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Suppose $E=\cup_{n=1}^{\infty}A_n$ and the cardinal number of $E$ is $c$ ($c$ represents the cardinal number of $(0,1)$). Prove that there exsits $n_0$ such that the cardinal number of $A_{n_0}$ is also $c$.

For now, I have proved if $E=A_1\cup A_2$ the conclusion holds. Then by induction the finite union case is also true. But I do not know how to prove the denumberable case stated above.

Appreciate any help or hint!

user823011
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  • Hint: show that the union won't be countable at a certain index. As an uncountable union does not produce a countable number. –  Sep 03 '21 at 07:21
  • Did you secrestly use the continuum hypothesis? Can you be more specific? I don't really follow. – user823011 Sep 03 '21 at 07:27

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Hint:

Are you aware that the cofinality of $\mathfrak{c}$ must be uncountable? This is a corollary of König's theorem.

Next, if you want to hit $\kappa$ by unioning stuff smaller than $\kappa$, you'll need to union at least $\text{cf}(\kappa)$ many things (do you see why?).

Putting these two facts together: if we try to union up countably many things of size $\lt \mathfrak{c}$, can we hit $\mathfrak{c}$?

As an aside, notice that we need choice in order to run this argument. It's consistent without choice that $\mathbb{R}$ is a countable union of countable sets!

I hope this helps ^_^

Chris Grossack
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    We can apply König's theorem right away: suppose on the contrary that each $A_n$ has smaller cardinality than $c$. Then $$|E|\le\sum_n|A_n|\ <\ c^\omega=(2^\omega)^\omega=2^\omega=c$$ – Berci Sep 03 '21 at 10:18