Probability of observing no heads in infinite coin tosses In class, the professor went through a sample solution of showing why the probability of observing 0 heads in infinite coin tosses is zero, using the limit of decreasing sequences.
She defined $A_i$ to be the cone set of obtaining no heads in the first i tosses, and showed that $A_i$ is a decreasing sequence.
She then said that
$\lim_{i\to \infty} A_i := \cap_{i\in \Bbb{N}} A_i =$ {no heads ever}
I can see why the above is intuitively true, but is there a rigorous proof for this? To my understanding, $\lim_{i\to \infty} A_i$ is just a notation which is defined as $\lim_{i\to \infty} A_i := \cap_{i\in \Bbb{N}} A_i$, so then what I need to figure out is why
$\cap_{i\in \Bbb{N}} A_i =$ {no heads ever}
The way I tried to prove this is by showing that an element in one set is in another, but I could only do this for one direction. The direction is proved is that if $x$ is an element of the LHS, then it is an element of the RHS via contrapositive.
To do this, I said that if $x$ is not in the LHS, then it has a "head" at some point $n$. So then $x \in A_i^c$ and hence $x \notin \cap_{i\in \Bbb{N}} A_i$.
However, I'm not sure if this approach is correct, and I have no idea how to prove the other direction. Any help would be great!
Thank you!
 A: For the definition of set-theoretic limit I would just refer you to this wiki page.
We want to prove $\bigcap_{i\in\Bbb N}A_i=\{\text{no heads ever}\}$. Let us denote $\omega=\{\omega_i\}_{i=1}^{\infty}$, denote the outcome of a particular trial consisting of an infinite coin tosses, where $\omega_i$ is the outcome of the $i^{th}$ toss. Then $$\{\text{no heads ever}\}=\{\omega\mid\omega_i\ne\mathsf{H}\ \forall i\}$$ Let us call this set $S$.
Also $$A_n=\{\omega\mid\omega_i\ne\mathsf{H}\ \forall i\in\{1,2,\cdots,n\}\}$$ Then we really need to prove $\bigcap_{i\in\Bbb N}A_i=S$.
$\bigcap_{i\in\Bbb N}A_i\subseteq S$: Let us take $\omega\not\in S$. Then there exists $n\in\Bbb N$, such that $\omega_n=\mathsf{H}$. But then $\omega\not\in A_n$. Hence $\omega\not\in\bigcap_{i\in\Bbb N}A_i$. Thus any $\omega$ that doesn't belong to $S$, does not belong to $\bigcap_{i\in\Bbb N}A_i$ either. Hence $\bigcap_{i\in\Bbb N}A_i\subseteq S$.
$\bigcap_{i\in\Bbb N}A_i\supseteq S$: Let us take $\omega\in S$. Fix any $n\in\Bbb N$. We know that $\omega_i\ne\mathsf{H}$ for all $i\in\Bbb N$, and hence obviously $\omega_i\ne\mathsf{H}$ for all $i\in\{1,2,\cdots,n\}$. Hence $\omega\in A_n$. Since $n$ was arbitrary we can say that $\omega\in A_n$ for all $n\in\Bbb N$. Hence $\omega\in\bigcap_{i\in\Bbb N}A_i$.
