SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Lemma 16. Fix a sequence  $b = (b_{t})$  where  $b_{t + 1} = f_{A_t}(b_t)$  such that the sequence  $A_{t}$  is  $T$ -live. Then, it holds that  $\lim_{t\to \infty}d(b_t,ME) = 0$ .
Proof. By lemma 1 we have that  $\Phi(b^{t+1}) \geq \Phi(b^t)$  making the sequence  $\Phi(b^t)$  monotone and bounded from above ( $\Phi(\cdot)$  is a bounded function). Hence it converges to some limit  $L$ . Either  $L = \Phi^*$  or  $L < \Phi^*$ . In the former case, the result is immediate by lemma 12 and  $b^t$  approaches the set  $ME$ . We show that the latter yields a contradiction. If  $L < \Phi^*$ , this implies that  $b^t$  is bounded away from  $ME$ , i.e., there exists  $\epsilon_0 > 0$  such that for all  $td(b^t, ME) \geq \epsilon_0$ . To see why this is true for all  $t$  and not just in the limit, we observe that since the sequence  $\Phi(b^t)$  is monotone, if we have  $d(b^T, ME) = 0$  at some time  $T$ , then we have  $\Phi(b^t) = \Phi^*$  for all  $t > T$ , which we currently assume is not the case. Therefore, every subsequence of  $b^t$  is bounded away from  $ME$ , implying that every condensation point of  $b^t$  is not in  $ME$ . The sequence  $b^t$  is bounded and therefore has a converging subsequence with a condensation point not in  $ME$ , which is a contradiction to the previous lemma.
Regarding convergence of prices, as stated in section 2, equilibrium prices for each Fisher market are unique and attained by any bid profile  $b^{*} \in ME$ , thus, since the prices are a continuous function of the bids, the convergence of bids to this set implies the convergence of prices  $p^t \to p^*$ .
The last lemma concludes our proof of the Theorem 1.
Appendix E Generic Markets
Proof. (Theorem 5): Assume by way of contradiction that a generic linear Fisher market has two distinct market equilibrium bid profiles  $b^{*} \neq b^{**}$ . For any market equilibrium  $b$  it must hold that: (1)  $\forall j \sum_{i} b_{ij} = p_{j}^{*}$  since equilibrium prices are unique, and (2)  $\forall i \sum_{j} b_{ij} = B_{i}$  by budget feasibility. As  $b^{*} \neq b^{**}$ , there exists a pair  $(i,j)$  with  $b_{ij}^{*} \neq b_{ij}^{**}$ , meaning that buyer  $i$  has a different bid on good  $j$  between  $b^{*}$  and  $b^{**}$ , and so by (1) it must be that exists a buyer  $k$  whose bid on good  $j$  was also changed so that the price  $p_{j}^{*}$  remains fixed; formally,  $b_{kj}^{*} \neq b_{kj}^{**}$ . In such case, by (2) there must be a good  $\ell$  for which buyer  $k$  has a different bid as well, since its budget  $B_{k}$  is fixed and fully utilized; formally  $b_{k\ell}^{*} \neq b_{k\ell}^{**}$ . As the graph  $\Gamma = \{\mathcal{B} \cup \mathcal{G}, E\}$  with  $E = \{\{i,j\}|b_{ij}' \neq b_{ij}^{*}\}$  is finite, following the process described above while satisfying the constraints (1) and (2) must lead to a cycle in the graph  $\Gamma$ .
Finally, we will show that there exists a market equilibrium with a cycle in its corresponding graph. Define  $b' = \lambda b^* + (1 - \lambda) b^{**}$  for some  $\lambda \in (0, 1)$  and note that  $b'$  is also market equilibrium as the set of market equilibria is a convex set (see the model section in the main text). Let  $\Gamma(b') = \{\mathcal{B} \cup \mathcal{G}, E(b')\}$  with  $E(b') = \{\{i, j\} | b_{ij}' > 0\}$  be the corresponding graph of  $b'$ . Observe that  $E \subseteq E(b')$  since if  $b_{ij}' \neq b_{ij}^{**}$  then it must be that  $b_{ij}' > 0$  or  $b_{ij}^{**} > 0$  and in any such case  $b_{ij}' > 0$ . Thus, the graph  $\Gamma(b')$  contains a cycle, contradicting Lemma 2 from the main text
Proof. (Lemma 2): Assume for the sake of contradiction that exists a cycle  $C$  in  $\Gamma(b^*)$ , w.l.o.g. name the vertices of buyers and goods participating in the cycle in an ascending order; that is,  $C = b_1g_1b_2g_2\ldots b_{k-1}g_kb_1$ , where  $b_i$  and  $g_i$  represent buyers and goods  $i$ , respectively. Recall that for any market equilibrium if  $x_{ij}^* > 0$  then  $\frac{a_{ij}}{p_j^*} = c_i$  for some constant  $c_i$  (see the model section in the main text). Applying this to the cycle  $C$  yields the following equations. (1) By considering edges from buyers to goods  $b_i \to g_i$  we obtain for  $i \in [k-1]$ $a_{i,i} = c_i p_i^*$ , and (2) by considering edges from goods to buyers  $g_i \to b_{i+1}$  we obtain for  $i \in [k-1]$ $a_{i+1,i} = c_{i+1} p_i^*$  and the edge closing the cycle yields  $a_{1,k} = c_1 p_k^*$ . Finally, by considering the product of ratios between valuations of buyers participating in the cycle we have the following condition.
$$
\begin{array}{l} \frac {a _ {2 1}}{a _ {1 1}} \frac {a _ {3 2}}{a _ {2 2}} \frac {a _ {4 3}}{a _ {3 3}} \dots \frac {a _ {i + 1 , i}}{a _ {i , i}} \dots \frac {a _ {k , k - 1}}{a _ {k - 1 , k - 1}} \frac {a _ {1 , k}}{a _ {k , k}} = \frac {c _ {2} p _ {1} ^ {*}}{c _ {1} p _ {1} ^ {*}} \frac {c _ {3} p _ {2} ^ {*}}{c _ {2} p _ {2} ^ {*}} \frac {c _ {4} p _ {3} ^ {*}}{c _ {3} p _ {3} ^ {*}} \dots \frac {c _ {i + 1} p _ {i} ^ {*}}{c _ {i} p _ {i} ^ {*}} \dots \frac {c _ {k} p _ {k - 1} ^ {*}}{c _ {k - 1} p _ {k - 1} ^ {*}} \frac {c _ {1} p _ {k} ^ {*}}{c _ {k} p _ {k} ^ {*}} \\ = \frac {c _ {2}}{c _ {1}} \frac {c _ {3}}{c _ {2}} \frac {c _ {4}}{c _ {3}} \dots \frac {c _ {i + 1}}{c _ {i}} \dots \frac {c _ {k}}{c _ {k - 1}} \frac {c _ {1}}{c _ {k}} \\ = 1, \\ \end{array}
$$
which contradicts the genericity condition.