# Infinite sum of cardinals

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).

• The cardinal after $\color{red}{\le}$ should simply be $\aleph_0 \times \aleph_0 = \aleph_0$ as we can estimate each $\operatorname{card}({A_i})$ by $\aleph_0$. "Clearly they cannot all be countable" is false, it's exactly what follows if you assume $\forall i \operatorname{A_i} < \aleph_i$. In fact that's where we get our contradiction from. – Henno Brandsma Sep 27 '20 at 16:35

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.
• @JeffreyWang We're not using CH: $\aleph_1$ is by definition the first/smallest uncountable ordinal, so having a strictly smaller cardinality than $\aleph_1$ implies immediately that it must be at most countable. – Henno Brandsma Sep 27 '20 at 16:37
• @JeffreyWang The continuum hypothesis says that $\operatorname{card}(\Bbb R)= \aleph_1$ but we're not talking about the size of $\Bbb R$ here. CH is irrelevant. – Henno Brandsma Sep 27 '20 at 16:39
• @JeffreyWang From minimality of $\aleph_1$ as I said! – Henno Brandsma Sep 27 '20 at 16:42