increasing sequence in $\omega_1$

Question

I want to prove that a (countable) increasing sequence in $\omega_1$ converges to some point. Here is my idea:

I first try to find the limit point. We know any point in the increasing sequence (say $x_n$) is also a point in $\omega_1$, which means only countably many points in $\omega_1$ is smaller than (or equal to) $x_n$. So the set of points in $\omega_1$, which is also smaller than some point in the sequence, is countable. $\omega_1$ is uncountable and well-ordered, so we can take the complement and get a minimum point. I claim the minimum point after the sequence is the limit point.

However, I am not sure whether the minimum point after the sequence can be an open singleton set, since I have asked some related questions, like the following one:

For each $\alpha\in\omega_1$ note that $\{\min(\omega_1\setminus I_\alpha)\}$ is an open set where $I_\alpha:=\{x\in\omega_1:x\le \alpha\}.$

So how can I proceed my proof (i.e. show that the minimum point can not be an open singleton) or my proof is fundamentally not correct?

Answer 1

Go ahead and let $\alpha$ be the minimum element of $\omega_1$ greater than all $\alpha_n,$ as you described it (which exists, if we assume, as I suspect you are assuming, that a countable union of countable sets is countable). First, note that $\alpha$ cannot be the minimum element of $\omega_1.$ We show now that $\alpha$ is not an immediate successor.

Suppose that $\alpha$ is an immediate successor--that is, there is some $\beta\in\omega_1$ with $\beta<\alpha$ such that there exists no $\gamma\in\omega_1$ with $\beta<\gamma<\alpha.$ By definition of $\alpha,$ we know that $\beta$ is not an upper bound on the sequence. Hence, there is some $n\in\omega$ such that $\beta\le a_n.$ Since we have an increasing sequence, then $\beta<a_{n+1},$ and by definition of $\beta,$ we must have $\alpha\le a_{n+1}.$ But then $\alpha$ cannot be an upper bound of the sequence, either, contradicting our choice of $\alpha.$

The upshot is this: given any $a,b\in\omega_1$ with $\alpha\in(a,b)$--that is, with $a<\alpha<b$--there will always be some $c\in\omega_1$ with $a<c<\alpha,$ so that $(c,b)$ is a strict subset of $(a,b),$ and still a neighborhood of $\alpha.$ Hence, $\{\alpha\}$ is not open.

Answer 2

Let $\alpha = \min(\omega_1 - \bigcup_{n\in \Bbb{N}} [0,a_n] )$. (Existence of $\alpha$ is guaranteed by axiom of choice.) To prove $\alpha$ is a limit point of $\langle a_n\rangle$, we only prove following proposition:

For each $\beta<\alpha$, there is $n$ s.t. $\beta< a_n<\alpha$.

Note that $\{(\beta,\alpha] : \beta<\alpha\}$ is a local basis of $\alpha$. (That is for each open set $V$ containing $\alpha$, there is $\beta<\alpha$ s.t. $(\beta,\alpha]\subset V$.)

Its proof is simple: if there is $\beta<\alpha$ s.t. $\beta\ge a_n$ for all $n$, then $\beta\in \omega_1 - \bigcup_{n\in \Bbb{N}}[0,a_n]$ so $\alpha\le\beta$, a contradiction.

Answer 3

it is a known fact that ω1 is sequence-compact. because:

1.the cofinality of ω1 is not smaller then ω,because if it was ,ω1 woulde be a countable union of countable sets.

2.every ordinal has a max limit-ordinal smaller-or-equil to it.and between them are a finity number of ordinals.(if not,we get a dicreasing sequence).

so we can build a converging sub-sequence,by looking at some up-bound,that is small enoght that there is an sequence-element between it and the limit-ord from 2,and checking if:

i. the sequence converges to it.

ii.there are infinity elements between it and its limit by claim 2,and then a converging sub-sequence.(infinity element in finity ords.

iii .if not,look at the sub-sequence that all elements are small then thet limit ordinal,and do it over again.

the loop will stop in finity step(if not,we ger a decreasing sequence of limit ordinals),and so there are infinity elements in the finity*finity ordinals we looked at.

in a increasing sequence,we get that the hole sequence converges,because ii is alwyas right,and because of that i also