If a countable union of sets has card $\mathfrak{c}$, prove at least one of them has card $\mathfrak{c}$

Question

If $A=\bigcup_{n=1}^{\infty}A_n$ and $A$ has cardinality $\mathfrak{c}$, where $\mathfrak{c}$ is the cardinal of the continuum, prove that at least one of the $A_n$ has cardinality $\mathfrak{c}$.

Answer 1

This follows from König's theorem, which reads

Theorem (König). If for $i \in I$ we have cardinals $\def\a{\mathfrak a}\def\b{\mathfrak b}\a_i < \b_i$, then $$ \sum_i \a_i < \prod_i \b_i $$

Now suppose for $I = \omega$, we have sets $A_i$ with cardinals $\a_i := |A_i| < \def\c{\mathfrak c}\c$, then $$ \left|\bigcup_i A_i\right| \le \sum_i \a_i < \prod_i \c = \c^{\aleph_0} = \c $$ Hence: If $\bigcup_i A_i$ has cardinality $\c$, one of the $A_i$ must also have cardinality $\c$.