Club, Stationary Sets, and Pressing Down: Examples and Counterexamples

Tails are unbounded; a nonzero limit point of a tail lies above its cutoff and hence in the tail. Nonzero limits are unbounded since $\beta +\omega <\kappa$ for $\beta <\kappa$; a nonzero limit point of limit ordinals is itself a limit.

No alpha lies in $C_{\alpha }$, so the full intersection is empty. But for every $\xi <\alpha$ we have $\alpha \in C_{\xi }$, including the vacuous test at zero. Thus the diagonal intersection is all of kappa, consistently with its club theorem.

The set $[\omega +1,\kappa )\cup \{1,2,3,\dots \}$ contains a club tail, but omits its nonzero limit point omega, and is therefore not closed.

In ZFC omega-one and omega-two are regular: a cofinal family of at most omega ordinals below omega-one has countable union; a cofinal family of at most omega-one ordinals below omega-two has union of size at most ${\mathrm{\aleph }}_1\cdot {\mathrm{\aleph }}_1={\mathrm{\aleph }}_1$. Both would contradict the cardinality of the ambient ordinal. For the second union, AC chooses injections of its at most aleph-one members into omega-one, so the union injects into the product of the index set with omega-one. These estimates use countable choice and infinite well-ordered cardinal multiplication.

Every nonzero countable limit has cofinality omega: enumerate it and take successive finite maxima to obtain a cofinal sequence; a finite subset cannot be cofinal in a limit. Thus $E_{\omega }^{{\omega }_1}$ is exactly the nonzero limits, a closed unbounded set.

Apply the stratum theorem at omega-two with lambda equal to omega and omega-one. The resulting stationary sets are disjoint since an ordinal has only one cofinality. A club contained in either would miss the other, contradicting stationarity.

S consists of successors, is unbounded in kappa, and omits zero. The uniquely defined predecessor map satisfies $f(\alpha +1)=\alpha <\alpha +1$, so it is regressive.

Each fibre is a singleton, hence misses a club tail and is nonstationary. Also S itself misses the club of nonzero limits: these are unbounded because $\beta +\omega <\kappa$, and closed because limits of limit ordinals are limits. Thus stationarity, the missing hypothesis of pressing down, cannot be replaced by unboundedness.

Let S be the nonempty-arrival stops. Zero is not in S. If S were stationary, choose one departing passenger at each of its stops, and send that stop to the chosen passenger's boarding index. This is strictly below the arrival stop because departures precede new boarding. Pressing down gives a stationary set of such stops with one common boarding index beta.

The selected passengers at distinct stops are distinct: each leaves only once and cannot reboard. The stationary fibre has cardinality aleph-one, although at most countably many passengers boarded at beta, a contradiction. Hence S is nonstationary, and its complement contains a club by definition of nonstationarity.

A passenger remaining after boarding at beta would make every arrival after beta nonempty. That would exclude empty arrivals on an entire tail, contradicting the unbounded club of empty arrivals.

For alpha below omega-two, any uncountable cofinality must equal omega-one: cofinality is a cardinal at most the cardinality of alpha, and this is at most aleph-one. If alpha has this cofinality, the stratum theorem with theta equal to alpha and lambda equal to omega says $E_{\omega }^{{\omega }_2}\cap \alpha$ is stationary. This proves the first equality.

Fix such an alpha and an increasing cofinal sequence $(a_{\xi }{)}_{\xi <{\omega }_1}$ in alpha. Recursively define a strictly increasing cofinal sequence c: take $c_0=a_0+1$, at successors take $c_{\xi +1}=\max (c_{\xi },a_{\xi +1})+1$, and at nonzero limits take the supremum of prior values. Countable initial segments remain bounded since alpha has cofinality omega-one. The range is unbounded and closed: a limit point below alpha corresponds to a bounded limit set of indices, whose supremum is below omega-one, and continuity includes that value.

At zero and successor indices the c values are successors and have cofinality one. At nonzero limit indices below omega-one, continuity and strict increase give a countable cofinal sequence with no last point, so the value has cofinality omega. Thus this club avoids $E_{{\omega }_1}^{{\omega }_2}\cap \alpha$. No eligible alpha belongs to its trace, giving the second equality.

At theta equal to omega, each tail $C_n=\{m<\omega :n\le m\}$ is unbounded and vacuously closed, because omega has no nonzero limit ordinal below it. Their intersection is empty and hence not club.

Even at the uncountable singular ordinal $\theta ={\mathrm{\aleph }}_{\omega }$, the tails $D_n=[{\mathrm{\aleph }}_n,{\mathrm{\aleph }}_{\omega })$ are club, but their intersection is empty since the alephs indexed by natural numbers are cofinal in theta. In both cases the countable index size equals, rather than lies strictly below, the ambient cofinality omega. Thus neither example meets the intersection theorem's hypotheses.

For countable alpha, omega times alpha is the order type of alpha many consecutive countable blocks, hence is countable in ZFC. Multiplication by omega on the left is strictly increasing: appending one nonempty omega block strictly increases the order type, and the recursive definition is monotone in the right argument. It is continuous at nonzero limits by the same definition. Thus f is a normal self-map of omega-one.

Associativity and induction give $f({\omega }^n)={\omega }^{n+1}$ for each finite n, starting from $\omega \cdot 1=\omega$. The countable supremum of these countable ordinals is ${\omega }^{\omega }<{\omega }_1$. Continuity gives $f({\omega }^{\omega })={\sup }_n{\omega }^{n+1}={\omega }^{\omega }$.

The normal fixed-point theorem makes the whole fixed-point set club (including zero, since $f(0)=0$). The normal-enumeration theorem applies to this club and provides its normal increasing enumeration.