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General equilibrium framework; two individuals, two goods represented in an Edgeworth box.

Is it true that if the preference are both strongly monotonic the Pareto set will go from the origin of an individual's axis to the one of the other's axis? By origin, I mean the points of the box where an individual has everything and the other nothing. The reason would be that the allocation of all goods to one of the single individuals is Pareto optimal.

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  • $\begingroup$ Could you please define what exactly you mean by "go from one origin to the other"? E.g. if all the points in the box are in the Pareto set, does the set go from one origin to the other? $\endgroup$ – Giskard Dec 12 '18 at 18:04
  • $\begingroup$ @denesp edited. $\endgroup$ – PhDing Dec 12 '18 at 18:52
  • $\begingroup$ I know what the word origin means, please answer my question: "...if all the points in the box are in the Pareto set, does the set go from one origin to the other?" $\endgroup$ – Giskard Dec 12 '18 at 19:17
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    $\begingroup$ By saying that the Pareto set goes from one origin to the other, you are presupposing that the set can be represented by a curve (e.g. the contract curve). But a set is a more general concept than a curve. While it's intuitive to describe a curve that connects the two origins as "going from one to the other", it's less intuitive to use the same language on a (connected) set that contains the two origins. This is perhaps why @denesp was having trouble understanding your question. $\endgroup$ – Herr K. Dec 12 '18 at 23:53
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    $\begingroup$ Alessandro, I was indeed asking for clarifications because these details matter when you make mathematical claims. Please believe me, I am not trying to be bothersome. Multiple times you got upset with me on this site for the same reason. Also, downvote was not by me. $\endgroup$ – Giskard Dec 13 '18 at 4:20
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The Pareto optimal set should always pass through the origin because it's one of the point that makes it impossible to better off a consumer without harming the other.

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    $\begingroup$ This doesn't answer OP's question, which asks whether strict monotonicity is sufficient to guarantee the existence of a "generalized contract curve". $\endgroup$ – Herr K. Dec 17 '18 at 17:24

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