Question: Compare the cardinality between the sets $\Bbb R$ and $\Bbb N^{\Bbb N}$ I was comparing the sets $\mathbb{N}^{\mathbb{N}}$ and $\mathbb{R}$ for a fun practice because I didn't know at first that $\mathbb{N}^{\mathbb{N}}$ was uncountable. But how can I compare two uncountable sets and see which has the larger cardinality?
 A: A way to approach this might be to show that $\mathbb{R}$ and $\lbrace0,1\rbrace^{\mathbb{N}}$ have the same cardinality, and then find a way to compare the cardinality of $\lbrace0,1\rbrace^{\mathbb{N}}$ and $\mathbb{N}^{\mathbb{N}}$ (which seems to be easier).
The intuition is that the base-2 representation of any real nomber can be considered as an element of $\lbrace0,1\rbrace^{\mathbb{N}}$.
(But you'll see that in practice, althought we intuitively understand why $\mathbb{R}$ and $\lbrace0,1\rbrace^{\mathbb{N}}$ have the same cardinality, it's not that simple to build a bijection between the 2 and requires a good amount of work.)
Just a suggestion thought.
A: Here's an explicit bijection $f$ from $\mathbb N^{\mathbb N}$ to the half-open interval $[0,1)$ of the real line. I'll write elements of $\mathbb N^{\mathbb N}$ as infinite sequences of natural numbers, and I'll adopt the convention that $0\in\mathbb N$. Define $f(a_0,a_1,a_2,\dots)$ to be the real number with binary expansion
$$
.\underbrace{11\cdots1}_{a_0\text{ ones}}0 \underbrace{11\cdots1}_{a_1\text{ ones}}0 
\underbrace{11\cdots1}_{a_2\text{ ones}}0\cdots.
$$
Notice that, although some numbers have two binary expansions, one ending with all zeros and one ending with all ones, only the former will be produced by $f$, so $f$ is one-to-one. Its image clearly consists of all reals in the unit interval, including the endpoint $0=f(0,0,0,\dots)$ but excluding the endpoint $1$.
To get an explicit bijection from $\mathbb N^{\mathbb N}$ to all of $\mathbb R$ rather than just the unit interval, combine $f$ with your favorite bijection $g:\mathbb N\to\mathbb Z$, and send $(b_0b_1,b_2,\dots)$ to $g(b_0)+f(b_1,b_2,b_3,\dots)$.
A: $\mathbb{N}^\mathbb{N}$ is indeed not countable, as it includes a representation of $2^\mathbb{N}$,
the power set of $\mathbb{N}$, and power sets have a larger cardinality than the base set.
The continuum hypothesis is that there is no cardinality between the one of $\mathbb{N}$ and the one of $\mathbb{R}$, so 
$$
\DeclareMathOperator{card}{card}
\card(\mathbb{N}) = \aleph_0
< \card(2^\mathbb{N}) 
= \card(\mathbb{N}^\mathbb{N})
= \card(\mathbb{R})
= \mathfrak{c}
$$
if one accepts the CH.
A: Given two sets $A$ and $B$, if you have a bijection $f:A \rightarrow B$ then we say that they have the same cardinality. If $f$ is an injection then this tells us that $|A| \leq |B|$ where, for any $X$, $|X|$ represents the cardinality of $X$.
So, to answer your question, to show that say, $|A| < |B|$, we need to show that there is an injection $f \rightarrow B$ this is often not too difficult but we also need to show that there is no injection $g: B \rightarrow A$. This is usually a bit harder.
When we show that the reals are bigger than the natural numbers we use the diagonal argument to show that no such injection exists although I realise this example doesn't quite work as you asked about uncountable sets but the same principle applies: it's hard to show that an injection does not exist. It's a bit like in topology when, to show that two spaces are not equivalent, we have to show that no homeomorphism exists (that is why topological invariants are so useful in topology, they help with this.
