Find a bijection to show $\left|B\right| = \mathfrak{c}$. Let $B = \left\{ A \cup \mathbb{N}_\text{even} : A\subseteq \mathbb{N}_\text{odd} \right\}$
I need to show  $\left|B\right| = \mathfrak{c}$ by using an equivalence function (bijection) to another set with the same cardinality.
Any idea?
Thanks.

Let $\lambda: 2^{\mathbb{N}} \to B$ be defined by $\lambda(X) = (2X+\{1\}) \cup 2 \mathbb{N}$. It is fairly straightforward to show that this is a bijection, let $\phi_1 = \lambda^{-1}$.
Now let $\Omega = \{ X \subset \mathbb{N} | \exists n_0 \text{ such that } \forall n \ge n_0, \ n \in X \} \cup \{ \emptyset \}$. 
Note that $\Omega$ is countable, let $\omega_k$ be an enumeration.
It is straightforward to check that $\phi_3:2^{\mathbb{N}} \setminus \Omega \to (0,1)$ defined by $\phi_3(X) = \sum_{k \in X} {1 \over 2^k}$is a bijection.
Now define the bijection $\phi_2: 2^{\mathbb{N}}  \to 2^{\mathbb{N}} \setminus \Omega $ as follows:
$\phi_2(X) = \begin{cases} \{ 2n-1 \}, & X=\{n\} \\
\{2 n \}, & X = \{\omega_n\} \\
X, & \text{otherwise}
\end{cases}$.
The map $\phi_4: (0,1) \to \mathbb{R}$ given by $\phi_4(x) = \tan ((2x-1){ \pi \over 2} )$ is also a bijection.
Hence we have a bijection $\eta: B \to \mathbb{R}$ given by
$\eta = 
\phi_4 \circ\phi_3 \circ \phi_2 \circ \phi_1$.
