I'm studying set theory and met with the following problem: Let set $E=\bigcup_{i=1}^\infty A_i$ be of cardinality $\aleph_1$; show that one of $A_i$ must also have cardinality $\aleph_1$.
I've come up with a proof using basic definitions. But I heard from a friend a very short proof via cardinality arithmetic:
Suppose each $A_i$ has cardinality $<\aleph_1$. Clearly they cannot all be countable. Therefore, we have
$$
\aleph_1=\operatorname{card}\left(\bigcup_{i=1}^\infty A_i\right)\leq\operatorname{card}\left(\bigcup_{i=1}^\infty\bigcup_{a\in A_i}(i,a)\right)\color{red}{\leq}\aleph_0\times\max_i\operatorname{card}(A_i)\leq\max_i\operatorname{card}(A_i)\times\max_i\operatorname{card}(A_i)<\aleph_1,
$$
which is a contradiction.
Is the proof correct? I don't believe in this proof myself because I highly doubt the validity of the less-or-equal sign marked red.
Edit: I find this less-or-equal sign correct now, because there's clearly an injective function from $\bigcup_{i=1}^\infty\bigcup_{a\in A_i}(i,a)$ to $\mathbb N\times\operatorname{argmax}\operatorname{card}(A_i)$. However, the proof is incorrect, since we are not sure whether there is a maximum cardinality among $\operatorname{card}(A_i)$. Henno Brandsma also points out some important points here (see comments).
Best Answer
Well, if $\operatorname{card}(A_i) < \aleph_1$ we know $\operatorname{card}(A_i) \le \aleph_0$. So if the conclusion would not hold all $A_i$ would be at most countable but then $\bigcup_{i=1}^\infty A_i$ would also be at most countable which is a contradiction with it having cardinality exactly $\aleph_1$. That's all.