From Jech's Set Theory, we have the following theorems on cardinal exponentiation (a Corollary on page 49):
Theorem. For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$ is always either:
- $2^{\aleph_{\beta}}$; or
- $\aleph_{\alpha}$; or
- $\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.
Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.
Corollary. If the Generalized Continuum Hypothesis holds, then
$$\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl}
\aleph_{\alpha} &\quad & \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\
\aleph_{\alpha+1} &&\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\
\aleph_{\beta+1} &&\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.}
\end{array}\right.$$
So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is not the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).
Then again, GCH is usually considered "boring" by set theorists, from what I understand.
A binary sequence $(x_n)$ is just a function $x: \mathbb{N} \to \{0,1\}$. The $x_n$ is an alternative notation for $x(n)$.
In cardinal arithmetic $\kappa^\lambda$, for two cardinals $\kappa,\lambda$, is defined as the cardinal number of the set of all functions from a set of size $\lambda$ to a set of size $\kappa$.
So the size of your $B$ (all binary sequences) is, by this definition, $|\{0,1\}|^{|\mathbb{N}|} = 2^{\aleph_0}$
Best Answer
You have $|\mathbb{N}|$ options for as many slots, so $|\mathbb{N}|^{|\mathbb{N}|}$ different sequences.
There are several ways to prove that this is $|\mathbb{R}|$. I'll give one, but try to come up with another (hint, power sets).
Continued Fractions. Given a sequence $a_i$ we can map it to a continued fraction by $f(\{a_i\})=[0;a_1,a_2,\ldots]$. When the sequences are taken over the naturals, this is a bijection with the $(\mathbb{R}-\mathbb{Q})\cap[0,1]$. Since there are only countably many rationals, the irrationals in $[0,1]$ and $[0,1]$ itself have the same cardinality by the properties of cardinal arithmetic. It's well-known that $|[0,1]|=|\mathbb{R}|$, and several proofs are given here.