I missed two lectures in my set theory course, and now I don't understand the homework problems.

One is this: let $\kappa$ be a regular uncountable cardinal. Show that the following sets are closed and unbounded in $\kappa.$

  • $\{\alpha<\kappa\,:\,\alpha \text{ is a limit ordinal}\},$
  • $\{\alpha<\kappa\,:\,\mathrm{cf}(\alpha)=\omega\}$ for $\kappa=\omega_1,$
  • $\{\lambda<\kappa\,:\,\lambda\text{ is a cardinal}\}$ for $\kappa$ inaccessible.

I don't understand what closed or unbounded subsets of $\kappa$ is. Could you explain this to me and give some pointers on solving the problem?

  • $\begingroup$ Should this question be tagged "topology"? $\endgroup$ – Bartek May 11 '13 at 14:13
  • 2
    $\begingroup$ In the third one, don't you need to assume that $\kappa$ is limit? If $\kappa=\mu^+$, then the set of cardinals under $\kappa$ is bounded by $\mu+1$... $\endgroup$ – tomasz May 11 '13 at 14:21
  • $\begingroup$ Sorry, $\kappa$ should be assumed to be inaccessible. I'm not sure how to interpret this. (Wikipedia gives more than one notion of inaccessibility.) $\endgroup$ – Bartek May 11 '13 at 14:28
  • $\begingroup$ Weak inaccessibility will suffice. $\endgroup$ – tomasz May 11 '13 at 14:30

An unbounded subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $\beta<\kappa$, there is some $\alpha\in A$ such that $\beta<\alpha$.

A closed subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $0<\alpha<\kappa$, if $\sup(A\cap\alpha)=\alpha,$ then $\alpha\in A$.

For the first one, do you know the result that all limit ordinals are of the form $\omega\cdot\alpha$ for some non-zero $\alpha$? In particular, you'll want to show that the limit ordinals less than $\kappa$ are those of the form $\omega\cdot\alpha$ for $0<\alpha<\kappa$. Use continuity of ordinal multiplication to show that this set is closed, and the fact that $\kappa$ is a cardinal to show that it's unbounded.

The second one is just a special case of the first one. (Why?)

The third one won't be too tricky. Do you know the recursive definition of the alephs? (Also, don't forget that the natural numbers are cardinals, too.)

  • $\begingroup$ Thanks. Could you explain why you changed the definition of a closed subset? Why doesn't the previous one work? $\endgroup$ – Bartek May 11 '13 at 14:59
  • $\begingroup$ @Bartek: The previous one didn't work because $\kappa$ is closed in itself. I'm used to working in closed unbounded classes (sometimes proper) of ordinals (that is, treating $\kappa$ as the class of all ordinals), in which case the other definition makes sense. As an alternate definition, a closed subset of $\kappa$ is some $A\subseteq\kappa$ such that for all non-empty subsets $B\subseteq A$, either $\sup B=\kappa$ or $\sup B\in A.$ Yet another is that every limit point of $A$ (where $\kappa$ is considered in the order topology) in $\kappa$ is in $A$. $\endgroup$ – Cameron Buie May 11 '13 at 15:05
  • $\begingroup$ OK, thanks. What is the continuity of ordinal multiplication? I interpret the term as "$\sup \{\omega\cdot\alpha\,:\,\alpha<\kappa\}=\omega\cdot\sup\{\alpha\,:\,\alpha <\kappa\}.$ " Is this correct? $\endgroup$ – Bartek May 11 '13 at 21:21
  • $\begingroup$ A more general statement than the above would mean, I think, that I don't need the result about $\omega\cdot\alpha$ for $0<\alpha<\kappa.$ Let $B\subseteq\{\alpha<\kappa\,:\,\alpha \text{ is a limit ordinal}\}.$ Then clearly $\sup B\leq\kappa.$ So it's enough to show that $\sup B$ is a limit ordinal. Suppose that for any set $X$ of ordinals we have $\sup\{\omega\cdot\alpha : \alpha\in X\}=\omega\cdot\sup X.$ Then clearly $\sup B$ is a limit ordinal. Is this statement false? $\endgroup$ – Bartek May 12 '13 at 1:09
  • $\begingroup$ @Bartek: Nicely done! That's the way you want to go. $\endgroup$ – Cameron Buie May 12 '13 at 16:08

Let $\kappa$ be a regular uncountable cardinal. $X \subseteq \kappa$ is unbounded if $\sup X = \kappa$.

A $\alpha$ is a limit point of $X$ if and only if $\alpha$ is a limit ordinal and $\sup(X \cap \alpha) = \alpha$.

$X \subseteq \kappa$ is closed if it contains all its limit points less than $\kappa$.

1 Let $X = \{\alpha < \kappa : \alpha \text{ is a limit ordinal}\}$. Limit points of $X$ must be limit ordinals. $X$ is closed. Now to show that $X$ is unbounded. Let $\alpha < \kappa$. Since $\kappa$ is a cardinal, $|\alpha| < \kappa$. Then $\alpha + \omega$ is a limit ordinal with cardinality less than $\kappa$. So $\alpha + \omega \in X$. $\alpha + \omega > \alpha$. $X$ is unbounded.

2 Let $X' = \{\alpha < \omega_1 : cf(\alpha) = \omega\}$. At least in Jech, the cofinality is only defined for limit ordinals. Using this convention, $X' \subseteq X$, where $X$ is the set of ordinals form part 1. The cofinality of any limit ordinal is always a cardinal (in fact regular cardinal). The only cardinality less than $\omega_1$ is $\omega$. It has been shown that if $\alpha < \omega_1$ is a limit ordinal, then $cf(\alpha) = \omega$. So $X' = X$. It is closed and unbounded by part 1.

3 Suppose $\kappa$ is (weakly) inaccessible. Let $X'' = \{\alpha < \kappa : \alpha \text{ is a cardinal}\}$. It is unbounded since by definition of weakly inaccessible, $\kappa$ is a limit cardinal. Suppose $\alpha$ is a limit point of $X''$. $\alpha$ is a cardinal since $\sup$ of a set of cardinals is always a cardinal.

  • $\begingroup$ Your limit point definition makes $0$ a limit point of every set of ordinals. $\endgroup$ – Cameron Buie May 12 '13 at 16:11
  • $\begingroup$ @CameronBuie I forgot something. Jech definition is "if $\alpha > 0$ is a limit ordinal, then $\alpha$ is a limit point of $X$ if $\sup(X \cap \alpha) = \alpha$". So according to his definition limit point should be nonzero limit ordinals. Nevertheless, I do not see anything in the theory of club and stationary sets that would go wrong if all club sets had $0$. $\endgroup$ – William May 12 '13 at 21:27
  • $\begingroup$ Well, for one thing, there re a great many clubs that are no longer clubs--for example, the first two of the sets mentioned by the OP will not be clubs if we require all clubs to have $0$. I don't know if you consider this to be "wrong", but it's certainly a departure from the standard theory. Also, it means that we're no longer considering the ordinals in the order topology. $\endgroup$ – Cameron Buie May 12 '13 at 23:23

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