Infinite intersection of sets, induction

Question

I'm looking for quick clarification on the use of induction, as I'm confused about when it can and can't be applied to claims involving $\infty$.

First, the definition of $\bigcap^ \infty_{n=1} A_n$: the set containing all elements that are members of $A_n \forall n\in N.$ Is this correct?

If the above definition is accurate, can't we use induction to show that an element belongs to the infinite intersection of sets, since we're just making an argument of what's true for all natural numbers? If not, what am I missing? I've read many answers on why induction can't be used for arguments of infinity, but it seems as if the definition of an infinite intersection doesn't use infinity in the way a limit would.

Answer 1

Induction could potentially be useful in showing that $x \in \bigcap_{n=1}^\infty A_n$, but it might not be.

It would be helpful in the case that you can easily prove $x \in A_n \Rightarrow x \in A_{n+1}$, but you can not easily prove directly that $x \in A_n$ for arbitrary $n$. I can't immediately think of an example where this holds, but surely one exists.

It would not be necessary if you can just directly show that $x \in A_n$ for all $n$. For instance, if $A_n = (-1/n,1/n)$, then it's very easy to directly show $0 \in A_n$ for all $n$. Induction is not helpful in this case.

Answer 2

Important! Induction lets you say that something is true for every finite natural number, but you can't say anything about any infinite value.

It's a subtle difference.

For instance. If $A_n = (0, \frac 1n)$ then $\cap_{n=1}^{k}A_n = (0, \frac 1n)$ and $\cap_{n=1}^M$ is non-empty for any $M$ but $\cap_{n=1}^{\infty} A_n$ IS empty.

This is because although something is true up to all possible finite $M$ it isn't true for the infinite value $\infty$.

Another example is $\sum_{k=0}^N a_i \frac 1{10^k}$ is rational number (it's a terminating decimal). But $\sum_{k=0}^{\infty} a_i\frac 1{10^k}$ might not be. It could be an infinite non-repeating decimal.

....

So....

If you can find an $x$ so that if $x \in A_k$ than $x\in A_{k+1}$ and that $x \in A_1$ then by induction $x \in $ every possible $A_n$ and $x \in \cap_{n=1}^{\infty} A_n$.

And if you can prove that if $x \in \cap_{n=1}^k A_n$ implies that $x\in \cap_{n=1}^{k+1} A_n$ then (because that means $x \in A_{k+1}$) that $x \in \cap_{n=1}^{\infty} A_n$.

BUT if yo can prove that if $\cap_{n=1}^k A_n$ is not empty implies that $\cap_{n=1}^{k+1} A_n$ is non empty, you have proven by induction that $\cap_{n=1}^M A_n$ is non empty for any $M \in \mathbb N$. !!!!BUT!!! you did NOT prove that $\cap_{n=1}^{\infty} A_n$ is non empty because $\infty$ is not a natural number you can ever reach. Induction says you can reach every finite natural number but it doesn't say anything about reach any infinite value.