Cardinality of the union of disjoint sets, each of which have a cardinality of reals What can be the Cardinality of the union of disjoint sets, each of which have a cardinality of reals? How should this be proved.
I know using Schroder bernstein theorem, it is easy to see that the cardinality of the union must be equal to the cardinality of reals. But without using it, how should this be proved?
I have an idea:
[0,1) has the same cardinality as that of reals. Similarly, [1,2) has the same cardinality as reals. Similarly a semi-open interval [n,n+1) has the same cardinality and they are all disjoint. So, can I combine them and simply say that their cardinality is equal to reals? Is this correct. I am looking at a more formal proof along this line.
 A: Map one of the sets to
$(-\infty, 0)$,
one to
$[0, 1]$,
and one to
$(1, \infty)$.
(added as requested)
Therefore,
each element of 
each of these sets
gets mapped into
a unique real,
so the cardinality of their union
is that of the reals.
A: One (extreme) option is to use the Axiom of Choice to prove that $\kappa+\lambda=\max\{\kappa,\lambda\}$ when at least one of $\kappa$ or $\lambda$ are infinite cardinals.  From this, it follows that given $A$ and $B$ disjoint sets with $|A|=|B|=|\mathbb{R}|$ that
$$|A|+|B|=\max\{|A|,|B|\}=\max\{|\mathbb{R}|,|\mathbb{R}|\}=|\mathbb{R}|$$
More generally, given any family $\mathcal{A}$ whose elements have cardinality of the continuum and where $|\mathcal{A}|\leq 2^{\aleph_0}$, then
$$\left|\coprod_{A\in \mathcal{A}}{A}\right|=\sum_{A\in \mathcal{A}}{|A|}=\sum_{A\in \mathcal{A}}{2^{\aleph_0}}=|\mathcal{A}|\cdot 2^{\aleph_0}=2^{\aleph_0}$$
using the related fact that $\kappa\cdot \lambda=\max\{\kappa,\lambda\}$ when at least one of $\kappa$ or $\lambda$ are infinite cardinals.
When $|\mathcal{A}|>2^{\aleph_0}$, the result no longer holds true, for then 
$$\left|\coprod_{A\in \mathcal{A}}{A}\right|=\sum_{A\in \mathcal{A}}{|A|}=\sum_{A\in \mathcal{A}}{2^{\aleph_0}}=|\mathcal{A}|\cdot 2^{\aleph_0}=|\mathcal{A}|$$
A: If you have $2^{\mathbb{R}}$ disjoint sets with at least 1 element then their union will have a cardinality of at least $2^{\mathbb{R}}$.
