Iterated dominance revisited

Abstract

Epistemic justifications of solution concepts often refer to type structures that are sufficiently rich. One important notion of richness is that of a complete type structure, i.e., a type structure that induces all possible beliefs about types. For instance, it is often said that, in a complete type structure, the set of strategies consistent with rationality and common belief of rationality are the set of strategies that survive iterated dominance. This paper shows that this classic result is false, absent certain topological conditions on the type structure. In particular, it provides an example of a finite game and a complete type structure in which there is no state consistent with rationality and common belief of rationality. This arises because the complete type structure does not induce all hierarchies of beliefs—despite inducing all beliefs about types. This raises the question: Which beliefs does a complete type structure induce? We provide several positive results that speak to that question. However, we also show that, within ZFC, one cannot show that a complete structure induces all second-order beliefs.

Appendices

Appendix A. Mathematical preliminaries

Lemma A1

Let $\varOmega$ and $\varPhi$ be metrizable and $f: \varOmega \rightarrow \varPhi$ be measurable. Then ${\underline{f}}$ is measurable.

Proof

Note, an open sub-basis for ${\mathcal {P}}(\varPhi )$ is given by the family of sets of the form

$$\begin{aligned} U(\overline{\nu },G,\varepsilon ) = \{ \nu \in {\mathcal {P}}(\varPhi ) : \nu (G) > \overline{\nu }(G) - \varepsilon \}, \end{aligned}$$

for $\overline{\nu } \in {\mathcal {P}}(\varOmega )$, G open in $\varPhi$, and $\varepsilon >0$. (See Billingsley 1968, page 236.) It suffices to show that, for each set $U=U(\overline{\nu },G,\varepsilon )$ in this open sub-basis, ${\underline{f}}^{-1}(U)$ is Borel in ${\mathcal {P}}(\varOmega )$.

Fix $U=U(\overline{\nu },G,\varepsilon )$. Let $r = \overline{\nu }(G)-\varepsilon$ and note that $f^{-1}(G)$ is Borel in $\varOmega$. With this, $\mu \in \underline{f}^{-1}(U)$ if and only if $\underline{f}(\mu ) \in U$ if and only if $\underline{f}(\mu )(G) > r$ if and only if $\mu (f^{-1}(G)) > r$. So, by Lemma 15.16 in Aliprantis and Border (2007), ${\underline{f}}^{-1}(U)$ is Borel in ${\mathcal {P}}(\varOmega )$.$\square$

Lemma A2

Let $\varOmega$ and $\varPhi$ be Polish and $f: \varOmega \rightarrow \varPhi$ be measurable and bijective. Then ${\underline{f}}$ is bijective.

Proof

The follows from the proof of Theorem 14.14 (2)-(3) in Aliprantis and Border (2007).$\square$

Appendix B. Proofs for Sect. 5

Lemma 1 and Proposition 1 are well-known in environments with stronger topological structure. We show here that the stronger topological structure is not important.

Remark B1

A strategy $s_{c} \in Y_{c}$ is undominated given $Y_{a} \times Y_{b}$ if and only if there exists some $\sigma _{d} \in {\mathcal {P}}(S_{d})$ with (i) $\pi _{c}(s_{c},\sigma _{d}) \ge \pi _{c}(r_{c},\sigma _{d})$, for all $r_{c} \in Y_{c}$, and (ii) $\sigma _{d}(Y_{d})=1$.

Proof of Lemma 1

Part (ii) is immediate from part (i). We show part (i) by induction on m.

$\mathbf {m=1:}$ If $(s_{c},t_{c}) \in \mathrm {R}_{c}^{1}$ then $s_{c}$ is optimal under $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c})$. So, by Remark B1, $s_{c}$ is undominated.

$\mathbf {m} \varvec{\ge } \mathbf {2:}$ Assume the result for m. If $(s_{c},t_{c}) \in \mathrm {R}_{c}^{m+1}$ then $s_{c}$ is optimal under $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c})$ and $\beta _{c}(\mathrm {R}_{d}^{m})=1$. It follows that $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c})(\mathrm {proj\,}_{S_{d}} \mathrm {R}_{d}^{m})=1$. So, by the induction hypothesis, $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c})(S_{d}^{m})=1$. Now the claim follows from Remark B1.$\square$

Proof of Proposition 1

Part (i) follows from Theorem 3(i) and Theorem 4(i). Part (ii) follows from Corollary 1(ii) and Friedenberg (2010).$\square$

Appendix C. Proofs for Sect. 6

The following lemmas are used in the proof of Lemma 3, which was in turn used in the proof of Theorem 2. We follow the notation introduced in Sect. 6.

Lemma C1

The collection

$$\begin{aligned} \{ \mathrm {T\,}[Q_{c},m] \backslash \mathrm {T\,}[Q_{c},m+1] : Q_{c} \in \mathbb {S}_{c}^{1} \text { and } m \ge 1 \} \end{aligned}$$

partitions $T_{c}$.

Proof

Clearly, elements of the collection are pairwise disjoint. It suffices to show that each $t_{c}$ is contained in $\mathrm {T\,}[Q_{c},m] \backslash \mathrm {T\,}[Q_{c},m+1]$, for some $Q_{c} \in \mathbb {S}_{c}^{1}$ and some $m \ge 1$. Certainly, each $t_{c} \in \mathrm {T\,}[Q_{c},1]$, for some $Q_{c}$. Moreover, using the fact that $\bigcap _{m}\mathrm {T\,}[Q_{c},m] = \emptyset$ (Property 2(ii)), it follows that there is some $m^*$ so that $t_{c} \not \in \mathrm {T\,}[Q_{c},m^*]$ and, so, there exists some m so that $t_{c} \in \mathrm {T\,}[Q_{c},m] \backslash \mathrm {T\,}[Q_{c},m+1]$.$\square$

Lemma C2

The collection of sets of probabilities

$$\begin{aligned} \{ (\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c}) : Q_{c} \in \mathbb {S}_{c}^{1} \text { and } m \ge 1 \}. \end{aligned}$$

partitions ${\mathcal {P}}(S_{d} \times T_{d})$.

Proof

Clearly, elements of the collection are pairwise disjoint. It suffices to show that each $\psi _{d}$ is contained in $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ for some $Q_{c} \in \mathbb {S}_{c}^{1}$ and some $m \ge 1$. To show this, note that each $\psi _{d} \in (\mathbb {BR}^{1}_{c})^{-1}(Q_{c})$ for some $Q_{c}$. Moreover, using Property 2(ii), it can be seen that $\bigcap _{m}P_{d}^{m} = \emptyset$. So, no $\psi _{d}$ can believe each $P_{d}^{m}$, i.e., there must be some $m^{*}$ so that $\psi _{d} \not \in (\mathbb {BR}^{m^{*}}_{c})^{-1}(Q_{c})$. So, for each $\psi _{d}$, there exists some m with $\psi _{d}$ contained in $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$.$\square$

Lemma C3

Fix some $Q_{c} \in \mathbb {S}_{c}^{1}$ and some $m \ge 1$. The set $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c})$ is closed.

Proof

Note,

$$\begin{aligned} (\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) = \{ \psi _{d} : \mathbb {BR}^{1}_{c} (\psi _{d}) = Q_{c} \} \cap \{ \psi _{d} : \psi _{d}(P_{c}^{m-1}) = 1 \}. \end{aligned}$$

A standard application of the Portmanteau Theorem (Kechris 1995, Theorem 17.20(i)-(ii)) gives that the set $\{ \psi _{d} : \mathbb {BR}^{1}_{c} (\psi _{d}) = Q_{c} \}$ is closed. It follows from Property 2(iii) that $P_{c}^{m-1}$ is clopen. So, the Portmanteau Theorem (Kechris 1995, Theorem 17.20(i)-(v)) gives that the set $\{ \psi _{d} : \psi _{d}(P_{c}^{m-1}) = 1 \}$ is closed. This establishes the result.$\square$

Lemma C4

Fix $Q_{c} \in \mathbb {S}_{c}^{1}$ and $m \ge 1$. The set $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ is a Polish space.

Proof

It suffices to show that $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ is a $G_{\delta }$ set, i.e., a countable intersection of open sets. If so, then the claim follows from Kechris (1995), Theorem 3.11.

Note,

$$\begin{aligned} (\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c}) = (\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \cap [ {\mathcal {P}}(S_{d} \times T_{d}) \backslash \{ \psi _{d} : \psi _{d}(P_{c}^{m}) = 1 \} ]. \end{aligned}$$

By Lemma C3 and Proposition 3.7 in Kechris (1995), $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c})$ is a $G_{\delta }$ set. The Portmanteau Theorem (Kechris 1995, Theorem 17.20(i)-(v)) gives that the set $\{ \psi _{d} : \psi _{d}(P_{c}^{m}) = 1 \}$ is closed, i.e., ${\mathcal {P}}(S_{d} \times T_{d}) \backslash \{ \psi _{d} : \psi _{d}(P_{c}^{m}) = 1 \}$ is open. So, $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ is a countable intersection of open sets, as required.$\square$

Lemma C5

For each $m \ge 0$ and each $s_{c} \in S_{c}^{m}$, there exists a type $t_{c}$ so that $(s_{c},t_{c}) \in P_{c}^{m} \backslash P_{c}^{m+1}$.

Proof

Throughout, fix $s_{c} \in S_{c}^{m}$. We break the proof into two cases.

$\mathbf {m=0:}$ Since the game is non-trivial, there exists some $\sigma _{d}$ so that $s_{c}$ is not optimal under $\sigma _{d}$, i.e., $s_{c} \notin \mathrm {BR\,}_{c}^{1}(\sigma _{d})$. For any $t_{c} \in \mathrm {T\,}[\mathrm {BR\,}_{c}^{1}(\sigma _{d}),1]$, $(s_{c},t_{c}) \in P_{c}^{0} \backslash P_{c}^{1}$.

$\mathbf {m} \varvec{\ge } \mathbf {1:}$ Using Property 1, there exists some $Q_{c} \in \mathbb {S}_{c}^{m}$ so that $s_{c} \in Q_{c}$. It follows from the construction that there is some $t_{c} \in \mathrm {T\,}[Q_{c},m]\backslash \mathrm {T\,}[Q_{c},m+1]$. So, $(s_{c},t_{c}) \in P_{c}^{m} \backslash P_{c}^{m+1}$.$\square$

Lemma C6

For each $Q_{c} \in \mathbb {S}_{c}^{1}$ and m, $\mathrm {T\,}[Q_{c},m] \backslash \mathrm {T\,}[Q_{c},m+1]$ is nonempty if and only if $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ is nonempty.

Proof

Note, by construction, $\mathrm {T\,}[Q_{c},m] \backslash \mathrm {T\,}[Q_{c},m+1]$ is nonempty if and only if $\mathrm {T\,}[Q_{c},m]$ is nonempty or, equivalently, if and only if $Q_{c} \in \mathbb {S}_{c}^{m}$. So, it suffices to show that $Q_{c} \in \mathbb {S}_{c}^{m}$ if and only if $(\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$ is nonempty.

First, suppose that $Q_{c} \in \mathbb {S}_{c}^{m}$. Then, there exists some $\sigma _{d} \in {\mathcal {P}}(S_{d})$ so that $\mathrm {BR\,}_{c}^{1}(\sigma _{d})=Q_{c}$ and $\sigma _{d}(S_{d}^{m-1})=1$. We will show that we can find some $\psi _{d}$ with $\mathbb {BR}^{m}_{c}(\psi _{d})=Q_{c}$ but $\psi _{d}(P_{d}^{m}) \ne 1$. This will establish the result.

Note, by Lemma C5, there exists a mapping $g: S_{d} \rightarrow T_{d}$ so that, for each $s_{d} \in S_{d}^{m}$, $g(s_{d}) \in P_{d}^{m-1} \backslash P_{d}^{m}$. Construct $\psi _{d}$ so that $\psi _{d}(\{s_{d},g(s_{d})\}) = \sigma _{d}(\{ s_{d} \})$, for all $s_{d}$. Then $\mathrm {marg\,}_{S_{d}}\psi _{d} = \sigma _{d}$, $\psi _{d}(P_{d}^{m-1})=1$, and $\psi _{d}(P_{d}^{m}) \ne 1$, as required.

Now suppose that $\psi _{d} \in (\mathbb {BR}^{m}_{c})^{-1}(Q_{c}) \backslash (\mathbb {BR}^{m+1}_{c})^{-1}(Q_{c})$. Then, $\mathrm {BR\,}^{1}_{c}(\mathrm {marg\,}_{S_{d}}\psi _{d}) =Q_{c}$ and $\psi _{d}(Q_{d}^{m-1})=1$. Using Lemma 2, $\mathrm {marg\,}_{S_{d}}\psi _{d}(S_{d}^{m-1})=1$. This establishes that $Q_{c} \in \mathbb {S}_{c}^{m}$.$\square$

Lemma C7

Fix Polish spaces $\varOmega$ and $\varPhi$. Let $\{ \varOmega ^{1}, \varOmega ^{2},\ldots \}$ be a partition of $\varOmega$ and $\{ \varPhi ^{1}, \varPhi ^{2},\ldots \}$ be a partition of $\varPhi$ so that each $\varOmega ^{n}$ and $\varPhi ^{n}$ is clopen. Suppose, for each n, there is a continuous map $f^{n} : \varOmega ^{n} \rightarrow \varPhi ^{n}$. Let $f:\varOmega \rightarrow \varPhi$ be such that $f(\omega )=f^{n}(\omega )$ if $\omega \in \varOmega ^{n}$. Then, f is continuous.

Proof

Since $\varOmega$ is Polish, it suffices to show that, if $\omega _{k}$ converges to $\omega$ in $\varOmega$, then $f(\omega _{k})$ converges to $f(\omega )$ in $\varPhi$. To show this, note that there exists some n so that $\omega \in \varOmega ^{n}$ and $f(\omega ) \in \varPhi ^{n}$. Since $\varOmega ^{n}$ is clopen, $\omega _{k} \in \varOmega ^{n}$, for k large. So, by construction, $f(\omega _{k}) \in \varPhi ^{n}$ for k sufficiently large. Using the fact that f is continuous on $\varOmega ^{n}$, it follows that $f(\omega _{k})$ converges to $f(\omega )$, as required.$\square$

Appendix D. Proofs for Sect. 8

This appendix first proves Theorem 3 and then turns to prove Remark 3.

1.1 D.1. Proof of Theorem 3.

Let us give the idea of the proof. We begin with an argument from Brandenburger and Dekel (1987): There is an epistemic game $(G,{\mathcal {T}})$ with a finite type structure ${\mathcal {T}}$ so that the RCBR prediction is the set of IU strategies.¹⁶ (See Lemma D1.) The key is that the RmBR set is preserved when going from ${\mathcal {T}}$ to any type structure ${\mathcal {T}}^{*}$ that is finitely terminal for ${\mathcal {T}}$. (See Lemma D2.) Taken together, this implies that, if a type structure is finitely terminal (resp. terminal) for all finite structures, then the RmBR (resp. RCBR) prediction contains the IU strategy set. (See Lemma D3.)

Lemma D1

There exists an $(S_{b},S_{a})$-based type structure ${\mathcal {T}}$ with $|T_{a}| \le |S_{a}|$ and $|T_{b}| \le |S_{b}|$ so that

• (i) For each m, $\mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{m} \times \mathrm {proj\,}_{S_{b}} \mathrm {R}_{b}^{m} = S_{a}^{m} \times S_{b}^{m}$.

• (ii) $\mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{\infty } \times \mathrm {proj\,}_{S_{b}} \mathrm {R}_{b}^{\infty } = S_{a}^{\infty } \times S_{b}^{\infty }$.

To prove Lemma D1, it will be useful to introduce some notation. Because the game is finite, we can find an $M<\infty$ so that $S_{a}^{M} \times S_{b}^{M}= S_{a}^{M-1} \times S_{b}^{M-1} = S_{a}^{\infty } \times S_{b}^{\infty }$. Fix one such M. For each $s_{c} \in S_{c}^{1}$, let

$$\begin{aligned} \overline{m}(s_{c})= {\left\{ \begin{array}{ll} m &{} \text { if } s_{c} \in S_{c}^{m} \backslash S_{c}^{m+1} \\ M &{} \text { if } s_{c} \in S_{c}^{\infty } . \end{array}\right. } \end{aligned}$$

Now construct a mapping $p_{c}: S_{c}^{1} \rightarrow {\mathcal {P}}(S_{d})$ so that, for each $s_{c} \in S_{c}^{1}$, (i) $s_{c}$ is optimal under $p_{c}(s_{c})$, and (ii) $p_{c}(s_{c})$ assigns probability 1 to $S_{d}^{\overline{m}(s_{c})-1}$. (See Remark B1.) Note, two features of this construction. First, if $s_{c} \in S_{c}^{2}$, $p_{c}(s_{c})(S_{d}^{1})=1$. Second, if $s_{c} \in S_{c}^{\infty }$, then $p_{c}(s_{c})(S_{d}^{\infty })=p_{c}(s_{c})(S_{d}^{M-1})=1$.

Proof of Lemma D1

Construct a type structure ${\mathcal {T}}$ as follows: Take each $T_{c} =S_{c}^{1}$. For each $s_{c} \in T_{c} =S_{c}^{1}$, let $\beta _{c}(s_{c}) \in {\mathcal {P}}(S_{d} \times S_{d}^{1})$ so that (i) for each $s_{d} \in S_{d}^{1}$, $\beta _{c}(s_{c})(s_{d},s_{d}) = p_{c}(s_{c})(s_{d})$, and (ii) for each $s_{d} \not \in S_{d}^{1}$, $\beta _{c}(s_{c})(\{s_{d}\} \times S_{d}^{1}) = p_{c}(s_{c})(s_{d})$.

It suffices to show the following:

• (i) If $s_{c} \in S_{c}^{m}$, then $(s_{c},s_{c}) \in R_{c}^{m}$.

• (ii) If $(s_{c},t_{c}) \in R_{c}^{m}$, then $s_{c} \in S_{c}^{m}$.

The proof is by induction on m.

$\mathbf {m=1}$: Part (i) is immediate. For Part (ii), fix $(s_{c},t_{c}) \in R_{c}^{1}$. By construction, $s_{c}$ is optimal under $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c}) = p_{c}(t_{c})$. So $s_{c} \in S_{c}^{1}$.

$\mathbf {m}{\varvec{\ge }} \mathbf {2}$: To show Part (i), fix $s_{c} \in S_{c}^{m+1}$. By the induction hypothesis, $(s_{c},s_{c}) \in R_{c}^{m}$. So, it suffices to show that $s_{c} \in T_{c}$ believes $R_{d}^{m-1}$. Toward that end note that $p_{c}(s_{c})(S_{d}^{m-1})=1$. So, by construction, $\beta _{c}(s_{c})(s_{d},t_{d})>0$ if and only if $s_{d}=t_{d}$. Then, by the induction hypothesis, $\beta _{c}(s_{c})(R_{d}^{m-1})=1$.

To show Part (ii), fix $(s_{c},t_{c}) \in R_{c}^{m+1}$. By construction, $s_{c}$ is optimal under $\mathrm {marg\,}_{S_{d}}\beta _{c}(t_{c}) = p_{c}(t_{c})$. Moreover, since $\beta _{c}(t_{c})(R_{d}^{m})=1$, the induction hypothesis gives that $p_{c}(t_{c})(S_{d}^{m})=1$. Thus, $p_{c}(t_{c})(S_{d}^{n})=1$ for each $n=0,\ldots ,m$. From this, $s_{c} \in S_{c}^{m+1}$.$\square$

Lemma D2

Fix $(S_{b},S_{a})$-based type structures

$$\begin{aligned} {\mathcal {T}}=( S_b,S_a;T_a,T_b;\beta _a,\beta _b )\quad \text {and} \quad {\mathcal {T}}^*= ( S_b,S_a;T_a^*,T_b^*;\beta _a^*,\beta _b^* ), \end{aligned}$$

with $T_{a},T_{b}$ are finite. If $\delta _{c}^{m}(t_{c})=\delta _{c}^{*,m}(t_{c}^{*})$ and $(s_{c},t_{c}) \in R_{c}^{m}$, then $(s_{c},t_{c}^{*}) \in R_{c}^{*,m}$.

In the proof of Lemma D2, we will make use of the following:

Remark D1

Types induce so-called coherent hierarchies of beliefs. In particular, for each $t_{c} \in T_{c}$ and each $m \ge n \ge 1$, $\mathrm {marg\,}_{Z_{c}^{n}} \delta _{c}^{m}(t_{c}) = \delta _{c}^{n}(t_{c})$.

Proof of Lemma D2

It suffices to show the following:

• (i) If $\delta _{c}^{m}(t_{c})=\delta _{c}^{*,m}(t_{c}^{*})$, then $(s_{c},t_{c}) \in R_{c}^{m}$ only if $(s_{c},t_{c}^{*}) \in R_{c}^{*,m}$.

• (ii) $(\rho _{d}^{*,m+1})^{-1}(\rho _{d}^{m+1}(\mathrm {R}_{c}^{m})) \subseteq \mathrm {R}_{c}^{*,m}$.

The proof is by induction on m.

$\mathbf {m = 1:}$ Begin with part (i). Suppose $\delta _{c}^{1}(t_{c})=\delta _{c}^{*,1}(t_{c}^{*})$. Note, $\delta _{c}^{1}(t_{c})=\mathrm {marg\,}_{S_{d}} \beta _{c}(t_{c})$ and $\delta _{c}^{*,1}(t_{c}^{*}) = \mathrm {marg\,}_{S_{d}} \beta _{c}^{*}(t_{c}^{*})$. So, $(s_{c},t_{c}) \in \mathrm {R}_{c}^{1}$ if and only if $(s_{c},t_{c}^{*}) \in \mathrm {R}_{c}^{*,1}$.

For part (ii), fix $(s_{c},t_{c}^{*}) \in (\rho _{d}^{*,2})^{-1}(\rho _{d}^{2}(\mathrm {R}_{c}^{1}))$. Then, there exists some $t_c$ such that $(s_{c},t_{c}) \in \mathrm {R}_{c}^{1}$ and $\rho _{d}^{*,2}(s_{c},t_{c}^{*}) = \rho _{d}^{2}(s_{c},t_{c})$. So, $\delta _{c}^{*,1}( t_{c}^{*} ) = \delta _{c}^{1}( t_{c} )$. Then, by part (i) (established for $m=1$), $(s_{c},t_{c}^{*}) \in \mathrm {R}_{c}^{*,1}$.

$\mathbf {m} \varvec{\ge } \mathbf {2:}$ Assume the result holds for m and we will show that it also holds for $m+1$.

To prove part (i), suppose $\delta _{c}^{m+1}(t_{c})= \delta _{c}^{*,m+1}(t_{c}^{*})$ and $(s_c,t_c)\in \mathrm {R}_c^{m+1}$. By Remark D1, $\delta _{c}^{m}(t_{c})= \delta _{c}^{*,m}(t_{c}^{*})$. So, by part (i) of the induction hypothesis, $(s_{c},t_{c}^{*}) \in \mathrm {R}_{c}^{*,m}$. As such, it suffices to show that $t_{c}^{*}$ believes $\mathrm {R}_{d}^{*,m}$.

Since $(s_{c},t_{c}) \in \mathrm {R}_{c}^{m+1}$, $\beta _{c}(t_{c})(\mathrm {R}_{d}^{m})=1$. Note, $\mathrm {R}_{d}^{m} \subseteq S_{d} \times T_{d}$ is finite and, so, $\rho _{c}^{m+1}(\mathrm {R}_{d}^{m})$ is Borel. As such,

$$\begin{aligned} \delta _{c}^{m+1}(t_{c})(\rho _{c}^{m+1}(\mathrm {R}_{d}^{m})) = \beta _{c}(t_{c})((\rho _{c}^{m+1})^{-1}(\rho _{c}^{m+1}(\mathrm {R}_{d}^{m}))) \ge \beta _{c}(t_{c})(\mathrm {R}_{d}^{m}) =1. \end{aligned}$$

Using the fact that $\delta _{c}^{*,m+1}(t_{c}^{*})=\delta _{c}^{m+1}(t_{c})$,

$$\begin{aligned} \delta _{c}^{*,m+1}(t_{c}^{*})(\rho _{c}^{m+1}(\mathrm {R}_{d}^{m}))=1. \end{aligned}$$

So,

$$\begin{aligned} \beta _{c}^{*}(t_{c}^{*})((\rho _{c}^{*,m+1})^{-1}(\rho _{c}^{m+1}(\mathrm {R}_{d}^{m})))=1. \end{aligned}$$

By part (ii) of the induction hypothesis, $(\rho _{c}^{*,m+1})^{-1}(\rho _{c}^{m+1}(\mathrm {R}_{d}^{m})) \subseteq \mathrm {R}_{d}^{*,m}$, so that $\beta _{c}^{*}(t_{c}^{*})(\mathrm {R}_{d}^{*,m})=1$, as desired.

To prove part (ii), fix $(s_{c},t_{c}^{*}) \in (\rho _{d}^{*,m+2})^{-1}(\rho _{d}^{m+2}(\mathrm {R}_{c}^{m+1}))$. There exists some $t_{c} \in T_{c}$ so that $(s_{c},t_{c}) \in \mathrm {R}_{c}^{m+1}$ and $\rho _{d}^{*,m+2}(s_{c},t_{c}^{*}) = \rho _{d}^{m+2}(s_{c},t_{c})$. So, $\delta _{c}^{*,m+1}( t_{c}^{*} ) = \delta _{c}^{m+1}( t_{c} )$. By part (i) established for $m+1$, $(s_{c},t_{c}^{*}) \in \mathrm {R}_{c}^{*,m+1}$.$\square$

Lemma D3

Fix an epistemic game $(G,{\mathcal {T}}^{*})$, where ${\mathcal {T}}^{*}$ is finitely terminal for all finite structures.

• (i) For each m, $S_{a}^{m} \times S_{b}^{m} \subseteq \mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{*,m} \times \mathrm {proj\,}_{S_{b}} \mathrm {R}_{b}^{*,m}$.

• (ii) If ${\mathcal {T}}^{*}$ is terminal for all finite structures, then $S_{a}^{\infty } \times S_{b}^{\infty } \subseteq \mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{*,\infty } \times \mathrm {proj\,}_{S_{b}}\mathrm {R}_{b}^{*,\infty }.$

Proof

Suppose ${\mathcal {T}}^{*}=(S_{b},S_{a};T_{a}^{*},T_{b}^{*},\beta _{a}^{*},\beta _{b}^{*})$ is finitely terminal for all finite type structures. Observe, by Lemma D1, there is a type structure ${\mathcal {T}}=(S_{b},S_{a};T_{a},T_{b},\beta _{a},\beta _{b})$ with finite type spaces so that

• For each m, $\mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{m} \times \mathrm {proj\,}_{S_{b}} \mathrm {R}_{b}^{m} = S_{a}^{m} \times S_{b}^{m}$.

• $\mathrm {proj\,}_{S_{a}} \mathrm {R}_{a}^{\infty } \times \mathrm {proj\,}_{S_{b}} \mathrm {R}_{b}^{\infty } = S_{a}^{\infty } \times S_{b}^{\infty }$.

Part (i): Fix $s_c\in S_c^m$. There exists some $t_c\in T_c$ so that $(s_c,t_c)\in \mathrm {R}_c^{m}$. Since $T_a$ and $T_b$ are finite and ${\mathcal {T}}^{*}$ is finitely terminal for all finite structures, there is some $t_{c}^{*}\in T_c^{*}$ with $\delta _{c}^{*,m}(t_{c}^{*}) = \delta _{c}^{m}(t_{c})$. By Lemma D2, $(s_c,t_c^{*})\in \mathrm {R}_c^{*,m}$.

Part (ii): Suppose ${\mathcal {T}}^{*}$ is terminal for all finite structures. Fix $s_c\in S_{c}^{\infty }$. There exists some $t_c\in T_c$ so that $(s_c,t_c)\in \mathrm {R}_{c}^{\infty }$. Since $T_a$ and $T_b$ are finite and ${\mathcal {T}}^{*}$ is terminal for all finite structures, then there exists $t_c^{*}\in T_c^{*}$ with $\delta _{c}^{*}(t_{c}^{*}) = \delta _{c}(t_{c})$. By Lemma D2, $(s_c,t_c^{*})\in \mathrm {R}_{c}^{*,\infty }$.$\square$

Proof of Theorem 3

Immediate from Lemmas 1 and D3.$\square$

1.2 D.2. Complete and not finitely terminal

We now turn to prove the claim in Remark 3:

Proposition D1

Suppose $X_{a}$ and $X_{b}$ are Polish and non-degenerate. Then there exists an $(X_{a},X_{b})$-based Polish complete type structure ${\mathcal {T}}$ that is finitely terminal but not terminal for all finite structures.

Proof We begin by constructing a game G: Let $S_{a}=X_{b}$ and $S_{b}=X_{a}$. Since each $X_{d}$ is non-degenerate, there are distinct strategies $s_{c}',s_{c}'' \in S_{c} = X_{d}$. Construct $\pi _{c}$ as follows:

$$\begin{aligned} \pi _{c}(s_{c},s_{d}) = {\left\{ \begin{array}{ll} 1 &{} \quad \text { if } (s_{c},s_{d}) \in \{ (s_{c}',s_{d}''), (s_{c}'',s_{d}') \} \\ 2 &{} \quad \text { if } (s_{c},s_{d}) \in (\{ s_{c}',s_{c}'' \} \times S_{d}) \backslash \{ (s_{c}',s_{d}''), (s_{c}'',s_{d}') \} \\ 0 &{} \quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Note, for each c and each $m\ge 1$, $S_{c}^{m} = \{ s_{c}',s_{c}'' \}$ and ${\mathbb {S}}_{c}^{m}=\{ \{ s_{c}' \},\{ s_{c}'' \},\{ s_{c}',s_{c}'' \} \}$.

Since, for $m\ge 1$, the sets $S_{c}^{m}$ and ${\mathbb {S}}_{c}^{m}$ are each finite, there exists a Polish and complete type structure ${\mathcal {T}}=(S_{b},S_{a};T_{b},T_{b};\beta _{a},\beta _{b})$ so that, in the epistemic game $(G,{\mathcal {T}})$, $\mathrm {R}_{a}^{\infty } \times \mathrm {R}_{b}^{\infty }=\emptyset$. (This amounts to repeating the proof of Theorem 1 line-by-line.) By Theorem 3.1 in Friedenberg (2010), ${\mathcal {T}}$ is finitely terminal. It remains to show that ${\mathcal {T}}$ is not terminal for all finite structures.

To do so, we show: If $(G,{\mathcal {T}}^{*})$ is an epistemic game where ${\mathcal {T}}^{*}$ is terminal for all finite structures, then $\mathrm {R}_{a}^{*,\infty } \times \mathrm {R}_{b}^{*,\infty } \ne \emptyset$. For that, we note that there exists an epistemic game $(G,\hat{{\mathcal {T}}})=(S_{b},S_{a};{\hat{T}}_{b},{\hat{T}}_{b};{\hat{\beta }}_{a},{\hat{\beta }}_{b})$ with $|{\hat{T}}_{a}| = |S_{a}^{1}| =2$ and $|{\hat{T}}_{b}|=|S_{b}^{1}|=2$ so that ${\hat{\mathrm {R}}}_{a}^{\infty } \times {\hat{\mathrm {R}}}_{b}^{\infty }= S_{a}^{\infty } \times S_{b}^{\infty }$. (Repeat the proof of Lemma D1 which takes each $T_{c}=S_{c}^{1}$.) Then the proof of Lemma D3 applies and we obtain that $S_{a}^{\infty } \times S_{b}^{\infty } \subseteq \mathrm {R}_{a}^{*,\infty } \times \mathrm {R}_{b}^{*,\infty }$.$\square$

Appendix E. Proofs for Sect. 9

We divide this appendix into three parts. First, we record a mathematical fact that will be useful. Second, we show the positive result mentioned in the main text; in fact, we generalize the result. Third, we complete the proof of the negative result.

1.1 E.1. Preliminary results

We begin with preliminary results that will be of use throughout this appendix.

Lemma E1

If $\beta _{c}(t_{c})$ has a countable set of measure one then, for each m, $\delta _{c}^{m}(t_{c})$ is atomic.

Proof Let $t_{c} \in T_{c}$ so that there is a countable set $E \subseteq X_{c} \times T_{d}$ with $\beta _{c}(t_{c})(E)=1$. Then, the image $\rho _{c}^{m}(E)$ is countable and $E \subseteq (\rho _{c}^{m})^{-1}(\rho _{c}^{m}(E))$. So,

$$\begin{aligned} \delta _{c}^{m}(t_{c})(\rho _{c}^{m}(E)) = \beta _{c}(t_{c})((\rho _{c}^{m})^{-1}(\rho _{c}^{m}(E))) \ge \beta _{c}(t_{c})(E) =1. \end{aligned}$$

As such, $\delta _{c}^{m}(t_{c})$ is atomic.$\square$

Corollary E1

Let ${\mathcal {T}}$ be a type structure such that, for each c and each $t_{c}$, $\beta _{c}(t_{c})$ has a countable set of measure one. Then ${\mathcal {T}}$ is atomic.

Corollary E2

If $X_{a},X_{b}$ are at most countable, then every countable $(X_{a},X_{b})$-based type structure ${\mathcal {T}}$ is atomic.

We will need the following concept:

Definition E1

Fix two $(X_{a},X_{b})$-based type structures, viz.

$$\begin{aligned} {\mathcal {T}}=(X_{a},X_{b};T_{a},T_{b};\beta _{a},\beta _{b}) \text{ and } {\mathcal {T}}^{*}=(X_{a},X_{b};T_{a}^{*},T_{b}^{*};\beta _{a}^{*},\beta _{b}^{*}). \end{aligned}$$

Say $(\tau _{a},\tau _{b})$ is a type morphism from ${\mathcal {T}}$ to ${\mathcal {T}}^{*}$ if each $\tau _{c}: T_{c} \rightarrow T_{c}^{*}$ is a measurable map with

$$\begin{aligned} (\underline{\mathrm {id\,}_{c} \times \tau _{d}}) \circ \beta _{c} = \beta _{c}^{*} \times \tau _{c}, \end{aligned}$$

where $\mathrm {id\,}_{c} \times \tau _{d} : X_{c} \times T_{d} \rightarrow X_{c} \times T_{d}^{*}$ satisfies $(\mathrm {id\,}_{c} \times \tau _{d})(x_{c},t_{d}) = (x_{c},\tau _{d}(t_{d}))$.

Lemma E2

Fix two $(X_{a},X_{b})$-based type structures, viz. ${\mathcal {T}}=(X_{a},X_{b};T_{a},T_{b};\beta _{a},\beta _{b})$ and ${\mathcal {T}}^{*}=(X_{a},X_{b};T_{a}^{*},T_{b}^{*};\beta _{a}^{*},\beta _{b}^{*})$. If $(\tau _{a},\tau _{b})$ is a type morphism from ${\mathcal {T}}$ to ${\mathcal {T}}^{*}$, then $\delta _{c}(t_{c}) = \delta _{c}^{*}(\tau (t_{c}))$ for each $t_{c} \in T_{c}$.

Lemma E2 is standard (see, e.g., Heifetz and Samet 1998, Proposition 5.1) and so the proof is omitted.¹⁷$\square$

1.2 E.2. Positive result

Lemma E3

Fix $(X_{a},X_{b})$-based type structures ${\mathcal {T}}$ and ${\mathcal {T}}^{*}$. If $\delta _{d}^{m}(t_{d})=\delta _{d}^{*,m}(t_{d}^{*})$ then, for each $x_{c} \in X_{c}$, $\rho _{c}^{m+1}(x_{c},t_{d})=\rho _{c}^{m+1}(x_{c},t_{d}^{*})$.

Proof

Let $\delta _{d}^{m}(t_{d})=\delta _{d}^{*,m}(t_{d}^{*})$ and note, by Remark D1, $(\delta _{c}^{1}(t_{c}),\ldots ,\delta _{c}^{m}(t_{c}))=(\delta _{c}^{*,1}(t_{c}^{*}),\ldots ,\delta _{c}^{*,m}(t_{c}^{*}))$. Now observe that

$$\begin{aligned} \rho _{c}^{m+1}(x_{c},t_{d})= & {} (\rho _{c}^{m}(x_{c},t_{d}),\delta _{d}^{m}(t_{d})) \\= & {} (x_{c}, \delta _{d}^{1}(t_{d}),\ldots ,\delta _{d}^{m}(t_{d})) \\= & {} (x_{c}, \delta _{d}^{*,1}(t_{d}^{*}),\ldots ,\delta _{d}^{*,m}(t_{d}^{*})) = \rho _{c}^{*,(m+1)}(x_{c},t_{d}^{*}), \end{aligned}$$

as required.$\square$

In the proof below, we will apply Lemma E3 both to the case where ${\mathcal {T}}$ and ${\mathcal {T}}^{*}$ are distinct type structures and to the case where they are the same type structure.

Proof of Theorem 4(i)

Fix $(X_{a},X_{b})$-based type structures ${\mathcal {T}}$ and ${\mathcal {T}}^{*}$, where ${\mathcal {T}}$ is countable and ${\mathcal {T}}^{*}$ is complete. We will show that, for each $t_{c} \in T_{c}$, there exists $t_{c}^{*} \in T_{c}^{*}$ so that $\delta _{c}^{*,m}(t_{c}^{*})=\delta _{c}^{m}(t_{c})$. Then, by Remark D1, $(\delta _{c}^{*,1}(t_{c}^{*}),\ldots ,\delta _{c}^{*,m}(t_{c}^{*}))=(\delta _{c}^{1}(t_{c}),\ldots ,\delta _{c}^{m}(t_{c}))$, as required. The proof is by induction on m.

$\mathbf {m=1:}$ Fix a type $t_{c} \in T_{c}$. By completeness, there exists a type $t_{c}^{*} \in T_{c}^{*}$ so that $\mathrm {marg\,}_{X_c}\beta _c^*(t_c^*)=\mathrm {marg\,}_{X_c}\beta _c(t_c)$. It follows that for each Borel $E \subseteq Z_{c}^{1}=X_{c}$,

$$\begin{aligned} \delta _{c}^{*,1}(t_{c}^{*})(E)=\mathrm {marg\,}_{X_c}\beta _c^*(t_c^*)(E)=\mathrm {marg\,}_{X_c}\beta _c(t_c)(E)=\delta _{c}^1(t_{c})(E), \end{aligned}$$

as required.

$\mathbf {m} \varvec{\ge } \mathbf {2:}$ Assume the result holds for m. By the induction hypothesis, there is a mapping $\tau _{d}^{m} : T_{d} \rightarrow T_{d}^{*}$ so that $\delta _{d}^{*,m}(\tau _{d}^{m}(t_{d}))= \delta _{d}^{m}(t_{d})$. By Lemma E3, for each $x_{c} \in X_{c}$, $\rho _{c}^{m+1}(x_{c},t_{d})=\rho _{c}^{*,(m+1)}(x_{c},\tau _{d}^{m}(t_{d}))$. Write $[t_{d}] := (\tau _{d}^{m})^{-1}(\{ \tau _{d}^{m} (t_{d})\})$. (So, each $t_{d}' \in [t_{d}]$ induces the same $m^{th}$-order belief as $t_{d}$.) Write ${\hat{T}}_{d} = \{ [t_{d}]: t_{d} \in T_{d} \}$.

Fix a type $t_{c}$. Construct $\psi \in {\mathcal {P}}(X_{d} \times T_{d}^{*})$ to be the image measure of $\mathrm {id\,}\times \tau _{d}^{m}$ under $\beta _{c}(t_{c})$. By completeness, there exists a type $t_{c}^{*} \in T_{c}^{*}$ with $\beta _{c}^{*}(t_{c}^{*}) = \psi$. We will show that $\delta _{c}^{*,m+1}(t_{c}^{*})=\delta _{c}^{m+1}(t_{c})$.

Fix some Borel $G \subseteq Z_{c}^{m+1}$. Since $T_{d}$ is countable,

$$\begin{aligned} \delta _{c}^{m+1}(t_{c})(G)= & {} \beta _{c}(t_{c})((\rho _{c}^{m+1})^{-1}(G) )\\= & {} \sum _{t_{d} \in T_{d}} \beta _{c}(t_{c})( F_{c}[t_{d}] \times \{t_{d} \} ), \end{aligned}$$

where $F_{c}[t_{d}] := (\rho _{c}^{m+1})^{-1}(G) \cap (X_{c} \times \{ t_{d} \} )$. Observe that, if $\tau _{d}^{m}(t_{d})=\tau _{d}^{m}(t_{d}')$, then $F_{c}[t_{d}]=F_{c}[t_{d}']$; this follows from Lemma E3. As such,

$$\begin{aligned} \delta _{c}^{m+1}(t_{c})(G)= & {} \sum _{[t_{d}] \in {\hat{T}}_{d}} \beta _{c}(t_{c})( F_{c}[t_{d}] \times [t_{d}] ). \end{aligned}$$

By construction,

$$\begin{aligned} \delta _{c}^{m+1}(t_{c})(G)= \sum _{[t_{d}] \in {\hat{T}}_{d}} \beta _{c}(t_{c})( F_{c}[t_{d}] \times [t_{d}] ) = \sum _{ t_{d} \in T_{d}} \psi _{c}( F_{c}[t_{d}] \times \{ \tau _{d}^{k}(t_{d}) \} ). \end{aligned}$$

Applying Lemma E3 again, $F_{c}[t_{d}] = (\rho _{c}^{*,m+1})^{-1}(G) \cap (X_{c} \times \{ \tau _{d}^{k}(t_{d}) \} )$. Thus,

$$\begin{aligned} \delta _{c}^{m+1}(t_{c})(G)= & {} \sum _{ t_{d} \in T_{d}} \psi _{c}( F_{c}[t_{d}] \times \{ \tau _{d}^{k}(t_{d}) \} ) \\= & {} \psi _{c}(( \rho _{c}^{*,m+1})^{-1}(G) \cap (X_{c} \times \tau _{d}^{k}(T_{d})) )\\= & {} \psi _{c}(( \rho _{c}^{*,m+1})^{-1}(G) )\\= & {} \delta _{c}^{*,m+1}(t_{c}^{*})(G), \end{aligned}$$

as desired.$\square$

We will prove a stronger version of Theorem 4(ii). To do so, we will need to introduce the following terminology.

Definition E2

Call a cardinal $\kappa$ is large if $\kappa$ is uncountable and there is a set X and a probability measure $\nu$ on $(X,2^{X})$ such that

• (i) $|X|=\kappa$,

• (ii) $\nu (E) \in \{ 0,1 \}$ for each subset E of X, and

• (iii) $\nu (F)=0$ for each finite subset F of X.

Call a cardinal $\lambda$ small if $\lambda$ is not large.

Theorem E1

Fix a complete $(X_{a},X_{b})$-based type structure ${\mathcal {T}}$. If $|X_a|$ and $|X_b|$ are small, ${\mathcal {T}}$ is finitely terminal for all atomic structures.

Note, ${\mathfrak {c}}$ is small (Fremlin 2008) and so Theorem E1 implies Theorem 4(ii). More generally, it cannot be proved in ZFC that large cardinals exist—but if they do exist, the first large cardinal is much greater than ${\mathfrak {c}}$, and even much greater than the first uncountable inaccessible cardinal. (See Fremlin 2008.)

Remark E1

In the literature, a cardinal $\kappa$ is called measurable if a set X of cardinality $\kappa$ has a $\kappa$-additive probability measure $\nu$ on $(X,2^X)$ such that Definition E2 (i)–(iii) holds. The existence of an uncountable measurable cardinal cannot be proved in ZFC. A cardinal $\kappa$ is large in the sense of Definition E2 if and only if $\kappa$ is $\ge$ the first uncountable measurable cardinal. So $\kappa$ is small if and only if there is no uncountable measurable cardinal below $\kappa$.

Say ${\mathcal {T}}$ is finitely terminal for all countable atomic structures if it is finitely terminal for all countable structures and it is finitely terminal for all atomic structures. To show Theorem E1, it suffices to show the following:

Lemma E4

Fix an $(X_{a},X_{b})$-based type structure ${\mathcal {T}}$. If $|X_a|$ and $|X_b|$ are small, then ${\mathcal {T}}$ is finitely terminal for all atomic structures if and only if ${\mathcal {T}}$ is finitely terminal for all countable atomic structures.

Proof of Theorem E1

The result follows from Theorem 4(i) and Lemma E4.$\square$

With the above in mind, it suffices to show Lemma E4. To do so, it will be useful to begin with the following:

Lemma E5

(Fremlin 2008).

• ${\mathfrak {c}}$ is small.

• If $\kappa$ is small and $\lambda \le \kappa$, then $\lambda$ is small.

• If $\kappa$ is small, then ${\mathfrak {c}}^\kappa$ is small.

Lemma E6

Suppose $|X_{a}|$ and $|X_{b}|$ are small. Then, for each c and each m, $|Z_{c}^{m}|$ is small.

Proof

The proof is by induction on m. The case of $m=1$ is immediate from the fact that $|X_{a}|$ and $|X_{b}|$ are small. Assume the result holds for m. Then, by the induction hypothesis, $|Z_{a}^{m}|$ and $|Z_{b}^{m}|$ are small. Each probability measure $\mu \in {\mathcal {P}}(Z_c^m)$ is a mapping from a subset of the power set of $Z_c^m$ into [0, 1]; so $|{\mathcal {P}}(Z_c^m)|\le {\mathfrak {c}}^\kappa$, where $\kappa =2^{|Z_c^m|}\le {\mathfrak {c}}^{|Z_c^m|}$. By Lemma E5, $\kappa$ is small, so $|{\mathcal {P}}(Z_c^m)|$ is small. Thus,

$$\begin{aligned} |Z_c^{m+1}|=|Z_c^m\times {\mathcal {P}}(Z_d^m)|=\max (|Z_c^m|,|{\mathcal {P}}(Z_d^m)|), \end{aligned}$$

and so $|Z_c^{m+1}|$ is also small.$\square$

Lemma E7

Let $\varOmega$ be a metrizable space. The following are equivalent:

• (i) Every discrete subset of $\varOmega$ has small cardinality.

• (ii) For every $\mu \in {\mathcal {P}}(\varOmega )$, every atom of $\mu$ contains a point mass of $\mu$.

• (iii) If $\mu \in {\mathcal {P}}(\varOmega )$, $\mu$ is atomic if and only if $\mu (D)=1$ for some countable $D\subseteq \varOmega$.

Proof

It is easily seen that (iii) implies (ii). It is also easily seen that the reverse direction of (iii) always holds: if $\mu (D)=1$ for some countable $D\subseteq \varOmega$ then $\mu$ is atomic.

(ii) implies (i): Assume that (i) fails, so there is a discrete subset E of $\varOmega$ of large cardinality. Then there is a Borel probability measure $\mu$ on $\varOmega$ such that $\mu (F)=0$ for every finite $F\subseteq E$, $\mu (E)=1$, and E is an atom of $\mu$. Then E is an atom that does not contain a point mass, so (ii) fails.

(i) implies (iii): Suppose (i) holds. Let $\mu$ be an atomic Borel probability measure on $\varOmega$. Then $\mu$ has a separable support. (See Billingsley 1968, Theorem 2, page 235.) This means that there is a closed separable set $C\subseteq \varOmega$ such that $\mu (C)=1$. Let D be a countable dense subset of C, and for each $n>0$ let $D_n$ be the union of all $\frac{1}{n}$-balls centered at elements of D. Then $C\subseteq D_n$ for each $n>0$, and $\bigcap _n D_n=D$. Therefore $\mu (D)=1$.$\square$

Corollary E3

If the cardinality of $\varOmega$ is small, then $\mu \in {\mathcal {P}}(\varOmega )$ is atomic if and only if $\mu (D)=1$ for some countable $D\subseteq \varOmega$.

To prove Lemma E4, we will take the union of pairwise disjoint type structures.

Definition E3

A family of $(X_{a},X_{b})$-based type structures $\{{\mathcal {T}}^i: i\in I\}$ is pairwise disjoint if, for each player c and each $i,j \in I$, $T_{c}^{i}$ is disjoint from $T_{c}^{j}$.

Definition E4

Let $\{{\mathcal {T}}^{i}: i\in I\}$ be a pairwise disjoint family of countable $(X_{a},X_{b})$-based type structures. The disjoint union, viz. ${\mathcal {T}}^{*}=\bigsqcup _{i\in I}{\mathcal {T}}^i$, is some ${\mathcal {T}}^*=(X_a,X_b;T_a^*,T_b^*;\beta _a^*,\beta _b^*)$ so that:

• (i) $T_{c}^{*}=\bigcup _{i\in I} T_{c}^{i}$;

• (ii) $T_{c}^*{}$ has the discrete topology; and

• (iii) $\beta _{c}^{*}:T_{c}^{*} \rightarrow {\mathcal {P}}(X_{c}\times T_{d}^{*})$ is such that, for each $i\in I$, $t_{c}^{i} \in T_{c}^{i}$, and Borel set $E \subseteq X_{c} \times T_{d}^{i}$, $(\beta _{c}^{*}(t_{c}^{i}))(E)=(\beta _{c}^{i}(t_{c}^{i}))(E)$.

So defined, ${\mathcal {T}}^{*}=\bigsqcup _{i\in I}{\mathcal {T}}^{i}$ is itself a type structure.¹⁸

Lemma E8

Let $\{{\mathcal {T}}^i:i\in I\}$ be a family of pairwise disjoint family of countable $(X_{a},X_{b})$-based type structures and let ${\mathcal {T}}^*=\bigsqcup _{i\in I} {\mathcal {T}}^i$ be the disjoint union of $\{{\mathcal {T}}^i:i\in I\}$.

• (i) For each $i\in I$ and each $t_{c} \in T_{c}^{i}$, $\delta _{c}^{i}(t_{c}) = \delta _{c}^{*}(t_{c})$.

• (ii) If the index set I is countable and ${\mathcal {T}}^i$ is atomic for each $i\in I$, then ${\mathcal {T}}^*$ is countable.

Proof

Part (ii) follows immediately from part (i). With this in mind, we show part (i). To do so, fix some ${\mathcal {T}}^{i}=(X_a,X_b;T_a^i,T_b^i;\beta _a^i,\beta _b^i)$ and write $\mathrm {id\,}_{c}^{i} : T_{c}^{i} \rightarrow T_{c}^{*}$ for the identity maps. Note, $(\mathrm {id\,}_{a}^{i},\mathrm {id\,}_{b}^{i})$ is a type morphism from ${\mathcal {T}}^{i}$ to ${\mathcal {T}}^{*}$. Then the claim follows from Lemma E2.$\square$

Lemma E9

Fix an atomic $(X_{a},X_{b})$-based type structure ${\mathcal {T}}=(X_{a},X_{b};T_{a},T_{b};\beta _{a},\beta _{b})$, where $|X_{a}|$ and $|X_{b}|$ are small. For each type $t_{c} \in T_{c}$ and each m, there exists a countable atomic $(X_{a},X_{b})$-based type structure $\overline{{\mathcal {T}}}=(X_{a},X_{b};{\overline{T}}_{a},{\overline{T}}_{b};{\overline{\beta }}_{a},{\overline{\beta }}_{b})$ and a type ${\overline{t}}_{c} \in {\overline{T}}_{c}$ so that ${\overline{\delta }}_{c}^{m}({\overline{t}}_{c})=\delta _{c}^{m}(t_{c})$.

Proof

The proof is by induction on m.

$\mathbf {m=1:}$ Fix some type $t_{c} \in T_{c}$. Construct $\overline{{\mathcal {T}}}$ as follows. Take ${\overline{T}}_{c} = \{{\overline{t}}_{c}\}$ and ${\overline{T}}_{d} \ne \emptyset$ finite. Set ${\overline{\beta }}_{c}({\overline{t}}_{c})$ so that, for each Borel $E_{c} \subseteq X_{c}$,

$$\begin{aligned} \overline{\beta }_{c}(\overline{t}_{c})(E_{c} \times \overline{T}_{d}) = \beta _{c}(t_{c})(E_{c} \times T_{d}). \end{aligned}$$

Choose $\overline{\beta }_{d}$ so that each $\overline{\beta }_{d}(\overline{t}_{d})$ is atomic.

Note, for each Borel $E_{c} \subseteq X_{c}$

$$\begin{aligned} \overline{\delta }_{c}^{1}(\overline{t}_{c})(E_{c}) = \overline{\beta }_{c}(\overline{t}_{c})(E_{c} \times \overline{T}_{d}) = \beta _{c}(t_{c})(E_{c} \times T_{d}) = \delta _{c}^{1}(t_{c})(E_{c}). \end{aligned}$$

So $\overline{\delta }_{c}^{1}(\overline{t}_{c}) = \delta _{c}^{1}(t_{c})$. Moreover, since ${\mathcal {T}}$ is atomic, $\overline{\delta }_{c}^{1}(\overline{t}_{c}) = \delta _{c}^{1}(t_{c})$ is atomic. By construction, each $\overline{\beta }_{d}(\overline{t}_{d})$ is atomic and, so, by Lemma E1, each $\overline{\delta }_{d}^{1}(\overline{t}_{d})$ is atomic.

$\mathbf {m} \varvec{\ge } \mathbf {2:}$ Assume the result holds for each player, each type in ${\mathcal {T}}$, and m. We will show that the same holds for $m+1$.

Fix a type $t_{c} \in T_{c}$ and note, by assumption, $\delta _{c}^{m+1}(t_{c})$ is atomic. Since $|X_{a}|,|X_{a}|$ are small, $|Z_c^{m+1}|$ is small and, so, each discrete subset of $Z_{c}^{m+1}$ is small. So, by Lemma E7, there is a finite or countable set of distinct points $E \subseteq Z_{c}^{m+1}$ so that (i) $\delta _{c}^{m+1}(t_{c})(E) =1$, and (ii) $\delta _{c}^{m+1}(t_{c})(\{z\})> 0$ for each $z \in E$. Note, E is the set of point masses of $\delta _{c}^{m}(t_{c})$ and so depends on both $t_{c}$ and m. It will be convenient to describe E as

$$\begin{aligned} E = \left\{ z^{k}: k \in K \right\} \end{aligned}$$

for some finite or countable index set K.¹⁹

For each $k \in K$,

$$\begin{aligned} \alpha ^{k} := \delta _{c}^{m+1}(t_{c})(\{z^{k} \})= \beta _{c}(t_{c})( (\rho _{c}^{m+1})^{-1}(\{z^k\} ) > 0. \end{aligned}$$

So, for each $k \in K$ there exists a point $(x_{c}^{k},t_{d}^{k}) \in X_{c} \times T_{d}$ with $\rho _{c}^{m+1}(x_{c}^{k},t_{d}^{k}) = z^{k}$. By the induction hypothesis, for each $k\in K$, there is a countable atomic type structure

$$\begin{aligned} \overline{{\mathcal {T}}}^{k}= (X_{a},X_{b};\overline{T}_{a}^{k},\overline{T}_{b}^{k};\overline{\beta }_{a}^{k},\overline{\beta }_{b}^{k}) \end{aligned}$$

and a type $\overline{t}_{d}^{k} \in \overline{T}_{d}^{k}$ with $\overline{\delta }_{d}^{k,m}(\overline{t}_{d}^{k}) = \delta _{d}^{m}(t_{d})$. By renaming points, we can take the family of type structures $\{\overline{{\mathcal {T}}}^{k}:k\in K\}$ to be pairwise disjoint. Thus, we can construct the disjoint union ${\mathcal {T}}^{*} = \bigsqcup _{k} \overline{{\mathcal {T}}}^{k}$. By Lemma E8(ii), ${\mathcal {T}}^{*}$ is a countable atomic type structure. Note, by construction, we choose the list of $\overline{t}_{d}^{k}, k\in K$ to be distinct (i.e., even if two indices are associated with the same types in ${\mathcal {T}}$) and so they are distinct types in $T_{d}^{*}$.

Construct a new type structure, viz. $\overline{{\mathcal {T}}}$, as follows: Take a new point $\overline{t}_{c} \notin T_{c}^{*}$ and set $\overline{T}_{c}=T_{c}^{*} \cup \{\overline{t}_{c} \}$. Set ${\overline{T}}_{d}=T_{d}^{*}$. Endow $\overline{T}_{c}$ and $\overline{T}_{d}$ with the discrete topology. Choose the maps $\overline{\beta }_{c}$ and $\overline{\beta }_{d}$ so that the identity maps form a type morphism from $\overline{{\mathcal {T}}}$ to ${\mathcal {T}}^{*}$. For $\overline{t}_{c}$, let $\overline{\beta }_{c}(\overline{t}_{c})$ be an atomic probability measure on $X_{c} \times \overline{T}_{d}$ such that

$$\begin{aligned} \overline{\beta }_{c}(\overline{t}_{c}) ( \{ (x_{c}^{k}, \overline{t}_{d}^{k}) \} ) = \alpha ^{k} \end{aligned}$$

for each $k\in K$. By Lemma E1, each $\overline{\delta }_{c}^{m}(\overline{t}_{c})$ is atomic. Since ${\mathcal {T}}^{*}$ is countable and atomic and each $\overline{\delta }_{c}^{m}(\overline{t}_{c})$ is atomic, it follows that $\overline{{\mathcal {T}}}$ is countable and atomic.

It remains to show that $\overline{\delta }_{c}^{m+1}(\overline{t}_{c}) = \delta _{c}^{m+1}(t_{c})$: Note, since the identity maps are a type morphism from $\overline{{\mathcal {T}}}$ to ${\mathcal {T}}^{*}$, we have

$$\begin{aligned} \overline{\delta }_{d}^{m}(\overline{t}_{d}^{k}) = \overline{\delta }_{d}^{k,m}(\overline{t}_{d}^{k}) = \delta _{d}^{m}(t_{d}^{k}). \end{aligned}$$

So,

$$\begin{aligned} \overline{\rho }_{c}^{m+1}(x_{c}^{k},\overline{t}_{d}^{k}) = \rho _{c}^{m+1}(x_{c}^{k},t_{t}^{k}) = z^{k}. \end{aligned}$$

It follows that

$$\begin{aligned} \overline{\delta }_{c}^{m+1}(\overline{t}_{c})(\{z^{k} \}) = \overline{\beta }_{c}(\overline{t}_{c})( (\overline{\rho }_{c}^{m+1})^{-1} (\{ z^{k} \}) ) \ge \overline{\beta }_{c}(\overline{t}_{c})(\{ (x_{c}^{k},\overline{t}_{d}^{k}) \}) = \alpha ^{k}. \end{aligned}$$

Using the fact that $\sum _{k} \alpha ^k=1$, it follows that $\overline{\delta }_{c}^{m+1}(\overline{t}_{c})(\{z^{k} \}) = \alpha ^{k}$ for each $k\in K$. So, $\overline{\delta }_{c}^{m+1}(\overline{t}_{c}) = \delta _{c}^{m+1}(t_{c})$, as desired.$\square$

Proof of Lemma E4

Fix an $(X_{a},X_{b})$-based type structure, viz. ${\mathcal {T}}^{*}$, that is finitely terminal for all countable atomic structures. Let ${\mathcal {T}}$ be an atomic $(X_{a},X_{b})$. By Lemma E9, ${\mathcal {T}}^{*}$ is finitely terminal for ${\mathcal {T}}$.$\square$

1.3 E.3. Negative result

Proof of Lemma 8

Let $T_a=T_b=[0,1]$ and endow $T_a$ and $T_b$ with the discrete topology. Then for each player c, $X_c\times T_d$ is discrete and has cardinality ${\mathfrak {c}}$. By Lemma 7(ii), each probability measure in ${\mathcal {P}}(X_{c}\times T_{d})$ has a countable set of measure one, and thus is atomic and is determined by a countable sequence of elements of $X_{c}\times T_{d} \times [0,1]$, corresponding to a sequence of points and measures of points. Therefore, ${\mathcal {P}}(X_{c}\times T_{d})$ has cardinality $(2^{\aleph _0})^{\aleph _0}=2^{\aleph _0}={\mathfrak {c}}$. Hence there are bijective functions $\beta _{c}$ from $T_{c}$ onto ${\mathcal {P}}(X_{c}\times T_d)$. Let ${\mathcal {T}}=(X_a,X_b;T_a,T_b;\beta _a,\beta _b)$. Since each $T_{c}$ is discrete, each $\beta _{c}$ is continuous; as such, ${\mathcal {T}}$ is a continuous type structure. Since each $\beta _c$ is onto, ${\mathcal {T}}$ is a complete type structure.

It remains to show that ${\mathcal {T}}$ is atomic. By the preceding paragraph, for each $\beta _{c}(t_{c})$, there is a countable set $E\subseteq X_c\times T_d$ such that $\beta _c(t_c)(E)=1$. So, by Corollary E1, ${\mathcal {T}}$ is atomic.$\square$

Appendix F. Variants of terminality: existence

Heifetz and Samet’s (1998) construction of a so-called universal type structure shows that there exists an $(X_{a},X_{b})$-based type structure that is terminal. Theorem 3 raises the question: Does there exist an $(X_{a},X_{b})$-based type structure that is terminal for all finite structures but not the universal type structure? If the answer were “no,” then the epistemic conditions provided by Theorem 3 would coincide with those provided by Tan and Werlang (1988, Result 3). This appendix shows that this is not the case. In particular, when $(X_{a},X_{b})$ are finite, there exists an $(X_{a},X_{b})$-based type structure that is not terminal, but is terminal for all finite type structures.To state the result, it will be useful to introduce some terminology.

Definition F1

Fix a type structure ${\mathcal {T}}=(X_{a},X_{b};T_{a},T_{b};\beta _{a},\beta _{b})$.

• (i) Call ${\mathcal {T}}$ finite if $T_{a}$ and $T_{b}$ are finite.

• (ii) Call ${\mathcal {T}}$ simple if, for each c, each type $t_c\in T_c$, and each m, $\delta _{c}^{m}(t_{c})$ has finite support.

Note that a probability measure has finite support if and only if some finite set has measure one. Every probability measure with finite support is atomic, so every simple type structure is atomic.

In the spirit of Definition 10, call an $(X_{a},X_{b})$-based type structure ${\mathcal {T}}^{*}$ finitely terminal for all simple (resp. finite simple) structures if it is finitely terminal for each simple (resp. finite and simple) $(X_{a},X_{b})$-type structure ${\mathcal {T}}=( X_a,X_b;T_a,T_b;\beta _a,\beta _b )$.

Proposition F1

Fix $(X_{a},X_{b})$.

• (i) There is a simple $(X_{a},X_{b})$-based type structure that is terminal for all finite structures.

• (ii) If $|X_{a}|$ and $|X_{b}|$ are small, there is an atomic $(X_{a},X_{b})$-based type structure that is terminal for all countable structures.

From this, we can conclude:

Corollary F1

Fix $(X_{a},X_{b})$.

• (i) There is an $(X_{a},X_{b})$-based type structure that is terminal for all finite structures but not terminal for all countable structures.

• (ii) If $|X_{a}|$ and $|X_{b}|$ are small, there is an $(X_{a},X_{b})$-based type structure that is terminal for all countable structures but not terminal.

Corollary F1 follows from Proposition F1 and the fact that there exists some type structure that induces a countable support (resp. full support) second-order belief. (This can be constructed or taken to follow from Heifetz and Samet 1999.)

To prove Proposition F1, we begin with a preliminary result:

Proposition F2

Fix $(X_{a},X_{b})$.

• (i) There is a simple $(X_{a},X_{b})$-based type structure that is terminal for all finite simple structures.

• (ii) There is an atomic $(X_{a},X_{b})$-based type structure that is terminal for all countable atomic structures.

The idea of the proof is clear: Construct an $(X_{a},X_{b})$-based type structure that is the disjoint union of all countable atomic $(X_{a},X_{b})$-based type structures. But this does not work because the class of all countable atomic type structures is a proper class. Instead, we construct a type structure that is the disjoint union of a set of countable atomic type structures that contains a copy of each countable atomic type structure.

Proof of Proposition F2

Let ${\mathcal {N}}=\{ {\mathcal {T}}^{i}: i \in I \}$ be the set of all finite simple (resp. countable atomic) $(X_{a},X_{b})$-based type structures ${\mathcal {T}}^{i}=(X_{a},X_{b}; T_{a}^{i},T_{b}^{i}; \beta _{a}^{i}, \beta _{b}^{i})$ whose type spaces are subsets of ${\mathbb {N}}$ and are endowed with the discrete topology. This set of type structures is not pairwise disjoint, but we can replace it by a pairwise disjoint set in the following way: For each ${\mathcal {T}}^{i} \in {\mathcal {N}}$, let

$$\begin{aligned} \hat{{\mathcal {T}}}^{i} = (X_{a},X_{b};{\hat{T}}_{a}^{i},{\hat{T}}_{b}^{i};{\hat{\beta }}_{a}^{i},{\hat{\beta }}_{b}^{i}), \end{aligned}$$

where ${\hat{T}}_{c}^{i} = T_{c}^{i} \times \{ i \}$ and ${\hat{\beta }}_{c}^{i}(E \times \{ i \}) = \beta _{c}^{i}(E)$ for each Borel $E \subseteq X_{c} \times T_{d}^{i}$. Write $\hat{{\mathcal {N}}}$ for the set of all such $\hat{{\mathcal {T}}}^{i}$. Note that $\hat{{\mathcal {N}}}=\{ \hat{{\mathcal {T}}}^{i}: i \in I \}$ is a pairwise disjoint set of countable atomic $(X_{a},X_{b})$-based type structures. Moreover, for each $\hat{{\mathcal {T}}}^{i} \in \hat{{\mathcal {N}}}$ (resp. ${\mathcal {T}}^{i} \in {\mathcal {N}}$), there is a type morphism from $\hat{{\mathcal {T}}}^{i}$ to ${\mathcal {T}}^{i}$ (resp. ${\mathcal {T}}^{i}$ to $\hat{{\mathcal {T}}}^{i}$).

Let ${\mathcal {T}}^{*} = \bigsqcup _{ i \in I} \hat{{\mathcal {T}}}^{i}$. Note that, for each $(t_{c},i) \in {\hat{T}}_{c}^{i} \subseteq T_{c}^{*}$ (resp. $t_{c} \in T_{c}^{i}$), $\delta _{c}^{*}((t_{c},i)) = {\hat{\delta }}_{c}^{i}((t_{c},i))=\delta _{c}^{i}(t_{c})$. (This follows from Lemma E8(i) and Lemma E2.) Since each ${\mathcal {T}}^{i}$ is simple (resp. atomic), ${\mathcal {T}}^{*}$ is simple (resp. atomic). Moreover, ${\mathcal {T}}^{*}$ is terminal for each ${\mathcal {T}}^{i} \in {\mathcal {N}}$.

Finally, fix an $(X_{a},X_{b})$-based type structure ${\mathcal {T}}$ that is finite and simple (resp. countable and atomic), but necessarily in ${\mathcal {N}}$. There exists a type morphism from ${\mathcal {T}}$ to some ${\mathcal {T}}^{i} \in {\mathcal {N}}$. By Lemma E2, ${\mathcal {T}}^{i}$ is terminal for ${\mathcal {T}}$. As such, ${\mathcal {T}}^{*}$ is terminal for ${\mathcal {T}}$.$\square$

Proposition F1(ii) follows immediately from Proposition F2(ii) and Lemma E4. To show Proposition F1(i), we need the following analogue of Lemma E4.

Lemma F1

An $(X_{a},X_{b})$-based type structure is finitely terminal for all simple structures if and only if it is finitely terminal for all finite simple structures.

The remainder of this appendix to devoted to showing Lemma F1.

Lemma F2

If $\beta _{c}(t_{c})$ has a finite set of measure one then, for each m, $\delta _{c}^{m}(t_{c})$ is simple.

Proof

Let $t_{c} \in T_{c}$ so that there is a finite set $E \subseteq X_{c} \times T_{d}$ with $\beta _{c}(t_{c})(E)=1$. Then, the image $\rho _{c}^{m}(E)$ is finite and $E \subseteq (\rho _{c}^{m})^{-1}(\rho _{c}^{m}(E))$. So,

$$\begin{aligned} \delta _{c}^{m}(t_{c})(\rho _{c}^{m}(E)) = \beta _{c}(t_{c})((\rho _{c}^{m})^{-1}(\rho _{c}^{m}(E))) \ge \beta _{c}(t_{c})(E) =1. \end{aligned}$$

Hence, $\delta _{c}^{m}(t_{c})$ is simple.$\square$

Lemma F3

Fix a finite index set I and a family of pairwise disjoint family of finite and simple $(X_{a},X_{b})$-based type structures $\{{\mathcal {T}}^i:i\in I\}$. Then, for each $i \in I$ and each $t_{c} \in T_{c}^{i}$, $\delta _{c}^{i}(t_{c}) = \delta _{c}^{*}(t_{c})$.

Proof

Immediate from Lemma E8(i).$\square$

Lemma F4

Fix a simple $(X_{a},X_{b})$-based type structure ${\mathcal {T}}=(X_{a},X_{b};T_{a},T_{b};\beta _{a},\beta _{b})$. For each type $t_{c} \in T_{c}$ and each m, there exists a finite and simple $(X_{a},X_{b})$-based type structure $\overline{{\mathcal {T}}}=(X_{a},X_{b};{\overline{T}}_{a},{\overline{T}}_{b};{\overline{\beta }}_{a},{\overline{\beta }}_{b})$ and a type ${\overline{t}}_{c} \in {\overline{T}}_{c}$ so that ${\overline{\delta }}_{c}^{m}({\overline{t}}_{c})=\delta _{c}^{m}(t_{c})$.

Proof

The proof is the same as the proof of Lemma E9 except that the word “countable” is replaced with “finite,” “atomic” is replaced with “simple,” the index set K if finite, Lemma E1 is replaced with Lemma F2 and Lemma E8 is replaced with Lemma F3. Note, in this case, we need not require that $|X_{c}|$ is small: If ${\mathcal {T}}$ is simple, there is necessarily a finite set of points in $Z_{c}^{m}$ that gets probability one under $\delta _{c}^{m}(t_{c})$, irrespective of whether or not $|Z_{c}^{m}|$ is small.$\square$

Proof of Lemma F1

Suppose ${\mathcal {T}}$ is an $(X_{a},X_{b})$-based type structure that is finitely terminal for all finite simple type structures. If ${\mathcal {T}}^*$ be an $(X_{a},X_{b})$-based simple type structure, then ${\mathcal {T}}$ is finitely terminal for ${\mathcal {T}}^*$. (See Lemma F4.)$\square$