How many partitions of the set $\Bbb Q$ are there? 
Let $\mathcal S:=\{P\mid P\text{ is a partition of }\Bbb Q\}.$ Find $\operatorname{card}(\mathcal S).$

My attempt:
$\boxed{\leq}:$ I believe each $P\in\mathcal S$ may be determined by a surjection $f_P:\Bbb Q\to\Bbb N,$ that is, each $B\in P$ is of the form $f_P^{-1}(n)$ for some $n\in\Bbb N$ and, because $f_P$ is surjective, there is no empty set in the partition $P.$
Let $\mathcal T:=\{f:\Bbb Q\to\Bbb N\mid f\text{ is surjective }\}.$ Then $\mathcal T\subseteq\{f:\Bbb Q\to\Bbb N\}\implies\operatorname{card}(\mathcal T)\le\aleph_0^{\aleph_0}=\mathfrak c$ because
$$\mathfrak c=2^{\aleph_0}\le\aleph_0^{\aleph_0}\le\mathfrak c^{\aleph_0}=\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\mathfrak c.$$
Therefore, $\operatorname{card}(\mathcal S)\le\mathfrak c.$
$\boxed{\geq }:$ Let's define $\Phi:\mathcal P(\color{blue}{[0,1]\cap\Bbb Q})\to\mathcal S,\Phi: A\mapsto\Phi(A)$
$$\Phi(A):=\begin{cases}\boldsymbol{\{}\Bbb Q\setminus(\color{blue}{\Bbb Q\cap [0,1]}\cup\{2\}),A,\{2\}\cup \color{blue}{\Bbb Q\cap [0,1]}\setminus A\boldsymbol{\}}, &A\ne\emptyset,\\\boldsymbol{\{}\Bbb Q\boldsymbol{\}},& A=\emptyset.\end{cases}$$
I think $\Phi(A)$ is well defined since all its elements are disjoint, non-empty and their union is $\Bbb Q.$
I took out $\{2\}$ in order for $\Phi(A)$ and $\Phi([0,1]\cap\Bbb Q\setminus A)$ to be different according to the axiom of extensionality and for $\Phi$ to be injective.
Hence $\mathfrak c=2^{\aleph_0}\operatorname{card}\mathcal P(\Bbb Q\cap[0,1])\le\operatorname{card}(\mathcal S).$
Finally, $\operatorname{card}(\mathcal S)=\mathfrak c.$
I would like to verify my work. Is my answer valid?

I've found a post about the cardinality of the set of all partitions of $\Bbb N$ where some answers appear to use similar approaches for one inequality.
