Elementary Set Theory, cofinal subset, cofinality, ordinal, totally ordered set problem

Question

Definition 1. Let $\langle u,<\rangle$ be totally ordered set, $v\subset u$. $v$ is cofinal subset of $u$ means that for all $a\in u$, there exist $b\in v$ ($a\le b$).

Definition 2. Let $\alpha,\beta$ be ordinals. $\beta$ is cofinal of $\alpha$ means that there exist a cofinal subset of $\alpha$, $v$, and $\langle \beta,\in\rangle\cong\langle v,\in \rangle$.

Definition 3. Let $\alpha$ be ordinal. $\operatorname{cf}(\alpha)$ means that $$\min\{\beta\mid \beta\text{ is a ordinal and there exist a cofinal subset of } \alpha, v,\text{ and } \langle \beta,\in\rangle\cong\langle v,\in\rangle))\}$$

Question 1. If $\alpha$ be limit ordinal, then $\operatorname{cf}(\aleph_\alpha)=\operatorname{cf}(\alpha)$.

Question 2. Let $\alpha,\beta$ be ordinals. $\operatorname{cf}(\alpha+\beta)=\operatorname{cf}(\beta)$.

I tried to solve it like blow,

about question 1. $\alpha\le\aleph_\alpha$ and $\operatorname{cf}(\alpha)\le\alpha$ hence $\operatorname{cf}(\alpha)\le\aleph_\alpha$......[*1] $\operatorname{cf}(\operatorname{cf}(\alpha)) = \operatorname{cf}(\alpha)$　and [*1] hence $\operatorname{cf}(\aleph_\alpha)\ge\operatorname{cf}(\alpha)$

about question 2. $\alpha\le\alpha+\beta$ hence $\operatorname{cf}(\alpha)\le\operatorname{cf}(\alpha+\beta)$

Is it correct? and how can I show the invesrses of inequalities. Please help me.

Answer 1

Your approach implicitly assumes that cofinality is a non-decreasing function, which as you know is false.

Here’s an extended HINT for the second question, which I think is a little easier than the first. Once you’ve worked your way through that problem, we can worry about the other problem in a separate question if you’re still stuck on it.

Let $u$ be a cofinal subset of $\beta$.

• Let $v=\{\alpha+\xi:\xi\in u\}$. Show that $v$ is cofinal in $\alpha+\beta$, and that $\langle v,\in\rangle\cong\langle u,\in\rangle$.
• Deduce that $\operatorname{cf}(\alpha+\beta)\le\operatorname{cf}(\beta)$.

Now let $u$ be a cofinal subset of $\alpha+\beta$, and let $v=\{\xi:\alpha+\xi\in u\}$.

• Show that $v$ is cofinal in $\beta$.
• Show that $\operatorname{cf}(v)=\operatorname{cf}(\{\alpha+\xi:\xi\in v\})\le\operatorname{cf}(u)$.
• Deduce that $\operatorname{cf}(\beta)\le\operatorname{cf}(\alpha+\beta)$.

Answer 2

I actually slightly disagree with Brian re: easiness; I'll give a sketch of a solution to problem 1.

The principle behind the solution is what I think of as "cofinal in cofinal = cofinal": if I have

• an ordinal $\alpha$,

• a cofinal embedding $h$ of an ordinal $\beta$ into $\alpha$ (that is, a sequence $\alpha_\eta$ ($\eta<\beta$) of ordinals $<\alpha$ which is cofinal in $\alpha$ - think of $\alpha_\eta$ as $h(\eta))$,

• and a cofinal embedding $g$ of an ordinal $\gamma$ into $\beta$ (that is, a sequence $\beta_\theta$ ($\theta<\gamma$) of ordinals $<\beta$ which is cofinal in $\beta$ - think of $\beta_\theta$ as $g(\theta))$,

then I can combine these to get a cofinal embedding of $\gamma$ in $\alpha$: look at $h\circ g: \gamma\rightarrow\alpha$. Put another way, "thinning ordinals out" doesn't change their cofinality. Note that in particular, this means $cf(\beta)\le cf(\alpha)$, and it's not too hard to show that the converse also holds as long as $h$ is order-preserving.

So in problem (1), we want to find a cofinal, order-preserving embedding of $\alpha$ into $\aleph_\alpha$. Can you think of one? (HINT: $\aleph_0, \aleph_1, \aleph_2, . . .$)