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!
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 ^_^