1

Let $(J, <)$ be a well ordered set. Let $S_\alpha \equiv \{ j \in J: j < \alpha \}$ ( section by $\alpha$ of $J$). Transfinite induction says that for any subset $J_0$, if the following property holds (that of being an inductive set):

$$ \forall \alpha \in J, (S_\alpha \subseteq J_0 \implies \alpha \in J_0) $$

then $J_0 = J$.

This can be proven by contradiction: Suppose such an inductive set $J_0$ exists and $J_0$ is not equal to $J$. Let $W \equiv J / J_0$ (for wrong) be the set of elements of $J$ not in $J_0$. $W$ is non-empty since $J_0$ is not equal to $J$. Let $w \equiv \min(W)$ be the smallest element of $W$, which exists as $W$ is non-empty and is well ordered (as a subset of $J$, a well ordered set). Since $w$ by definition is the smallest element that does not belong to $J_0$, the set $S_w$ of elements smaller than $w$ is a subset of $J_0$. Thus, we know that $S_w \subseteq J_0$. This implies that $w \in J_0$ as $J_0$ was assumed inductive. This contradicts the definition of $w$ as an element that does not belong to $J_0$. Thus, $J/J_0$ must be empty, or $J = J_0$ $\square$.

Doesn't this also prove induction, automatically, since transfinite induction is stronger than induction? So why do we need the Axiom of infinity, which asserts that's there is a set we can perform induction on?

However, this answer on math.se explains why induction needs to be an axiom in Peano arithmetic. This suggests that we must have an Axiom of Infinity.

I now know three "facts" which seem to contradict each other:

  1. Transfinite induction is stronger than induction.
  2. Transfinite induction can be proven in ZF.
  3. Induction must be an axiom in ZF (cannot be proven).

One of these must be wrong. I suspect the first one. I'd like a clarification.

Siddharth Bhat
  • 8,832
  • 1
    How would this "prove induction", if you don't even know the set $\mathbb{N}$ exists? That's what the axiom of infinity gives you. – Eric Wofsey Jun 17 '21 at 05:40
  • 3
    "Transfinite induction", in the form you have stated it, tells you something you can do if you have a well-ordered set. It doesn't tell you that any interesting well-ordered sets actually exist. – Eric Wofsey Jun 17 '21 at 05:41

1 Answers1

6

Induction is an axiom (schema) for PA, not an axiom of ZF, rather it is just transfinite induction applied to a special set, and the INF axiom merely guarantees that this special set exists. In effect, it guarantees that we can model the subject matter of PA (the natural numbers) in ZF. Without the INF axiom, you still have transfinite induction and can in particular use it to prove things about the ordinals $\emptyset,\{\emptyset\}, \{\{\emptyset\},\emptyset\}, \ldots$, which are the go-to example of a well-ordered class. However, you cannot prove that any of these ordinals is infinite. In other words, you cannot prove that the (minimal) inductive set guaranteed by INF is in fact a set (and not a proper class).

Hagen von Eitzen
  • 382,479
  • Indeed, if you take $X_0 = \emptyset$, $X_n = \mathcal P(X_{n-1})$ and $X = \cup_{n\in\mathbb N } X_n$, you can check that $X$ is a model for ZF minus the infinity axiom, in which every set is finite. – SolubleFish Jun 17 '21 at 09:32
  • Perhaps I am confused. I thought the axiom of infinity assets that a set $I$ exists, which has the empty set $\phi$ and for every $x \in I$, it also has $x \cup { x } \in I$. This seems to assert the existence of this special infinite set $I$, and that it obeys the induction principle? To me, INF seems to assert both the existence of this "special set" and an induction principle on it? – Siddharth Bhat Jun 17 '21 at 10:21
  • 1
    Induction merely states uniqueness of subsets of a well-ordered set, not existence of any particular well-ordered sets. – BrianO Jun 21 '21 at 08:49