Prove the set is uncountable

Question

We have the set $\mathcal{M}$ of all functions $f: \mathbb{N} \rightarrow \mathbb{R}$. This set is uncountable. How to prove it?

To find the proof, I supposed the contrary, but I cannot find a contradiction so far. Is the proof similar to Cantor's diagonal method?

Answer 1

The constant functions among these are already in bijection with $\Bbb R$.

Answer 2

The proof is something along this line: $|\mathbb{R}^{\mathbb{N}}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot \aleph_0}=2^{\aleph_0}>\aleph_0$. You know that $\aleph_0\cdot \aleph_0=\aleph_0$ since $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$ are in bijection; in fact, the standard embedding of $\mathbb{N}$ into $\mathbb{N}\times \mathbb{N}$ is injective, and the function $$f:\mathbb{N}\times \mathbb{N} \longrightarrow \mathbb{N}$$ $$(n,m)\longmapsto 2^n\cdot 3^m$$ is injective for the fundamental theorem of arithmetic. Thus, the Cantor-Bernstein theorem grants that there is a bijection between the two sets.