# Are these sets stationary?

Assume $\kappa > \omega$ is a cardinal of uncountable cofinality and $S$ is a stationary set in $\kappa$. If $\alpha < \kappa$ is a successor ordinal then the set $S+\alpha=\{\sigma+\alpha : \sigma \in S\}$ is not stationary in $\kappa$ because it does not intersect the subset of limit ordinals in $\kappa$ (which is club in $\kappa$). Is $S + \alpha$ stationary in $\kappa$ if $\alpha < \kappa$ is a limit ordinal? What about the set $\{\sigma + \sigma : \sigma \in S\}$?

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You should make it that $\operatorname{cf}(\kappa)>\omega$, otherwise there are clubs without any limit points as well. – Asaf Karagila Jun 21 '11 at 18:17
I also believe that this needs to be modified that $\operatorname{cf}(\alpha)<\operatorname{cf}(\kappa)$ in this case. I will address both these cases in my answer (granted I will have one... feel free to change your question in the meantime). – Asaf Karagila Jun 21 '11 at 18:30
@Asaf Karagila: Indeed, if we don't assume that $\kappa$ is of uncountable cofinality, then the limit ordinals might not be club in $\kappa$. Then $S+\alpha$ might be stationary even though $\alpha$ is a successor. I added the assumption on uncountable cofinality of $\kappa$. – LostInMath Jun 21 '11 at 18:39
I believe that you can cook something for $\alpha>\operatorname{cf}(\kappa)$ so it doesn't work (e.g. take $C$ a club of limit ordinals of minimal order type and $\alpha>\operatorname{cf}(\kappa)$ we might be able to cook some stationary set meeting $C$ and $C$ has no point with cofinality same as $\alpha$) – Asaf Karagila Jun 21 '11 at 18:45
If I recall correctly, the set of limits of limits is a club. Then if $\alpha=\omega,\ S+\alpha$ does not intersect this club. – Ross Millikan Jun 21 '11 at 18:50

## 1 Answer

If $\kappa$ is regular and uncountable, then no set of the form $S = A + \alpha$ can be stationary (for any $A \subseteq \kappa$ and $\alpha \in \kappa$) by Fodor's lemma. The function which sends each $\beta \in S$ to the least $\gamma$ such that $\gamma + \alpha = \beta$ is regressive.

A similar argument shows that no set of the form $\{\alpha + \alpha : \alpha \in A\}$ is stationary (again, for regular $\kappa$).

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Thank you for the answer. I think the same argument works also in the case $\kappa$ is singular. In that case, assuming $f:S \to \kappa$ is a regressive function from a stationary set $S$, Fodor's lemma guarantees a stationary set $f^{-1}[0,\gamma)$ for some ordinal $\gamma < \kappa$. And if we take your function $f$, then the sets $f^{-1}[0,\gamma)$ are clearly bounded by $\gamma+\alpha$, thus nonstationary. – LostInMath Jun 21 '11 at 21:25
You have to work a little harder than that, since regularity of $\kappa$ is part of the hypothesis of Fodor's lemma. – Brian M. Scott Jun 22 '11 at 20:45
@Brian M. Scott: It's true that in most formulations of Fodor's lemma the regularity of $\kappa$ is assumed. However, the following proposition is also true: if $\alpha$ is a limit ordinal with uncountable cofinality and $f: S \to \alpha$ is a regressive function from a stationary set $S \subset \alpha$, then the set $f^{-1}[0,\beta)$ is stationary in $\alpha$ for some $\beta < \alpha$. – LostInMath Jun 23 '11 at 11:35