Prove that the cardinality of $\Bbb N$ is less than the cardinality of $\Bbb R$ I understand there are different ways of showing this. One would be to use functions and mappings and another way is to express cardinality arithmetically? But if the sets are uncountable then it is not possible to express arithmetically?
 A: This is my favorite proof: 
Proposition: $\mathbb{R}$ is uncountable. 
Proof: Let define the map $f: \mathcal{P}(\mathbb{N}) \to \mathbb{R}$ by the formula $f(A):= \sum_{n\in A}10^{-n}$. Note that the map is well-define since $\sum_{n=0}^\infty 10^{-n}$  is absolutely convergent. 
We shall show that $f$ is one-to-one. Suppose to the contrary that there exists $A, B \in \mathcal{P}(\mathbb{N}) $ such that $f(A)=f(B)$ but $A \not=B$. Then the set $(A\setminus B )\cup (B\setminus A)$ is non-empty. Let $n_0$ be the least element and for concreteness assume that $n_0 \in A\setminus B $. So we have 
\begin{align}0=f(A)-f(B)=\sum_{n\in A}10^{-n}-\sum_{n\in B}10^{-n}=\sum_{n\in A:n<n_0}10^{-n}+10^{-n_0}+\sum_{n\in A:n>n_0}10^{-n}\\-\sum_{n\in B:n<n_0}10^{-n}-\sum_{n\in B:n>n_0}10^{-n}\\
=10^{-n_0}+\sum_{n\in A:n>n_0}-\sum_{n\in B:n>n_0}10^{-n}\\
\ge 10^{-n_0}-\sum_{n>n_0}10^{-n}\\
=10^{-n_0}-\frac{10^{-n_0}}{9}>0
\end{align}
a contradiction. Hence $f$ is an injection. 
Then $\mathcal{P}(\mathbb{N})\simeq f(\mathcal{P}(\mathbb{N}))$, i.e.,  $f(\mathcal{P}(\mathbb{N}))$  has the same cardinality as $\mathcal{P}(\mathbb{N})$ and by the Cantor's theorem  $\mathcal{P}(\mathbb{N})$ is uncountable. Since $f(\mathcal{P}(\mathbb{N}))$ is a subset of $\mathbb{R}$ then this forces $\mathbb{R}$ to be uncountable and in particular greater cardinality than $\mathbb{N}$. 
as was to be shown.
Edit: I've fixed some typos.
A: $\mathrm{card}(\mathbb{N}) < \mathcal{P}(\mathbb{N}) = \mathrm{card}(2^\mathbb{N}) = \mathrm{card}(\mathbb{R})$.  The first two are trivial.  For the last, there are two historical proofs (due to Cantor): the first one and the diagonal one.  Further discussion of the cardinality of the continuum is here.
