Is the following claim true?
If $\bigcup_{i=1}^\infty A_i$ has cardinality $\kappa$, then some $A_i$ has cardinality $\kappa$. Here $\kappa$ is uncountable.
I'm interested in the particular case $\kappa=c:=|\mathbb{R}|$. In other words, if each $A_i$ has cardinality strictly less than $\kappa$, can we deduce that $\bigcup_{i=1}^\infty A_i$ has cardinality less than $\kappa$?
I know very little cardinal arithmetic beyond the usual rules for sums and products. But I came across this in an analysis text, so I wonder whether it's true.
Counter-example:
$$\left| \bigcup_{i=1}^\infty \omega_i \right| = \omega_\omega$$
We have:
$$\kappa = \left| \bigcup_{i=1}^\infty A_i \right| \le \omega \cdot \sup_{i=1..\infty} |A_i| = \sup_{i=1..\infty} |A_i|$$
So in order to conclude that $\kappa = |A_i|$ for some $i$, we need $\operatorname{cof}(\kappa) = \omega$, otherwise there are coutner-examples. For $\kappa = \mathfrak c$ though, you are lucky, since $\operatorname{cof}(\mathfrak c) > \omega$, so you can always conclude that $\kappa = |A_i|$ for some $i$.
Specializing on the case $\kappa = \mathfrak c$:
Let $|A_i| < \mathfrak c$ for all $i$. By Konig's lemma: $$\left| \bigcup_{i=1}^\infty A_i \right| < \left| \prod_{i=1}^\infty \mathfrak c \right| = \mathfrak c^\omega = 2^{\omega \cdot \omega} = 2^\omega = \mathfrak c$$