Can we use Proof by Induction to extend DeMorgan's Law to Infinite Number of Sets?

Question

Please do not mark this as a duplicate, I am asking something different. So I find myself confused with two differing opinions on this topic. Consider the question below,

I'm well aware of the fact that the term sequence has been used in the question, implying a countably infinite number of sets. Let us narrow down the discussion for a countably infinite number of sets for now. Regarding this, I have two opposing viewpoints:

1. We cannot use Proof by Induction to prove the following simply because there is no integer $n$ s.t. $n + 1 \rightarrow \infty$.
2. We can use Proof by Induction because Induction is indexed on the set of Natural Numbers $\mathbb{N}$, and there is a simple bijection from $\mathbb{N}$ to all the $A_{i}$'s.

I am stuck here. Which viewpoint is correct, at least, with regards to this question (framed in terms of a sequence). Am I missing something?

Answer 1

First, I don't understand what you mean by "there is no integer s.t. $n+1\to \infty$". So I may be unable to answer correctly, but here is my point of view: Induction will not allow you to show what you want. It will allow you to show that $$ \forall N\geq 1,\quad \left(\bigcup_{i=1}^{N}A_i\right)^c=\bigcap_{i=1}^{N}A^c_i. $$ Ok, now so what ? Well you basically have two family of sets $(B_N)_{N\geq 1}$ and $(C_N)_{n\geq 1}$ s.t. $B_N=C_N$ for all $N\geq 1$. So, you would like in some sens take the limit, but what does this mean ? It is not clear a priori. For instance, if you assume that $(C_N)_{n\geq 1}$ is a drecreasing sequence, you may define $\lim\limits_{N \to \infty} C_N=\bigcap_{N=1}^{\infty}C_N$, but you would be back to your original problem doing that.

Anyway, you really don't need to bother with that to show the desired relation. It can be done directly, showing that both set are included within each other.

Answer 2

While I don't see how you could use induction non-trivially to prove $\ \left(\bigcap_\limits{i=1}^\infty A_i\right)^c=$$\,\bigcup_\limits{i=1}^\infty A_i^c\ $, the reason given in your statement $1$ doesn't seem coherent enough to me to be convincing. The reason I would give for induction being inapplicable is that induction is used to prove propositions of the form $\ \bigwedge_\limits{n\in\mathbb{N}}P(n)\ $ where $\ P(n)\ $ is a predicate on $\ \mathbb{N}\ $, and I don't see how $\ \left(\bigcap_\limits{i=1}^\infty A_i\right)^c=$$\,\bigcup_\limits{i=1}^\infty A_i^c\ $, can be regarded as a proposition of that form in any non-trivial way. Since it's equivalent to $$ \bigwedge_{n\in\mathbb{N}}\left[\left(\bigcap_{i=1}^\infty A_i\right)^c=\bigcup_{i=1}^\infty A_i^c\right]\ , $$ however, you could, of course, give a perfectly valid, but very silly, proof by induction by first proving that $\left(\bigcap_\limits{i=1}^\infty A_i\right)^c=$$\,\bigcup_\limits{i=1}^\infty A_i^c\ $ is true for $\ n=1\ $, assuming (redundantly, since you've just proved it) that it's true for $\ n=m\ $, for some arbitrary natural number $\ m\ $, and then reproving it for $\ n=m+1\ $. I'd regard that as a trivial (and very silly) use of proof by induction. I wouldn't be prepared to assert categorically that a non-trivial proof of the statement by induction can't exist, but I haven't the foggiest idea what such a proof would look like.