he infinite coin-toss space $\Omega_{\infty}$ of Example 1.1.4 is uncountably infinite. In other words, we cannot list all its elements in a sequence.
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1 General Probability Theory
To see that this is impossible, suppose there were such a sequential list of all elements of $\Omega_{\infty}$ :
$$
\begin{aligned}
& \omega^{(1)}=\omega_1^{(1)} \omega_2^{(1)} \omega_3^{(1)} \omega_4^{(1)} \ldots, \\
& \omega^{(2)}=\omega_1^{(2)} \omega_2^{(2)} \omega_3^{(2)} \omega_4^{(2)} \ldots \\
& \omega^{(3)}=\omega_1^{(3)} \omega_2^{(3)} \omega_3^{(3)} \omega_4^{(3)} \ldots,
\end{aligned}
$$
$\vdots$
An element that does not appear in this list is the sequence whose first component is $H$ if $\omega_1^{(1)}$ is $T$ and is $T$ if $\omega_1^{(1)}$ is $H$, whose second component is $H$ if $\omega_2^{(2)}$ is $T$ and is $T$ if $\omega_2^{(2)}$ is $H$, whose third component is $H$ if $\omega_3^{(3)}$ is $T$ and is $T$ if $\omega_3^{(3)}$ is $H$, etc. Thus, the list does not include every element of $\Omega_{\infty}$.
Now consider the set of sequences of coin tosses in which the outcome on each even-numbered toss matches the outcome of the toss preceding it, i.e.,
$$
A=\left\{\omega=\omega_1 \omega_2 \omega_3 \omega_4 \omega_5 \ldots ; \omega_1=\omega_2, \omega_3=\omega_4, \ldots\right\} .
$$
(i) Show that $A$ is uncountably infinite.
(ii) Show that, when $0<p<1$, we have $\mathbb{P}(A)=0$.
Uncountably infinite sets can have any probability between zero and one, including zero and one. The uncountability of the set does not help determine its probability.