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Let $n$ be the number of consumers and $m$ be the number of commodities.

The Arrow-Debreu theorem requires closed and convex consumption sets $X_i \subset \mathbb{R}^m$ for all buyers $i \in [n]$. Additionally, it requires the utility function of any consumer $i$, $u_i: \mathbb{R}^m \to \mathbb{R}$ to be continuous, quasi-concave and non-satiated over the consumption set $X_i$, where non satitation is defined as $\forall \mathbf{x} \in X_i, \exists \mathbf{y} \in X_i$ such that $u_i(\mathbf{y}) > u_i(\mathbf{x})$ (pages 268-269).

It seems to me like these assumptions are contradictory. How can the utility function be non-satiated if the consumption set is closed, i.e., compact since it is a closed subset in $R^m$. Doesn't compactness and continuity of the utility function guarantee that there exists a bundle within the consumption set that maximizes utility which implies that non-satiation cannot hold?

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  • $\begingroup$ Looking at the reference, the definition of nonsatiation is that $\forall x \in \hat{X}_i$ ... . This set $\hat{X}_i$ is defined (at 3.3.0) as a subset of $X_i$, which I believe resolves the contradiction. $\endgroup$
    – Max
    Mar 5 at 3:19
  • $\begingroup$ You're referring to the modified assumption III.b' they make later. But they state it in claim III.b the way I have it above and do actually use it. $\endgroup$
    – Denizalp
    Mar 5 at 3:30
  • $\begingroup$ At the bottom of p.268, the authors say "The set of consumption vectors $X_i$ available to individual $i$ $(=1,\cdots,m)$ is a closed convex subset of $R^l$ which is bounded from below". So $X_i$ isn't necessarily compact... $\endgroup$
    – Herr K.
    Mar 5 at 4:26
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    $\begingroup$ The Heine-Borel Theorem establishes that $S\subset \mathbb R^n$ is compact if and only if $S$ is both closed and bounded (from below and above). For example, $[0,\infty)$ is closed but not compact. $\endgroup$
    – Herr K.
    Mar 5 at 4:36
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    $\begingroup$ Oh makes sense!!! For some reason, I was thinking that bounded from below implied bounded. Thank you!! $\endgroup$
    – Denizalp
    Mar 5 at 4:45
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Converting my comments into an answer:

At the bottom of p.268, the authors say:

The set of consumption vectors $X_i$ available to individual $i$ $(=1,\cdots,m)$ is a closed convex subset of $R^l$ which is bounded from below.

[Emphasis added.]

Since the Heine-Borel Theorem establishes that $S\subset R^n$ is compact if and only if $S$ is both closed and bounded (from below and above), one cannot conclude necessarily that $X_i$ is compact. (A counterexample is that $[0,\infty)$ is closed but not compact.)

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The concepts of feasible set and utility should be thought of separately.

Non-satiation just means that there are no "thick" indifference curves - this guarantees that an agent is not indifferent between two close, but different bundles of goods.

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  • $\begingroup$ I am not sure I understand how this fixes the math they have. Also, I do not think your interpretation of non-satiation is correct. Additionally, uniqueness is never guaranteed by non-satiation, strict concavity of utility is required for that (among other things). $\endgroup$
    – Denizalp
    Mar 5 at 3:35
  • $\begingroup$ Yes, it is a requirement, not the only requirement, for well-behaved consumer theory - I'll edit above to make sure this is clear for future readers. What I mean to stress is that it is simply a property of the utility function, the feasible set is only relevant insofar as it determines what x or y you could have. If you look online, you will see some definitions of non-satiation involve a ball of arbitrary radius, if you think about it this way in your compact space, you can always shrink the distance arbitrarily small - hence no "thick" indifference lines. $\endgroup$
    – qwerty
    Mar 5 at 3:54
  • $\begingroup$ I normally think of non-satiation independently of the consumption sets as well but that is not how non-satiation is stated in the paper. This is a seminal paper and the math does not suggest that you can separate the definition of non-satiated from the consumption sets, so there must be something else going on, which I would like to figure out. $\endgroup$
    – Denizalp
    Mar 5 at 4:33

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