It may be pertinent to frame the answer defining a random variable as a measurable function $X$ from the sample space of a probability space $(\Omega, \mathcal F, \mathbb P)$ to the space $(\mathbb R, \mathcal B)$ of the real numbers endowed with the Borel sigma algebra (i.e. the smallest sigma algebra which contains the open sets of $\mathbb R$), such that
$$X^{-1}(A)=\{\omega \in \Omega \mid X(\omega)\in A\}\in \mathcal F \;\; \forall A \in \mathcal B.$$
Constructing the sigma-field (or sigma-algebra) generated by two random variables will bring home the idea as follows:
The random experiment is tossing a fair die once, which has as possible outcomes or sample space $\Omega=\{1,2,3,4,5,6\}.$
It is up to the definition of the probability space to determine which events will have assigned probabilities. We could consider the events even or odd outcome and contrapose it to the events prime or not-prime. Thus we define two random variables: $X: \Omega \to \{0,1\}$ maps the outcome to $1$ if the die lands on an odd number, and otherwise $0.$ On the other hand, the second random variable $Y: \Omega\to \{0,1\}$ maps to $1$ when the die lands on a prime number.
Now these two random variables will generate two different sigma algebras. In the first case, the pre-image of the random variable $X$ will include the set $V=\{1,3,5\}$ corresponding to the value of the random variable $1$ in the Borel set $A_1=\{1\}:$
$$V=X^{-1}(A_1)=\{\omega\in\Omega \mid X(\omega) \in A_1\}=\{1,3,5\}$$
This reads: The pre-image (not the inverse function!) of the set $A_1$ for the random variable $X$ is the set $\{\text{...}\}$ containing those outcomes ($\omega$) in the big set $\Omega$ such that (that is the little bar $\mid$) the random variable maps to elements in $A_1.$ And the fact that $\{1,3,5\}\in \mathcal F$ (see below) makes the function $X$ measurable.
Logically, there would be the set $W =\{2,4,6\}$ corresponding to the pre-image of the value $0$ in the Borel set $ A_2=\{0\},$ or
$$W=X^{-1}(A_2)=\{\omega\in\Omega \mid X(\omega) \in A_2\}=\{2,4,6\}$$
The sigma algebra will also contain complementary, union and finite intersections, ending up with the sigma algebra $\mathcal F=\{\{1,3,5\} , \{2,4,6\}, \emptyset,\Omega \},$ of which $\{1,3,5\}$ and $\{2,4,6\}$ are called atoms (an atom of $\mathcal F$ is a set $F \in \mathcal F$ such that the only subsets of $F$ which are also in $\mathcal F$ are the empty set $\emptyset$ and $F$ itself). And the probability space will determine (in a fair die) that $\frac12$ is assigned to each one of them by the PMF: $\mathbb P\left(\{1,3,5\}\in \mathcal F\right )= \sum_{\mathbf 1_{1,3,5}} p(\omega_i )=3\frac 1 6.$
Here is the construct:
All Borel sets of interest have pre-images corresponding to sets in the sigma algebra $\mathcal F.$
If instead the events of interest were powers of 2 and powers of 3, the sigma algebra would contain $\small \left\{\emptyset, \Omega,\{1,2,4\},\{3,5,6\},\{1,3\},\{2,4,5,6\},\{1,2,3,4\},\{1,3,5,6\},\{1,2,4,5,6\},\{2,3,4,5,6\}\right\}$, the function simply mapping the face of the die to the numbers $1,\dots,6$ would not be a random variable, because the set formed by the pre-image of the singleton Borel set $\small \{5\}$ is not in the sigma algebra (not measurable): $\small \text{pre-image}(\{5\})=\{5\}\notin \mathcal F$.
Going back to the random variable $X,$ i.e. (odd, even), it will not contain information allowing separation of $1$ from $3$ for example, because they are both outcomes belonging to the same event ("odd").
The same exercise for $Y$ will result in two atomic sets $\{2,3,5\}$ and $\{1,4,6\},$ and a sigma algebra $\mathcal G=\{\{2,3,5\} , \{1,4,6\}, \emptyset,\Omega \}.$
This random variable $Y,$ in contradistinction to $X,$ contains information allowing us to separate the outcomes $1$ and $3.$
Best Answer
Our sample space $\Omega$ is the set of all possible infinite sequences (tuples) of Heads (hereafter shorted as $H$) and Tails (hereafter shortened as $T$). A sequence can be thought of like a set with the following major differences: elements can be repeated and order of elements matters.
Consider a smaller example. Consider the following sample space of all possible results of three coin flips in sequence:
$$X = \{(\omega_1,\omega_2,\omega_3)~:~\omega_i\in \{H,T\}\}$$
The notation above is interpreted as meaning in words as "$X$ is the set of all possible triples of the form $(\omega_1,\omega_2,\omega_3)$ where each element of a triple is either an $H$ or a $T$. This could have been written explicitly as the following:
$$X = \left\{(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)\right\}$$
Note that for example $(H,H,T)$ is a valid triple despite the fact that there are multiple $H$'s appearing and further it is considered a different triple than $(H,T,H)$ despite the fact that both have two $H$'s and one $T$ since the order in which they occur is different. Since the elements of our sample space are themselves sequences or tuples, we do not care if there are repeats within the sequences or tuples, just so long as the sequences or tuples themselves aren't repeated (and even that wouldn't be such a bad thing., we would just consider them as occurring only once).
Note, the difference in how these are enclosed. Sets are enclosed with curly brackets like so: $\{~~~\}$ while tuples and sequences are enclosed with circular brackets like so: $(~~~)$
Going back to your original example,
$$\Omega = \{(\omega_1,\omega_2,\omega_3,\dots)~:~\omega_i\in\{H,T\}\}$$
This, similarly to before is the set of all possible sequences where each element in the sequences come from the set $\{H,T\}$.
One such sequence might begin $(H,T,H,T,H,T,H,T,\dots)$ while another might begin $(H,H,H,H,H,H,\dots)$, etc...
You are getting closer however there is a property about $\Omega$ which your notation above gets wrong. When we begin to write elements and taper off with ellipses (the three periods in a row
...) this implies that the list of elements is not only infinite but countably infinite. Your attempt at notating this would have people look at a glance and think there are only countably many such possible infinite sequences of heads and tails. This is incorrect.There are in fact uncountably infinitely many infinite sequences of heads and tails. As such, we cannot even begin to list them in a pattern which would eventually list them all. A variation on Cantor's Diagonal Argument will prove that. Alternatively, consider replacing $H$'s by $1$'s and $T$'s by $0$'s and interpret each sequence as a sequence of binary numbers occurring after the decimal. You will have described every possible real number between $0$ and $1$ (some of which twice), again showing $\Omega$ has cardinality at least as great as the continuum.
As such, when referring to the tuples in our set, we should avoid using the ellipses and instead just use set builder notation as we had before. In the original notation, we had not decided to give an arbitrary sequence a label, opting to just refer to it as $(\omega_1,\omega_2,\dots)$, but if you insist on giving these labels then we could do it as the following:
$$\Omega = \{\omega~:~\omega = (\omega_1,\omega_2,\omega_3,\dots),~\omega_i\in\{H,T\}\}$$
In words that is $\Omega$ is the set of all elements $\omega$, where $\omega$ is itself a countably infinite sequence of elements each of whom are either $H$ or $T$.
Compare this to the way the original was phrased: $\Omega$ is the set of all countably infinite sequences of elements each of whom are either $H$ or $T$.
The descriptions are hardly different and the rewriting you proposed was not necessary.