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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.

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    $\begingroup$ It's not clear to me how $A$ is defined. $\endgroup$ – user98602 Sep 1 '14 at 19:22
  • $\begingroup$ Oops. Let me correct this please. $\endgroup$ – Elimination Sep 1 '14 at 19:22
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    $\begingroup$ Show that the power set of $\mathbb{N}$ has cardinality $\mathfrak{c}$. $\endgroup$ – copper.hat Sep 1 '14 at 19:22
  • $\begingroup$ You use the same symbol $A$ for the set to be defined and for the definition $\endgroup$ – Andrea Mori Sep 1 '14 at 19:23
  • $\begingroup$ @copper.hat, I have to use an equivalence function (bijection) which is onto other set with the cardinality of the the continuum. $\endgroup$ – Elimination Sep 1 '14 at 19:24
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HINT: Note that there is a bijection between $B$ and $\mathcal P(\Bbb N_{\rm odd})$.

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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 $\eta = \phi_4 \circ\phi_3 \circ \phi_2 \circ \phi_1$.

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