For a proposition $P$, if one inducts using only standard induction (not transfinite induction) in order to show that $P(n)$ is true $\forall n \in \mathbb{Z}^{+}$, then can one claim $P(\infty)$ (where "$\infty$" represents countable infinity)?
For example, I know someone who was trying to prove: "Let $f: X \to Y$; let $A_i \subseteq X \forall i \in \mathbb{Z}^{+}$. Then $f(\cup_{i = 1}^{\infty}(A_i)) = \cup_{i = 1}^{\infty}(f(A_i))$". I know that this can be done for arbitrary unions, so I am not worried about actually proving the assertion. My problem is that they inducted over finite unions (in the foregoing statement, replace "$\infty$" with "$n$", where $n \in \mathbb{Z}^{+}: n > 1$) in order to get the desired infinite or limiting result (I am not sure that it is meaningfully limiting, so pardon the uncertain wording). They proved the analogous claim for any pair ($n=2$) of sets and then did a standard induction (not transfinite induction) argument in order to generalize to any positive integer $n$, other than $n = 1$ (which would be a trivial union anyway). They believe that this is enough to prove the claim for the infinite countable union. But, to me, this proves only the claim for finite unions. Which of us is correct?
How would one induct in order to get the analogous claim about countable unions? Does that require transfinite induction (treating $\infty$ here as $\omega$)?
Clearly, one cannot always do this. Just define $P(n)$ to be $n \in \mathbb{Z}^+$. Then clearly, we can show $\forall n \in \mathbb{Z}^+ (P(n))$ using induction, but obviously $P(\infty)$ doesn’t hold.
I don’t know the exact details of your acquaintance’s proof, but the approach you outlined is on its face not valid. Perhaps there is some clever argument which generalises from finite to infinite unions, but they’d have to articulate this.
In this case, the direct proof is quite trivial - in fact, probably more straightforward than an induction proof.