Let $(\Omega_\infty,\mathcal{F}_\infty,\Bbb P)$ be a probability space of infinitely coin-tosses. Let $A = \{\omega=\omega_1 \omega_2 \ldots \mid \omega_{2i-1} = \omega_{2i}, \forall i \in \Bbb N\}$. Show that $A$ is uncountable.
attempt: Suppose for contradiction, let $A$ be a countable set. Then, we can make a sequential list of all elements of $A$: \begin{align*} \omega^{(1)} &= \omega_1^{(1)} \omega_1^{(1)} \omega_3^{(1)} \omega_3^{(1)} \ldots, \\ \omega^{(2)} &= \omega_1^{(2)} \omega_1^{(2)} \omega_3^{(2)} \omega_3^{(2)} \ldots, \\ \omega^{(3)} &= \omega_1^{(3)} \omega_1^{(3)} \omega_3^{(3)} \omega_3^{(3)} \ldots, \\ \vdots \end{align*} Now, we construct an element \begin{equation*} \omega = \omega_1 \omega_2 \omega_3 \ldots, \end{equation*} where $\omega_{2i-1} \ne \omega_{2i}$, for some $i \in \Bbb N$. Clearly, $\omega \notin A$. On the other hand, $\omega$ is a sequence of infinitely outcomes of coin-tosses, a contradiction. Hence, $A$ is uncountable.
Does this proof look OK? In another way, I think we can make a bijection map, say, $f$, from $A$ to $\Omega_\infty$ defined by $f(\omega_1 \omega_2 \omega_3 \cdots) = \omega_1 \omega_3 \omega_5 \ldots$. But, I can't show that $f$ is a bijection mapping yet because a little bit confused. Any ideas? Thanks in advanced.
1.) Your proof seems good to me.
2.) Yes, you can. For the definition of $f$, it's clear that, $f$ is a one-to-one and onto map from $A$ to $\Omega_\infty$, since $\omega_{2i-1} = \omega_{2i}$ for all positive integer $i$. That is, \begin{equation*} f(\omega_1 \omega_2 \omega_3 \omega_4 \omega_5 \omega_6 \ldots) = f(\omega_1 \omega_1 \omega_3 \omega_3 \omega_5 \omega_5 \ldots) = \omega_1 \omega_3 \omega_5 \ldots. \end{equation*} Thus, $f$ is a bijection from $A$ to $\Omega_\infty$. Hence, the cardinality of $A$ is the same as that of $\Omega_\infty$. Since $\Omega_\infty$ is uncountably infinite, we must have that $A$ is also uncountably infinite.