Show that if $\mathcal{A}$ is a collection of inductive sets, then the intersection of the elements of $\mathcal{A}$ is an inductive set

Question

Show that if $\mathcal{A}$ is a collection of inductive sets, then the intersection of the elements of $\mathcal{A}$ is an inductive set

My Attempted Proof:

Let $\mathcal{A}$ be a collection of inductive sets. Then$$\mathcal{A}= \left\{A_i\right\}_{i\in I} $$

where $A_i$ is an inductive set and $I$ is an arbitrary indexing set.

Each $A_i$ contains $1$, and $\forall x \in A_i$ we have $x+1 \in A_i$.

$$\text{Put } \ \ \gamma = \bigcap_{A_i\in\mathcal{A}}A_i$$

Since $1 \in A_i$ for all $A_i\in \mathcal{A} \implies 1 \in \gamma$.

Now put $x = 1$. Since $1 \in \gamma \implies x+1 = 2 \in A_i$ for all $A_i \in \mathcal{A} \implies x+1 \in \gamma$

Now put $x = 2$. Since $2 \in \gamma \implies x+1 = 3 \in A_i$ for all $A_i \in \mathcal{A} \implies x+1 \in \gamma$

Continuing this process recursively we can see that $1 \in \gamma$, $x \in \gamma$ and $x+1 \in \gamma$ for all $x \in \gamma$. Thus $\gamma$ is an inductive set. $\square$

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Firstly is my proof correct? If so how rigorous is it? If it is correct and fairly rigorous, then any suggestions on how to improve it? I feel it is a bit hand-wavy at the moment

Answer 1

The idea is generally correct. Two points, and one caveat:

1. "Continuing this process recursively" is not particularly formal. It's fine when you're not fully expected to formalize something (e.g., in a paper when the claim is easy enough, or when a full formal statement will be unreadable). But since you've asked for feedback, this should be pointed out.

   The way to formalize "continuing this process recursively" is by induction. Which is a good segue into the next point.

2. Induction is not what you need here. You want to show that $1\in\gamma$, and that if $x\in\gamma$, then $x+1\in\gamma$. This is not a proof by induction, but rather proving directly from the definition of inductive sets and intersections.

One caveat, however, is that the intersection should be over a non-empty collection. If the context is bounded in $\Bbb R$, however, then the intersection over an empty collection is $\Bbb R$ in which case you're covered and everything is fine. It's just something to note, though.

Answer 2

OP asked for a non-hand-wavy improvement. Here is how I would prove the same statement without mention of recursion (which seems to be the core of your question):

Recall

• A set $A$ is inductive if $0\in A$ and for each $n\in A$, $n+1\in A$.

Proposition. For each nonempty collection $ \mathcal{C} $ of inductive sets, $ \bigcap\mathcal{C} $ is inductive.

Proof. Let $ \mathcal{C} $ be an arbitrary nonempty collection of inductive sets. Therefore $ 0\in\bigcap\mathcal{C} $. To prove $ \bigcap\mathcal{C} $ is inductive, let $ n\in\bigcap\mathcal{C} $ be arbitrary. Therefore for each $ A\in\mathcal{C} $, $ n\in A $. Therefore for each $ A\in\mathcal{C} $, $ n+1\in A $. Therefore $ n+1\in\bigcap\mathcal{C} $. Therefore $ \bigcap\mathcal{C} $ is inductive.