# Arrow-Debreu Theorem of Existence: Non satiation

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?

• 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.
– Max
Mar 5 at 3:19
• 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. Mar 5 at 3:30
• 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... Mar 5 at 4:26
• 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. Mar 5 at 4:36
• Oh makes sense!!! For some reason, I was thinking that bounded from below implied bounded. Thank you!! Mar 5 at 4:45

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.

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.)