Cardinality of union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$ I have a home work question which is: " what is the cardinality of the union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$?"
I believe somehow we can get to: cardinality = $({{\aleph }_{0}} )(\mathfrak{c})$ 
but how would we get there and what does that make the cardinality? Does it make it $\mathfrak{c}$? 
if so how could this be proven? 
Help is appreciated,
Thanks
EDIT: forgot to write disjoint
 A: The fact that the union of $\kappa$ sets, each of cardinality $\lambda$, is equal to $\kappa\lambda$ follows simply from the fact that $\kappa\lambda$ is defined to be the cardinality of $\kappa\times\lambda$, and there is an obvious bijection between the disjoint union of $\{X_i\mid i\in\kappa\}$ in which each $X_i$ comes equipped with a bijection $f_i\colon X_i\to\lambda$, and $\kappa\times\lambda$: map each $(a,i)\in \bigcup_{i\in\kappa}X_i\times\{i\}$ to $(f_i(a),i)$. This is easily seen to be a bijection.
If your $X_i$ don't "come equipped" with bijections to $\lambda$ then you may need the Axiom of Choice in order to select a bijection for each $X_i$. On the other hand, the cardinal equality
$$\sum_{i\in\kappa}\lambda = \lambda\kappa$$
does not require the Axiom of Choice, since we can pick $f_i$ to be the identity for each $i$.
A: Under the assumption that each set of the form $\mathbb R\times\{n\}$ (or can be made of this form in a definable way), then the union of $\aleph_0$ sets, each of cardinality $2^{\aleph_0}$ is again $2^{\aleph_0}$.
If the axiom of countable choice holds then we may choose a function from each disjoint set into $\mathbb R\times\{n\}$. Now we have
$$2^{\aleph_0}\cdot\aleph_0\le\aleph_0^{\aleph_0}\cdot\aleph_0=\aleph_0^{\aleph_0}=2^{\aleph_0}$$
(The cardinal arithmetic does not require the axiom of choice to hold in any form)

If, however, we do not impose any of the above limitations (the sets are not of a form similar to $X\times\{n\}$, and the axiom of countable choice is missing), then problems may arise which makes the question ill-defined. 
It is possible for the question to be interpreted as above, while on the other hand, let $V$ be a model of $ZFA+AC$ with continuum many atoms, write $A=\bigcup A_n$ where $A_n$ are a disjoint partition of $A$, each piece of cardinality continuum. Consider permutations of $A$ which preserve the partitioning, alongside the ideal of supports generated by finite unions of $A_n$'s.
In the resulting permutation model $U$ we have that each $A_n\in U$, and so is the partition (since it is fixed by all permutations). We also have that there is no choice function on the $A_n$'s.
Suppose $f\colon A\to\mathcal P(\omega)$ a bijection in $U$, let $E$ be a support for $f$, without loss of generality $E=A_0\cup\ldots\cup A_n$. Let $\pi$ be a permutation fixing $E$, and therefore $\pi f=f$, we have that $\langle x,\alpha\rangle\in f$ then $\langle \pi x,\pi \alpha\rangle\in\pi f=f$.
Take $a\in A$ such that $\pi a\neq a$. Since $\pi f = f$ we have that $(\pi f)(\pi a) = \pi (f(a))$, which contradicts the injectivity of $f$, since $\operatorname{Rng}(f)$ is a subset of ordinals, which is fixed by all permutations and in particular by $\pi$.
Now use Jech-Sochor to transfer this into a symmetric extension of a model of $ZFC$, up to $\omega$.
We have a model of $ZF$ in which $2^{\aleph_0}$ can be well ordered, but the sum of $\aleph_0$ many copies of a continuum does not make a continuum.

This is quite the necrobumping, but I came up with a much simpler example of a countable union of sets of size $\frak c$ which is not of size $\frak c$:
Suppose that $V$ is a model of $$ZF+\text{There is a countable family of disjoint pairs without a choice function}$$ such model is Cohen's second model. Let $\{P_n\mid n\in\omega\}$ be such countable family of pairs, and let $S$ be $\bigcup_n P_n$.

Fun fact: $S$ cannot be linearly ordered, if it could be then we could choose the minimal one of each $P_n$.

Without loss of generality $S\cap\mathbb R=\varnothing$. Let $A_n=\{n\}\times\mathbb R\cup P_n$, since we only add two elements to a size isomorphic to $\mathbb R$ we do not increase in cardinality.
Consider $A=\bigcup_n A_n=\mathbb N\times\mathbb R\cup S$. We observe that $A$ cannot be linearly ordered since $S$ cannot be linearly ordered, however if $A$ were of size $\frak c$ then it could be matched with the real numbers and therefore be linearly ordered.
Conclusion, $|A|\neq\frak c$. On the other hand, clearly there is an injective function from $\mathbb R$ into $A$ therefore ${\frak c}<|A|$.
