Does there exist an increasing unbounded sequence in $\omega_{1}$?

Question

I ask for some forgiveness, I am quite unfamiliar with actually working with ordinals. My question is: Does there exist a sequence $(\alpha_{k})_{k \in \mathbb{N}}$ such that $\alpha_{k} < \alpha_{k + 1}$ for all $k \in \mathbb{N}$, and such that $(\forall \beta \in \omega_{1})(\exists k \in \mathbb{N})(\alpha_{k} \geq \beta)$? I apologize if this is a fairly obvious question, but I'm not very familiar with ordinals. What I have/understand so far is that $\omega_{1}$ is the union of all countable ordinals. Moreover, $\omega_{1}$ is a limit ordinal, so $(\forall \alpha < \omega_{1})(\exists \beta < \omega_{1})(\alpha < \beta < \omega_{1})$, so my first thought is to define $(\alpha_{k})$ recursively by saying, "Set $\alpha = \alpha_{1}$. For each $\alpha_{k}$ exists $\beta_{k}$ such that $\alpha_{k} < \beta_{k} < \omega_{1}$; let $\alpha_{k + 1} = \beta_{k}$. But this is not enough. I could just as easily say $\alpha_{k} = k$, satisfying the recursion but only exhausting $\omega$. How might I go about constructing such a sequence?

Answer 1

What you ask is whether the cofinality $cf(w_1)$ is $w.$ The cofinality $cf(x)$ of a limit ordinal $x$ is the least cardinal of a $y\subset x$ such that $y$ is unbounded in $x$. Equivalently, $cf(x)$ is the least ordinal $d$ such that exists a strictly increasing $f:d\to x$ such that $\{f(e):e\in d\}$ is unbounded in $x.$

If $x$ is an infinite cardinal and $cf(x)=x$ then $x$ is called regular; if $cf(x)<x$ then $x$ is called singular. Note that $cf(x)$ itself is a regular cardinal

In ZFC it is a theorem that every infinite successor cardinal is regular. The successor cardinal $k^+$ is the least cardinal ordinal greater than the infinite cardinal $k.$ Hence in ZFC we have $cf(w_1)=cf(w^+)=w_1$ and the sequence you look for does not exist.

In 1972, Solovay showed, with the assumption that there is a measurable cardinal (formerly called $2$-measurable) that it is consistent with ZF that $cf(w_1)=w.$

In ZF it is sufficient to assume a corollary of the Axiom Of Choice :A countable union of countable sets is countable, in order to prove that $cf(w_1)=w_1$. Because, for $f:w\to w_1$ we have $\sup \{f(n):n\in w\}=\cup \{f(n):n\in w\}$ is a countable ordinal

Answer 2

Such a sequence doesn't exist. The cofinality of $\omega_1$ is $\omega_1$, so any increasing countable sequence of ordinals $<\omega_1$ must be bounded in $\omega_1$. To see this, let $(\alpha_n : n \in \mathbb{N})$ be any sequence of ordinals $<\omega_1$. Then $\alpha_{\omega} := \bigcup_{n \in \mathbb{N}} \alpha_n$ is a countable ordinal, so is $<\omega_1$. But then $\alpha_n < \alpha_{\omega}+1 < \omega_1$ for all $n \in \mathbb{N}$.