TITLE: Prove that the cardinality of the set of all transitive relation on $\Bbb N$ is $\mathfrak c$
QUESTION [3 upvotes]: How do I prove the following statement? 
I have an idea of how to go here, but need a further explanation - 
Firstly, I'll define the set of all transitive relation on $\Bbb N$ as $T$.

First of all, prove that the cardinal number of the set $\mathcal P(\Bbb N\times \Bbb N) = \mathfrak c$. That is easily done.
Define a function : $f\colon \mathcal P(\Bbb N\times\Bbb N) \to T$, as follows:
$R$ is in $\mathcal P(\Bbb N\times \Bbb N)$. If $R$ is transitive, $f(R) = R$, otherwise $f(R) =$ the closest transitive relation that contains $R$.
That function splits $\mathcal P(\Bbb N\times \Bbb N)$ into different departs, according to the result $f$ sends them to. I need to prove that the cardinal number of those results is $\mathfrak c$.
There is a $1\leftrightarrow1$  function between $T$ and the group of results of $F$. therefore, cardinal number of $T$ is equal and is $\mathfrak c$.

I can't figure out how to prove point number 3. 
(Sorry for my english)

REPLY [2 votes]: For each $A\in\mathcal P(\Bbb N)$ we can define the map $1_A\colon \Bbb N\to\{0,1\}$, $n\mapsto\begin{cases}1&n\in A\\0&n\notin A\end{cases}$ and then the transitive relation $R_A$ by $n\mathop{R_A}m\iff 1_A(n)<1_A(m)$. Note that $R_A\ne R_B$ for $A\ne B$, except that $R_\emptyset=R_{\Bbb N}$.
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