Continuity postulates and solvability axioms in economic theory and in mathematical psychology: a consolidation of the theory of individual choice

Abstract

This paper presents four theorems that connect continuity postulates in mathematical economics to solvability axioms in mathematical psychology, and ranks them under alternative supplementary assumptions. Theorem 1 connects notions of continuity (full, separate, Wold, weak Wold, Archimedean, mixture) with those of solvability (restricted, unrestricted) under the completeness and transitivity of a binary relation. Theorem 2 uses the primitive notion of a separately continuous function to answer the question when an analogous property on a relation is fully continuous. Theorem 3 provides a portmanteau theorem on the equivalence between restricted solvability and various notions of continuity under weak monotonicity. Finally, Theorem 4 presents a variant of Theorem 3 that follows Theorem 1 in dispensing with the dimensionality requirement and in providing partial equivalences between solvability and continuity notions. These theorems are motivated for their potential use in representation theorems.

Appendices

Appendix

In the Appendix, we present the proofs of the results and six technical examples.

Appendix A: Proofs of the Results

Proof of Theorem 1 (a) The implication that Wold-continuity implies weak Wold-continuity follows from their definitions. Uyanik and Khan (2022, Proposition 3) prove that continuity implies Wold-continuity for finite dimensional spaces, their arguments extend to the current setting trivially.

Next, we show that weak Wold-continuity implies restricted solvability. Towards this end assume $\succsim$ is a weakly Wold-continuous binary relation on a convex set $X \subseteq \prod _{i\in I}X_{i}$. Let $y=(a_{i}, y_{-i} )$, $z= (b_{i},y_{-i} )$ and $y\succsim x\succsim z$. If $y\sim x \sim z,$ or $y\succ x \sim z$, or $y\sim x \succ z,$ then restricted solvability trivially follows. Hence, assume $y\succ x \succ z$. From weak Wold-continuity there exists a straight line $y\lambda z$ intersecting the indifference class of x, i.e., $\lambda y + (1-\lambda ) z \sim x$, where $\lambda y + (1-\lambda ) z= (\lambda a_{i}+(1-\lambda ) b_{i}, y_{-i})$. Therefore, $\succsim$ is restricted solvable.

It remains to show that weak Wold-continuity implies Archimedean property. Towards this end assume $\succsim$ is Wold-continuity but not Archimedean. Then, there exists $x,y,z\in X$ such that $x\succ y$ but for all $\lambda \in (0,1)$, $x\lambda z\nsucc y$ (the proof of the case where $x\nsucc y\lambda z$ is analogous). By completeness, $y\succsim x\lambda z$. Pick $\lambda \in (0,1)$. If $y\succ x\lambda z$, then $x\succ y\succ x\lambda z$ and weak Wold-continuity imply there exist $\delta \in (0,1)$ such that $y\sim x\delta z$. If $y\sim x\lambda z$, then set $\delta =\lambda$. Then by transitivity, $x\succ x\delta z$. By weak Wold-continuity, there exists $\gamma \in (0,1)$ such that $x\succ x\gamma z\succ x\delta z\sim y$. Hence, by transitivity, $x\gamma z\succ y$. This furnishes us a contradiction with the assumption that for all $\lambda ' \in (0,1)$, $x\lambda ' z\nsucc y$.

(b) Continuity implies separate continuity follows from the definitions of the two continuity postulates. For the proof of the remaining implications except mixture-continuity implies separate continuity, see Uyanik and Khan (2022, Proposition 3).

It remains to show that mixture-continuity implies separate continuity. Assume $\succsim$ is separately continuous. Pick an index i, $x\in X$ and a sequence $(y_i^n, y_{-i})$ converging $(y_i, y_{-i})\in X$ such that $(y_i^n, y_{-i})\succsim x$. mixture-continuity implies that the set $A=\{\lambda \in [0,1]~|~(y_i, y_{-i})\lambda y\succsim x\}$ is closed. It is clear that $1\in A$ and for all $\lambda >0$, $\lambda \in A$. Hence $1\in A$. Therefore, The proof that for all $x\in X$ and all lines L parallel to any coordinate axis, $A_\succsim (x)\cap L$ is closed. The proof that for all $x\in X$ and all lines L parallel to any coordinate axis, $A_\precsim (x)\cap L$ is closed analogously follows. Since $\succsim$ is complete, $\succsim$ is separately continuous.

(c) Assume $\succsim$ is separately continuous. Let $(a_{i}, y_{-i})\succsim x\succsim (b_{i}, y_{-i})$. Separate continuity implies that the weakly worse-than and weakly better-than sets of x are closed in the subspace of X determined by the line $L_i$ parallel to the i-$\text {th}$ coordinate axis and passing through $y_{-i}$, i.e. $\succsim _{x}\upharpoonright _{L_i}$ and $\precsim _{x}\upharpoonright _{L_i}$ are closed in $X\cap L_i$. Since $X\cap L_i$ is convex, it is connected. Moreover, it follows from the completeness of the relation that the intersection of these two sets is non-empty, i.e., $\succsim _{x}\upharpoonright _{L_i}\cap \precsim _{x}\upharpoonright _{L_i}\ne \varnothing$. Hence, there exists ${\tilde{x}}\in X\cap L_i$ such that ${\tilde{x}}\sim x$. Since ${\tilde{x}}\in X\cap L_i$, there exists $c_i\in \mathbb {R}$ such that ${\tilde{x}}=(c_i,{\bar{y}}_{-i})$. Therefore, $\succsim$ is restricted solvable.

(d) Assume $\succsim$ is unrestricted solvable. And assume its is not restricted solvable. Then there exists no b such that $bq\sim ap$ whenever $\bar{b}q\succsim ap \succsim \underline{b}q$. But this contradicts unrestricted solvability.$\square$

Proof of Theorem 2 We provide the proof by induction. Note that for $n=1$, separate continuity is equivalent to continuity by definition. Now let $n=2$ and assume without loss of generality that $\succsim$ is weakly monotone in its first coordinate. We now show that $\succsim$ has closed lower sections. Towards this end, assume there exist $x\in X$ and a sequence $y^k\rightarrow y$ such that $y^k\precsim x$ for all k and . Then, either $y\succ x$ or $y\bowtie x$.

Assume $y\succ x$. Separate continuity implies that there exists $\varepsilon >0$ such that for all z in the $\varepsilon$-neighborhood of y, $(z_1, y_2)\succ x$. Pick $(z_1, y_2)$ in the $\varepsilon$-neighborhood of y such that $z_1<y_1-\varepsilon /2$. Separate continuity then implies that there exists $\delta _1>0$ such that for all $z'$ in the $\delta _1$-neighborhood of $(z_1, y_2)$, $(z_1, z'_2)\succ x$. Define $\delta =\min \{\varepsilon /2, \delta _1\}$. Since $y^k\rightarrow y$, there exists ${\hat{k}}\in {\mathbb {N}}$ such that $y^{{\hat{k}}}$ is in the $\delta$-neighborhood of y. Then, $(z_1, y_2^{{\hat{k}}})$ is in the $\delta _1$-neighborhood of $(z_1, y_2)$ and $z_1<y_1^{{\hat{k}}}$. Hence, by weak monotonicity, $y^{{\hat{k}}}\succsim (z_1, y_2^{{\hat{k}}})$. It follows from $(z_1, y_2^{{\hat{k}}})\succ x$ and transitivity of $\succsim$ that⁹$y^{{\hat{k}}}\succ x$. This furnishes us a contradiction.

When $y\bowtie x$, replacing $\succ$ with $\bowtie$ in the paragraph above yields $y^{{\hat{k}}}\succsim (z_1, y_2^{{\hat{k}}})$, as above. Then, $y^{{\hat{k}}}\succsim (z_1, y_2^{{\hat{k}}})\bowtie x$. If $x\succsim y^{{\hat{k}}}$, then transitivity of $\succsim$ implies that $x \succsim (z_1, y_2^{{\hat{k}}})$, yielding a contradiction. Hence . This furnishes us a contradiction.

Therefore, $\succsim$ has closed lower sections. The proof of the closed upper sections is analogous. Moreover, the proof of the open sections have a similar construction. Finally, completeness of $\succsim$ follows from (2021, Theorem 2).

Now assume the theorem is true for $n-1$ dimensional spaces, $n>2$. We next show that it is true for n dimensional spaces. Recall that $\succsim$ is weakly monotone in all coordinates (except possibly one). Pick a coordinate i in which $\succsim$ is weakly monotone. We now show that $\succsim$ has closed lower sections. Towards this end, assume there exist $x\in X$ and a sequence $y^k\rightarrow y$ such that $y^k\precsim x$ for all k and . Then, either $y\succ x$ or $y\bowtie x$.

Assume $y\succ x$. Separate continuity implies that there exists $\varepsilon >0$ such that for all z in the $\varepsilon$-neighborhood of y, $(z_i, y_{-i})\succ x$. Pick $(z_i, y_{-i})$ in the $\varepsilon$-neighborhood of y such that $z_i<y_i-\varepsilon /2$. By induction hypothesis, $\succsim$ is continuous on $n-1$ dimensional spaces. Hence, there exists $\delta _i>0$ such that for all $z'$ in the $\delta _i$-neighborhood of $(z_i, y_{-i})$, $(z_i, z'_{-i})\succ x$. Define $\delta =\min \{\varepsilon /2, \delta _i\}$. Since $y^k\rightarrow y$, there exists ${\hat{k}}\in {\mathbb {N}}$ such that $y^{{\hat{k}}}$ is in the $\delta$-neighborhood of y. Then, $(z_i, y_{-i}^{{\hat{k}}})$ is in the $\delta _i$-neighborhood of $(z_i, y_{-i})$ and $z_i<y_i^{{\hat{k}}}$. Hence, by weak monotonicity, $y^{{\hat{k}}}\succsim (z_i, y_{-i}^{{\hat{k}}})$. It follows from $(z_i, y_{-i}^{{\hat{k}}})\succ x$ and transitivity of $\succsim$ that $y^{{\hat{k}}}\succ x$. This furnishes us a contradiction.

The remaining part of the proof is analogous to the case $n=2$ and uses the modification of the construction above for general n, hence omitted.$\square$

Proof of Theorem 3 Let $\succsim$ be a complete, transitive, order dense and weakly monotonic binary relation on a convex and order-bounded set $X \subseteq {\mathbb {R}}^n$. Theorems 1 and 2 prove all the relationships except that restricted solvability implies separate continuity. Towards this end, assume $\succsim$ is restricted solvable but not separately continuous.

Assume there exists $x\in X$ and a line L parallel to a coordinate axis such that $A_\succsim (x)\cap L$ is not closed. Then, there exist $x\in X$, an index i, and a sequence $y^k \rightarrow y$ on the line $L_i$ parallel to coordinate i such that $y^k\succsim x$ for all k and $y\prec x$. Then it follows from transitivity and weak monotonicity assumptions that $y_i^k> y_i$ for all k. Pick $m\in {\mathbb {N}}$. Since $y\prec x$, order denseness implies that there exists $x'\in X$ such that $y\prec x'\prec x$. Then transitivity implies $y\prec x'\prec y^m$, and hence restricted solvability implies there exists $z\in L_i$ such that $z\sim x'$. It follows from weak monotonicity that $y_i<z_i<y^m_i$, and from transitivity that for all $z_i'\in (y_i, z_i]$, $(z'_i, z_{-i})\prec x$. However, since $y^k\rightarrow y$, there exists $z_i'\in (y_i, z_i]$, such that $z_i'=y_i^k$ for some k and $(z'_i, z_{-i})\succsim x$. This yields a contradiction.

The proof that for all $x\in X$ and all lines L parallel to any coordinate axis, $A_\precsim (x)\cap L$ is closed analogously follows. Since $\succsim$ is complete, $\succsim$ is separately continuous.$\square$

Proof of Theorem 4 We first show that Archimedean postulate implies mixture-continuity. Assume there exist $x,y,z\in X$ such that $\{\lambda : x\lambda y\succsim z\}$ is not closed. Then there exists $\lambda ^n\rightarrow \lambda$ such that $x\lambda ^n y\succsim z$ but $z\succ x\lambda y$. By strong order-boundedness assumption, there exists $\varepsilon >0$ such that $p=((x\lambda y)_i +\varepsilon )_{i\in I}\in X$. By weak monotonicity, $p\succsim p\delta (x\lambda y)$ for all $\delta \in [0,1]$. By Archimedean, there exists $\beta \in (0,1)$ such that $z\succ p\beta (x\lambda y)$. Note that there exists $\epsilon \in (0,\varepsilon )$ such that $(p\beta (x\lambda y))_i - (x\lambda y)_i > \epsilon$ for all $i\in I$. Therefore, there exists $N\in {\mathbb {N}}$ such that for all $n\ge N$, $(x\lambda ^n y)_i < (p\beta (x\lambda y))_i$ for all $i\in I$. By weak monotonicity, $p\beta (x\lambda y)\succsim x\lambda ^n y$ for all $n\ge N$. By transitivity, $z\succ x\lambda ^n y$ for all $n\ge N$. This yields a contradiction. Closedness of the lower sections follows analogously.

By the figure above, we now prove the equivalence of mixture continuity, Archimedean and weak Wold-continuity. The proof that restricted solvability implies separate continuity follows from the argument in the proof of Theorem 4 since it does not hinge on the dimension of the underlying space.

Appendix B: Examples

In this section we show that in Figure 1, the converse relationships among the continuity postulates are false. All the binary relations presented here satisfy completeness and transitivity.

The first example shows that continuity, and hence all other continuity postulates, do not imply unrestricted solvability even under weakly monotone and convex preferences.

1. Let $\succsim$ be a binary relation defined on ${\mathbb {R}}^{2}$ as $(x_{1},x_{2})\succsim (y_{1},y_{2})$ if and only if $f(x_{1},x_{2})\ge f(y_{1},y_{2})$ where $f(x_{1},x_{2})=x_{2}.$ To see that $\succsim$ does not satisfy unrestricted solvability, for all $x_1\in \mathbb {R}$, $(2,2)\succ (x_1,1)$, hence $\succsim$ is not unrestricted solvable in the second component. It is clear that $\succsim$ is continuous, hence it satisfies the remaining six continuity postulates, following from Theorem 1. $\square$

The second example shows that unrestricted solvability and restricted solvability do not imply separate continuity, Archimedean and weak Wold-continuity, and hence any of the other continuity postulates.

2. Let the binary relation $\succsim$ be defined on a bounded subset $X\subset {\mathbb {R}}^{2}$ as $(x_{1},x_{2})\succsim (y_{1},y_{2})$ if and only if $f(x_{1},x_{2})\ge f(y_{1},y_{2})$ where

$$\begin{aligned} f(x_{1},x_{2}) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x_{1}+x_{2}< 1 \\ 0.8, &{} \text { if }x_{1}+x_{2}= 1 \\ 1, &{} \text { if } x_{1}+x_{2}> 1 \end{array}\right. } \end{aligned}$$

To see why this is not separately continuous and hence not continuous, pick $(x_{1},x_{2})=(1,0).$ For the binary relation to be separately continuous, the restriction of (1, 0) to any line parallel to either $x_{1}$ or $x_{2}$ axis should be continuous. Let $x_{2}=0.5.$ Then $\succsim _{(1,0)}\upharpoonright _{x_{2}=0.5}=(0.5,b]$ where b is some bound on $x_{1}.$ This representation is neither Wold nor weak Wold-continuous (does not satisfy order-denseness). However, this binary relation is both restricted and unrestricted solvable.

This is not Archimedean. Pick $x=(0.6,0.4)$ and $y=z=(0.6,0.5).$ Then $y\succ x$ but $y\sim x\delta z$ for any $\delta \in (0,1).$ $\square$

The third example shows that separate continuity does not imply weak Wold-continuity, Archimedean and unrestricted solvability, hence does not imply mixture-continuity, Wold-continuity and continuity postulates. It also shows that restricted solvability does not imply any of these continuity postulates.

3. Let the binary relation $\succsim$ be defined on ${\mathbb {R}}^{2}$ as $(x_{1},x_{2})\succsim (y_{1},y_{2})$ if and only if $f(x_{1},x_{2})\ge f(y_{1},y_{2})$ where

$$\begin{aligned} f(x_{1},x_{2}) = {\left\{ \begin{array}{ll} \dfrac{x_{1}x_{2}}{x_{1}^{2}+x_{2}^{2}} &{} \text { if } (x_{1},x_{2})\ne (0,0), \\ 0 &{} \text { if } (x_{1},x_{2}) = (0,0). \\ \end{array}\right. } \end{aligned}$$

In this example, we have, $(1,1)\succ (3,1)\succ (0,0).$ There exists no $\lambda \in [0,1]$ such that $\lambda (1,1)+(1-\lambda )(0,0) \sim (3,1).$ Hence, this is not Weak Wold-continuous. Consequently, this is also not Wold-continuous.

It is easy to show that this binary relation is not unrestricted solvable. Pick a pair (1, 1) and let the third element of the to-be-determined pair is 0. Then unrestricted solvability claims that there exists q such that $(1,1)\sim (q,0).$ In this case, there exists no q such that the indifference holds.

Finally this example illustrates that a binary relation that is restricted solvable, but not Archimedean. The function is continuous on any line parallel to a coordinate axis, hence the induced binary relation is restricted solvable. However, it is not continuous on the 45-degree line. In particular it is defined on $\mathbb {R}^2_+$ and the function’s value is 1 on all point on the 45-degree line except 0, and at 0 its value is 0. Therefore if you pick $x=(0,0), y=z=(1,1)$, then $y\succ x$ but for all $\lambda \in (0,1)$, $y\sim x\lambda z$, hence Archimedean property fails. Similarly, it also shows that Archimedean does not imply mixture-continuity. $\square$

The fourth example illustrates a binary relation that is Archimedean and Wold-continuous, and hence weakly Wold-continuous and restricted solvable, but not mixture-continuous, separately continuous and continuous. It is originally presented in Uyanik and Khan (2022).

4. Let $X=\mathbb {R}_+$ and $f(x)=\text {sin}(1/x)$ if $x>0$ and $f(0)=1$. Then, it is clear that $f(x)\in [-1,1]$ for all $x\in X$. Define a binary relation $\succsim$ on X as $x\succsim y$ is and only if $f(x)\ge f(y)$. Pick ${\bar{x}}$ such that $f({\bar{x}})\in (0,1)$. The set $\{x'\in X|{\bar{x}}\succsim x'\}$ is not closed since it contains a sequence $x_n\rightarrow 0$ but $0\succ {\bar{x}}$. Therefore, $\succsim$ is not continuous. Moreover, since X is one dimensional, $\succsim$ is not mixture-continuous and separately continuous. See Uyanik and Khan (2022, Proof of Proposition 3) for a proof that $\succsim$ satisfies Wold-continuity and Archimedean properties. $\square$

The fifth example shows that Archimedean property does not imply restricted solvability, hence all other continuity postulates in Figure 1.

5. Let $\succsim$ be a binary relation defined on ${\mathbb {R}}^{2}_+$ as $(x_{1},x_{2})\succsim (y_{1},y_{2})$ if and only if $f(x_{1},x_{2})\ge f(y_{1},y_{2})$ where

$$\begin{aligned} f(x_{1},x_{2}) = {\left\{ \begin{array}{ll} \text {sin}(1/x_2) &{} \text { if } x_2>0,\\ x_1 &{} \text { if } x_1\in [0,1]\cap {\mathbb {Q}}, x_2=0,\\ 0 &{} \text { if } x_1\in [0,1]\cap {\mathbb {Q}}^c, x_2=0,\\ 1 &{} \text { if } x_1> 1, x_2=0.\\ \end{array}\right. } \end{aligned}$$

It is not difficult to show that $\succsim$ satisfies the Archimedean property. To see that $\succsim$ fails to satisfy restricted solvability, note that $(1,0)\succ (0,2) \succ (0,0)$ since $f(1,0)=1, f(0,0)=0$ and f(0, 2) is an irrational number in (0,1) interval (sin() of a non-zero rational number is irrational). Hence, for all $x_1\in \mathbb {R}_{+}$, either $(x_1,0)\succ (0,2)$ or $(x_1,0)\prec (0,2)$. $\square$

The sixth and final example illustrates the importance of the interiority assumption in our results by showing that Theorems 2 and 3 above are false on a general choice set; see also Karni and Safra (2015) for the use of the interiority assumption to show that the Archimedean property is equivalent to mixture-continuity under cone-monotonicity.

6. Let $X=\{x\in (-1,1)^2: x_1=x_2\}$. Then X is an order bounded and convex set in $\mathbb {R}^2$, it is not open in $\mathbb {R}^2$. (Note that X is open in the one dimensional subspace containing it.) Let $u: X\rightarrow \mathbb {R}$ be defined by $u(x)=0$ for all $x\le 0$ and $u(x)=1$ if $x_1>0$. Let $\succsim$ be the preference relation induced by u. Then, $A_\succsim (0.5, 0.5)=\{x\in X: x_1>0\}$ which is not closed in X, hence $\succsim$ is not continuous. It is clear that $\succsim$ is weakly monotone. Since the restriction of any line parallel to a coordinate axis is a singleton, $\succsim$ is trivially separately continuous.

Note that in the example above, we can replace X by $\{x\in [-1,1]^2: x_1=x_2\}$, hence boundaries can be included. Similarly, we can replace X by a set with non-empty interior: $\{x\in [0,1]^2: 2x_1\ge x_2\ge 0.5 x_1 \}$. Let $u: X\rightarrow \mathbb {R}$ be defined by $u(0)=0$ and $u(x)=1$ if $x \ne 0$. Let $\succsim$ be the preference relation induced by u. Then $\succsim$ is weakly monotone, separately continuous and discontinuous.