$$\{f \mid f : \mathbb{N} → \{0, 1\}\} - \{f\mid f : \{0, 1\} → \mathbb{N}\}$$ Where $f$ is a function. Prove that this difference is an uncountable set.
I am pretty stuck on how to start this problem. I understand that the set from $0$ to $1$ is uncountable while the set of natural numbers is countable, but I can't seem to start off the proof.
The comments point out that the two sets we have are, in fact, disjoint, so we have that the cardinality of their difference is just the cardinality of the first set. Even without using this fact, we can still show that we have an uncountable set.
The set $\{f\mid f:\Bbb N\to\{0,1\}\}$ has cardinality $2^{|\mathbb N|}=2^{\aleph_0}$. To see this, notice that each element of $\Bbb N$ can be mapped to one of two elements, either $0$ or $1$.
The set $\{f\mid f:\{0,1\}\to\Bbb N\}$ has cardinality $|\Bbb N|^2=\aleph_0$. To see this, notice that each element of $\{0,1\}$ can be mapped to one of $|\Bbb N|$ elements, either $0,1,2,3,\dots$.
We start with $2^{\aleph_0}$ (uncountably many) elements and remove at most $\aleph_0$ (countably many) elements, so our set difference has at least $2^{\aleph_0}-\aleph_0=2^{\aleph_0}$ elements, i.e., it is uncountable.
