Suppose you have infite countable events which depend on an integer $n$, i.e. $A_1(n),...,A_k(n),....$ And you know for each event $\lim_{n\rightarrow\infty }Pr(A_k(n))=1$ for every $k\in\mathbb{N}$. What is the probability that all events happen simulaneously i.e. $\lim_{n\rightarrow\infty}\Pi_{k=1}^{n}Pr(A_k(n))$ ?
The context of my question is this one here:
Assume that for a probabilityfunktion $p(n)$ and random graphs $G(n,p(n))$ and a first order statement $A$ $\lim_{n\rightarrow\infty}Pr(G(n,p(n))\models A)=1$.
Then for every $A$, $\lim_{m,n\rightarrow\infty}Pr[(G(n,p(n)\models A)\iff H(m,p(m))\models A)] = 1$.
I now want to know the probability that $G(n,p(n))$ is elementary equivalent to $H(m,p(m))$. I.e. that $G(n,p(n))\models A \iff H(m,p(m))\models A$. For a single $A$ the probabilty approaches $1$, but can we say the same thing for all $A$s at the same time? There are infinite many countable $A$s and that is the background of my question.
Ultimately I want to show that $\lim_{m,n\rightarrow\infty}Pr[\text{Duplicator wins EHR(G(n,p(n)),H(m,p(m)),t)}]=1, \forall t$
I have to show this statement above in order to use the lemma which states:
$\text{Duplicator wins every EHR(G,H,t) forall t}\iff \forall A\in FO, G\models A\iff H\models A$
It can be a lot of different things.
Let $p_k(n)$ be the probability of $A_k(n)$. If $p_k(n)=1$ for all $k$ and $n$, then clearly the product is $1$. However, say that $p_k(n)$ is $0$ if $n<k$ and $1$ otherwise. For each $k$, $$\lim_{n\to\infty} p_k(n)=1,$$ since $p_k(n)$ is eventually $1$. However, for each $n$, $$\prod_{k=1}^\infty p_k(n)=0,$$ since all but finitely many terms are $1$. So, the limit of the product is $0$.
More generally, if you set $p_k(n)=1$ whenever $n\neq k$ and $p_k(n)=q$ when $k=n$, then the product at each $n$ is $q$ and so the limit is always $q$. So, you can make the limit be anything between $0$ and $1$ (inclusive).