The Lumenary · Research pathways

Large cardinals

Name: ______________________

Date: ______________________

You will learn to

Define inaccessible and measurable cardinals and prove that $V_\kappa \models \mathsf{ZFC}$ when $\kappa$ is inaccessible
Explain why ZFC cannot prove that an inaccessible cardinal exists (if ZFC is consistent), and what consistency strength means
Prove that every measurable cardinal is inaccessible, and describe how measurable cardinals correspond to non-trivial elementary embeddings $j : V \to M$
Survey the hierarchy of large cardinals and its consequences for sets of reals, separating theorems from open problems

1.
Which of these cardinals are regular? (Choose all that apply; assume ZFC.)
- A. $\aleph_0 = \omega$
- B. $\aleph_1$
- C. $\aleph_\omega$
- D. $\beth_\omega$
2.
The cumulative hierarchy starts $V_0 = \varnothing$ , $V_{n+1} = \mathcal P(V_n)$ . How many elements does $V_4$ have?
3.
Why is $\omega$ not measurable, even though non-principal ultrafilters on $\omega$ exist? (Consider countable completeness.)
- A.A non-principal ultrafilter on $\omega$ contains every $\{n\}^c$ , and these countably many sets have empty intersection
- B.There are no non-principal ultrafilters on $\omega$
- C. $\omega$ is singular
- D. $\omega$ is not a strong limit
4.
In the proof that $V_\kappa \models \mathsf{ZFC}$ for inaccessible $\kappa$ , which axiom's verification uses the regularity of $\kappa$ in an essential way?
- A.Replacement
- B.Power Set
- C.Pairing
- D.Extensionality
5.
Suppose ZFC proved "there is an inaccessible cardinal". What would follow?
- A.ZFC would prove $\mathrm{Con}(\mathsf{ZFC})$ , so by Gödel's second incompleteness theorem ZFC would be inconsistent
- B.ZFC would decide the continuum hypothesis
- C.ZFC would prove $V \ne L$
- D.Nothing special: it would just be one more theorem
6.
Write IC for "there is an inaccessible cardinal". Which statements are correct? (Choose all that apply; assume ZFC + IC is consistent.)
- A. $\mathsf{ZFC} + \mathrm{IC} \vdash \mathrm{Con}(\mathsf{ZFC})$
- B.ZFC proves $\mathrm{Con}(\mathsf{ZFC} + \mathrm{IC}) \to \mathrm{Con}(\mathsf{ZFC})$
- C.ZFC proves $\mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathrm{IC})$
- D. $\mathsf{ZFC} + \mathrm{IC} \vdash \mathrm{Con}(\mathsf{ZFC} + \mathrm{IC})$
7.
Which Lean predicate correctly expresses that an ultrafilter U : Ultrafilter α is κ-complete?
- A.∀ S : Set (Set α), #S < κ → (∀ s ∈ S, s ∈ U) → ⋂₀ S ∈ U
- B.∀ S : Set (Set α), #S ≤ κ → (∀ s ∈ S, s ∈ U) → ⋂₀ S ∈ U
- C.∀ S : Set (Set α), (∀ s ∈ S, s ∈ U) → ⋂₀ S ∈ U
- D.∀ s t, s ∈ U → t ∈ U → s ∩ t ∈ U
8.
In the proof that a measurable $\kappa$ is a strong limit, we take an injection $g : \kappa \to \mathcal P(\lambda)$ with $\lambda < \kappa$ and sets $Y_i \in U$ deciding whether $i \in g(\alpha)$ . Why does $Y = \bigcap_{i < \lambda} Y_i$ have at most one element?
- A.For $\alpha, \beta \in Y$ , $g(\alpha)$ and $g(\beta)$ contain the same $i < \lambda$ , so $g(\alpha) = g(\beta)$ , and $g$ is injective
- B.Because $U$ is $\kappa$ -complete
- C.Because $\kappa$ is regular
- D.Because $\lambda < \kappa$
9.
In Scott's proof that a measurable cardinal implies $V \ne L$ , we assume $V = L$ and take $j : V \to M$ with critical point $\kappa$ , the least measurable. Why is $M = V$ ?
- A. $M$ is a transitive class model of ZF containing all ordinals, $L$ is the least such class, and $V = L$
- B.Every elementary embedding is surjective
- C.Ultrapowers of $L$ are always equal to $L$ by definition
- D.Because $\kappa$ is measurable in $M$
10.
Write laver_table(n) returning a dict rows with rows[p][q - 1] = p * q in the Laver table $A_n$ on $\{1, \dots, 2^n\}$ : row $2^n$ is $1, 2, \dots, 2^n$ , and for $p < 2^n$ fill the rows from $p = 2^n - 1$ down to $1$ using $p * 1 = p + 1$ and $p * (q + 1) = (p * q) * (p + 1)$ . Then write period(row) returning the least period of a list.
Computational — Practice

Answer key — Large cardinals

1.
$\omega$ and $\aleph_1$ are regular; $\aleph_\omega$ and $\beth_\omega$ are suprema of $\omega$ -sequences of smaller cardinals, so they have cofinality $\omega$ and are singular. An inaccessible must be regular and a strong limit and uncountable — none of these four qualifies.
2.
$|V_0| = 0$ , $|V_1| = 2^0 = 1$ , $|V_2| = 2^1 = 2$ , $|V_3| = 2^2 = 4$ , $|V_4| = 2^4 = 16$ (and $|V_5| = 65536$ ). Every $V_n$ is finite because $\omega$ is a strong limit — the finite shadow of Lemma 1.
3.
If $U$ is non-principal, $\{n\} \notin U$ , so $\{n\}^c \in U$ for every $n$ . Countable completeness would put $\bigcap_n \{n\}^c = \varnothing$ in $U$ , which is impossible. So a countably complete ultrafilter on $\omega$ is principal, and measurability — which needs a $\kappa$ -complete non-principal ultrafilter on $\kappa$ , hence at least a countably complete one — requires uncountable $\kappa$ ; the definition says so explicitly.
4.
Replacement: if $F$ maps $x \in V_\kappa$ into $V_\kappa$ , then $F[x]$ has fewer than $\kappa$ elements, and its rank is the supremum of fewer than $\kappa$ ordinals below $\kappa$ , which is below $\kappa$ exactly because $\kappa$ is regular. For the singular $\beth_\omega$ , the map $n \mapsto \beth_n$ has range of rank $\beth_\omega$ , so Replacement fails in $V_{\beth_\omega}$ .
5.
If ZFC proved that an inaccessible $\kappa$ exists, then ZFC would prove that $V_\kappa$ is a model of ZFC, hence that ZFC is consistent. By Gödel's second incompleteness theorem, a consistent theory like ZFC cannot prove its own consistency — so ZFC would be inconsistent. Hence, if ZFC is consistent, it does not prove that an inaccessible exists.
6.
ZFC + IC proves Con(ZFC) via $V_\kappa$ , and trivially Con(ZFC + IC) implies Con(ZFC). The converse implication is not provable in ZFC, and ZFC + IC does not prove its own consistency. So ZFC + IC is strictly stronger than ZFC in consistency strength.
7.
κ-complete means closed under intersections of fewer than κ members, which is the first predicate. With ≤ κ or with no bound, no non-principal ultrafilter on a set of size κ would qualify; the last option holds for every ultrafilter.
8.
Each $Y_i$ fixes whether $i \in g(\alpha)$ , the same answer for all $\alpha \in Y_i$ . So all $\alpha \in Y$ have the same $g(\alpha)$ , and as $g$ is injective, $|Y| \le 1$ . But $Y \in U$ by $\kappa$ -completeness, and sets in a $\kappa$ -complete non-principal ultrafilter have size at least $\kappa$ (Lemma 4). Contradiction — so $2^\lambda < \kappa$ .
9.
Since $j(\alpha) \ge \alpha$ , $M$ contains all ordinals; it is a transitive class model of ZF; and $L$ is contained in every such class. With $V = L$ , $L \subseteq M \subseteq V = L$ . Then $M \models$ " $j(\kappa)$ is the least measurable" says the same in $V$ , so $j(\kappa) = \kappa$ , contradicting $j(\kappa) > \kappa$ .

10.

def laver_table(n):
    size = 2 ** n
    rows = {size: list(range(1, size + 1))}
    for p in range(size - 1, 0, -1):
        row = [p + 1]
        for _ in range(size - 1):
            row.append(rows[row[-1]][p])
        rows[p] = row
    return rows


def period(row):
    for d in range(1, len(row) + 1):
        if len(row) % d == 0 and row == row[:d] * (len(row) // d):
            return d

The periods of row 1 begin $1, 1, 2, 4, 4, 8, 8, 8, 8, 16$ . Laver proved from the rank-into-rank axiom I3 that they tend to infinity; whether ZFC alone proves this is not known.