Proving cardinality of an uncountable sum

Question

I'm trying to prove the following thing:

For a family $\mathcal{A}$ of countable sets such that $|\bigcup\mathcal{A}|$ is uncountable and such that $\big|\{A\in\mathcal{A}: x\notin A\}\big| \leq\aleph_0$ for every $x\in\bigcup\mathcal{A}$, we have $|\bigcup\mathcal{A}|=\aleph_1$.

Of course cardinality of this sum is at least $\aleph_1$. To show the other inequality, I firstly tried to find an injection from $\mathcal{A}$ to the set of all countable ordinals thinking that I might assign a type for every $A$ in the family and get $|\mathcal{A}|\leq\aleph_1$ (and proceed with $|\bigcup\mathcal{A}|\leq|\mathcal{A}|\cdot|A|)$. Sadly that's a dead-end, because sets of this family can be of the same ordinal type.

Now I'm thinking of a contradiction if I assume $|\bigcup\mathcal{A}|=\aleph_{\alpha}$, but can't find any as of yet.

Could you help me, give a hint?

Answer 1

Suppose the size of the union is $\ge\aleph_2$. Then there are at least $\aleph_2$-many $A$s. Fix distinct elements $x_\alpha$ ($\alpha<\omega_1$) of the union $\bigcup \mathcal{A}$. By the hypothesis, each $x_\alpha$ is in all but countably many of the $A$s. But then:

• What can you say about the size of the set $\{A\in\mathcal{A}: \exists \alpha<\omega_1(x_\alpha\not\in A)\}$?

• How does this compare with the size of $\mathcal{A}$?

• Why does this violate the assumption that each $A\in\mathcal{A}$ is countable?

Answer 2

By contradiction, suppose $f:\cup \mathcal A\to \omega_2$ is an injection. Let $g:\omega_1\to \mathcal A$ also be an injection. For $x\in \omega_2$ let $h(x) =\min \{y\in \omega_1: f(x)\in g(y)\}.$ We know $h(x)$ exists because $\{y\in \omega_1: f(x)\not \in g(y)\}$ is countable.

Since $h:\omega_2\to \omega_1$, there exists $y\in \omega_1$ such that $|h^{-1}\{y\}|=\omega_2.$ But $h^{-1}\{y\}$ is countable because $$x\in h^{-1}\{y\}\implies h(x)=y\implies f(x)\in g(y)\implies x\in f^{-1}(g(y)),$$ while $f$ is an injection and $g(y)\in \mathcal A$ is countable.