Sandwiched sequence converge to same limit in $\omega_1$

Question

I am stuck on a question that might need a trick to crack, any help is appreciated

Problem statement

Let $(a_n), (b_n)$ be sequences on $\omega_1$ as a topological space, such that $a_n \leq b_n \leq a_{n+1}$, show that the sequences converge to the same limit

Proof attempt:

Let $a = a_n$ as $n \to \infty$, $b = b_n$ as $n \to \infty$. We want to show that $a = b$.

Let $P(a)$ be the predessor set of $a \in \omega_1$ defined by $P(a) = \{x| x < a, a \in \omega_1\}$

Then $a_n \leq b_n \leq a_{n+1} \leq b_{n+1} $ implies $P(a_n) \subseteq P(b_n) \subseteq P(a_{n+1})\subseteq P(b_{n+1})$

Then as $n\to \infty$...$P(a) = P(b)$ so $a=b$.

I know there are huge gaps here but I am completely lost

Answer 1

HINT: First let me correct your notation: ‘$a=a_n$ as $n\to\infty$’ doesn’t actually make sense. What you mean is that $a=\lim_na_n$ or, if you want to be a bit more specific, $a=\lim\limits_{n\to\infty}a_n$, and similarly for $b$. Also, once you’ve defined $a$ to be specifically the limit of the $a_n$, it’s a bad idea to use $a$ as the generic variable in your definition of $P(\cdot)$.

I suggest that you show that $a\le b$ and that $b\le a$. For each $n\in\Bbb N$ you have $a_n\le b_n$; use that to show that $a\le b$. You also have $b_n\le a_{n+1}$, and you can use that to show that $b\le a$. I’ve left a spoiler-protected further hint below.

If $b<a$, then $(b,a]$ is an open nbhd of $a$ in the topology of $\omega_1$, so there must be an $m\in\Bbb N$ such that $a_n\in(b,a]$ whenever $n\ge m$. What does this say about $b_{m+1}$, for instance?