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).

  • $\begingroup$ 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. $\endgroup$ Sep 27, 2020 at 16:35

1 Answer 1


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.

  • $\begingroup$ Yes, if the first statement is true then the problem is quite easy. But we are not allowed to use the continuum hypothesis here. $\endgroup$ Sep 27, 2020 at 16:36
  • $\begingroup$ @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. $\endgroup$ Sep 27, 2020 at 16:37
  • $\begingroup$ @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. $\endgroup$ Sep 27, 2020 at 16:39
  • $\begingroup$ Hmm... CH states that there is no set whose cardinality is strictly between that of the integers and the real numbers. I assume that your first statement follows from this. If not, then how? Thanks :) $\endgroup$ Sep 27, 2020 at 16:41
  • $\begingroup$ @JeffreyWang From minimality of $\aleph_1$ as I said! $\endgroup$ Sep 27, 2020 at 16:42

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