Show $\left| \{H,T\ \}^{\oplus \mathbb{N} }\right| = \left| \mathbb{R}^{\mathbb{N} }\right|$ 
Let $\mathbb{R}^{\mathbb{N}} = \{ f: \mathbb{N} \to \mathbb{R}\}$ with the product $\sigma$-algebra $\mathcal{B}^{\otimes \mathbb{N}} = \bigotimes\limits_{n \in \mathbb{N}} \mathcal{B}$, where $\mathcal{B}$ is the Borel-$\sigma$-algebra. 

In our measure theory lecture we defined $\Omega := \{ H, T\}$ (symbolising head and tails).
Now, let $\mathcal{F} := \mathcal{P}(\Omega)$ be a $\sigma$-algebra on $\Omega$.
Then, the measurable space $(\Omega^{\mathbb{N}}, \mathcal{F}^{\otimes \mathbb{N}})$ models throwing a coin a countable number of times.
Now, we only stated and didn't proof the following


*

*$\mathcal{F}^{\otimes \mathbb{N}} \subsetneq \mathcal{P}(\Omega^{\mathbb{N}})$

*$|\mathcal{P}(\Omega^{\mathbb{N}})| = | \mathcal{P}(\mathbb{R}) | > | \mathbb{R} |$, where $ | S |$ denotes the cardinality of a set $S$.


Can anybody help me prove this with measure-theoretic means?
Any help is greatly appreciated.
 A: I think I already proved the first part here: A question in the book 'probability and martingales'
For the second part, it is not really related to the measure theory. It is clear that there exist a one-to-one correspondence between $\Omega^\mathbb N$ and $\{0,1\}^\mathbb N$. Moreover, $f:(x_n)_{n\in\mathbb N}\in\{0,1\}^\mathbb N\mapsto\sum_{n\in\mathbb N}\frac{1}{2^n}x_n$ is one-to-one and onto from $\{0,1\}^\mathbb N$ to $[0,1]$. So there is a one-to-one correspondence between $\{0,1\}^\mathbb N$ and $\mathbb R$, hence $\vert \mathcal P(\Omega^\mathbb N)\vert=\vert\mathcal P(\mathbb R)\vert$.
Let us now compare $\mathcal P(\mathbb R)$ and $\mathbb R$. Suppose that there exists an onto map $f:\mathbb R\to\mathcal P(\mathbb R)$. Let $A=\{x\in\mathbb R\mid x\notin f(x)\}$. Since $f$ is onto, there exists $x\in\mathbb R$ such that $f(x)\in A$. By definition of $A$, we deduce that $x\in f(x)\iff x\in A\iff x\notin f(x)$, which is nonsense. Therefore, there is no onto map from $\mathbb R$ to $\mathcal P(\mathbb R)$, which means that $\vert\mathbb R\vert<\vert\mathcal P(\mathbb R)\vert$.
