What could be said about the cardinality of $\bigcup_{i\in I} A_i$ if $I$ and all the $A_i$ have cardinality $2^{\aleph_0}$

Question

The countable union of a countable set is countable. Does the same hold for sets with cardinality $|\mathbb R|$. More specifically, if $A_i$ are sets of the same cardinality as the real numbers, and $I$ is an index set also with cardinality $|\mathbb R|$, is $|\bigcup_{i\in I} A_i| = |\mathbb R|$?

Answer 1

This question can be reduced to the question "is it true there is a bijection between $\mathbb{R}$ and $\mathbb{R}^2$?" The answer is yes.

Answer 2

The question can be cast into cardinal arithmetic. We can prove the following:

If $X$ is an infinite set, then $|X|=|X^2|$.

Moreover, we can prove the following cardinal arithmetic theorem:

$\left|\bigcup_{i\in I}A_i\right|\leq|I|\cdot\sup\{|A_i|\mid i\in I\}$.

With equality in the case that the union is a union of pairwise disjoint sets. If $I$ and all the $A_i$'s have the same cardinality, then we get that the union has the same cardinality as them as well.

These proofs are not entirely trivial and require the axiom of choice.