supremum. – Daniel Fischer Feb 23 '17 at 17:02
  • @user419308 I literally just mean the set of $f_i$s. I mean exactly what I wrote. – Noah Schweber Feb 23 '17 at 17:04
  • @DanielFischer Ah, I was unaware of that usage - I've seen "inductive" used to denote what I described above, in English, although I've only seen it so used once or twice. – Noah Schweber Feb 23 '17 at 17:04
  • @DanielFischer: Why "when nonempty"? In the hypotheses of Zorn's lemma one does not have to require that something is nonempty. –  Feb 23 '17 at 17:05
  • @user419308 You are right. The empty poset (if you consider it a poset) fails the condition of Zorn's Lemma - the chain ${}$ doesn't have an upper bound, since there are no elements to bound it! But the empty poset also doesn't have any maximal elements, so it fails the conclusion of Zorn's lemma too; so no exception needs to be made. – Noah Schweber Feb 23 '17 at 17:08
  • @user419308 I was misled by the wikipedia article that explicitly said "when nonempty", I wasn't quick enough to think of the empty chain guaranteeing nonemptiness of an inductively ordered set before posting the comment. – Daniel Fischer Feb 23 '17 at 17:08