Folland ordinals exercise

Question

Consider following exercise from Folland's book:

I succesfully proved (a),(b),(c) and now I want to prove (d).

Attempt (outline): Put $E:= \bigcap_1^\infty E_n$. We may assume $E \subseteq \Omega$ by removing the element $\omega_1$ (this modified is uncountable iff the original set is uncountable). We use (c) to prove that $E$ is uncountable.

So, let $x \in \Omega$. Define a sequence $(x_n)_n$ by a diagonal argument:

Define $x_0:= x+1$ $x_1 \in E_1, x_2 \in E_2, x_3 \in E_1, x_4 \in E_2, x_5 \in E_3, x_6 \in E_1, x_7 \in E_2, x_8 \in E_3, x_9 \in E_4, x_{10}\in E_1$ and continue. We can also arrange that $x_1 < x_2 < x_3 <\dots$, again using (c).

Following the hint, $(x_n)_n$ converges. I proved that $\lim_n x_n = \sup_n x_n = \bigcup_n x_n$. By construction thus $y:=\lim_n x_n \geq x_0 =x+1> x $. Also, all subsequences of $(x_n)$ converge to $x$ so since $(x_n)_n$ contains subsequences consisting entirely from terms of each $E_n$, we see that $y=\lim_n x_n \in \overline{E_n} = E_n$ for all $n$ and hence $y \in E$, as desired.

Does the above proof outline look correct?

Answer 1

Yes, it’s correct. Another way is to show that the intersection of two uncountable closed sets is uncountable by using the same basic argument but simply alternating between the two sets. It immediately follows that the intersection of finitely many uncountable closed sets is uncountable, so you can let $F_n=\bigcap_{k=1}^nE_n$, choose $x_1>x$ in $F_1$, and for $n>1$ choose $x_n>x_{n-1}$ in $F_n$. This way you automatically get a sequence with subsequences — tails, in fact — in each $E_n$.

Answer 2

Yes, your proof is fine. Perhaps a neater way of defining the sequence is by requiring that:

1. $(x_n)$ is increasing, and
2. if $k$ is the largest such that $2^k\mid n$, then $x_n\in E_{k+1}$.

This way it is slightly clearer than "and continue", and more formal as to how to continue. Then it's also very easy to point at a subsequence in $E_{k+1}$ by simply taking $x_{2^k(2m+1)}$ for $m\in\Bbb N$.