on cardinality of a set Set $S$ is a collection of disjoint sets each having cardinality of that of $\mathbb{R}$.The cardinality of set $S$ is also that of $\mathbb{R}$. $F$ is the set which is union of all the elements of $S$. Is the cardinality of set $F$ is equal to that of $\mathbb{R}$ ?
 A: You are asking whether $\mathbb{R} \times \mathbb{R}$ has the same cardinality as $\mathbb{R}$.  The answer is yes, thanks to the fact that $|\mathbb{R}| = 2^{\aleph_0}$ and $2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}$.  More explicitly, $\mathbb{R}$ has the same cardinality as the set $S = \{ 0, 1 \}^{\mathbb{N}}$ of binary sequences, and there is an obvious bijection $S \times S \to S$ given by "interweaving" sequences: that is, sending
$$(a_1, a_2, a_3, ...) \times (b_1, b_2, b_3, ...) \to (a_1, b_1, a_2, b_2, a_3, b_3, ...).$$
In general, the statement that $A \times A$ has the same cardinality as $A$, for $A$ infinite, is true for all alephs.  I think I have been told that whether this statement is true for all infinite sets is equivalent to AC.  
A: Yes. The cardinality of $F$ is $\sum\limits_{x\in S}|x| = \sum\limits_{x\in S}\mathfrak{c} = \mathfrak{c}\mathfrak{c}=\mathfrak{c}$.
In cardinal arithmetic, if $\kappa$ and $\lambda$ are nonzero cardinals, and at least one is infinite, you have
$$\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}.$$
