In Mas-Colell et al.'s Microeconomic Theory Proposition 3.E.1(ii) (p. 58) states that if $\succsim$ is a rational (i.e. complete and transitive), continuous, and locally nonsatiated preference relation on ($X\equiv\mathbb{R}_{+}^L$) represented by continuous utility function $u(\cdot)$, then $x^*\in\text{arg}\min\{p\cdot x:u(x)\ge \underline{u}>u(0)\}$ implies $x^*\in\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$ (in words, that under the assumptions the solution to the EMP also solves the UMP).

Now I have trouble deciphering a step of the proof (bolded). The proof goes as follows.

Proof. Suppose $x^*\in\text{arg}\min\{p\cdot x:u(x)\ge u(x^*)>u(0)\}$ and $x^*\notin\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$. First we note that $x^*\ne0$. Now, by our hypothesis we have that there exists $x'\in X$ with $u(x')>u(x^*)$ and $p\cdot x'\le p\cdot x^*$. Consider a bundle $x''=\alpha x'$ where $\alpha\in(0,1)$ ($x''$ is a scaled version of $x'$). By continuity of $u(\cdot)$, if $\alpha$ is close enough to $1$, then we will have $u(x'')>u(x^*)$ and $p\cdot x''<p\cdot x^*$. And the conclusion clearly follows from the previous argument. $\square$

I don't understand the argument bolded (which I transcribed exactly as in the book), I suppose it has something to do with continuity of preferences that in the limit (i.e. "$\alpha$ close to $1$") $x''\succ x'$ but I don't see how, I think I should build some sort of sequence like $(1-1/(n+1))x'$ that in the limit has that property, but not sure how to see this argument.

I would be very grateful if anyone can point me the underlying argument of the reasoning mentioned.



1 Answer 1


By definition, a function is continuous if arbitrarily small changes in its value can be assured by choosing sufficiently small changes of its argument. Since $u(x')>u(x^*)$ and the utility function $u$ is continuous, a small enough change from $x'$ from $x''$ will ensure that $u(x'')>u(x^*)$ too.

Formally, let $\varepsilon=\frac{1}{2}[u(x')-u(x^*)]$. Then, by continuity of $u$, there exists a $\delta>0$ such that for all $x\in X$

$$|x-x'|<\delta \implies u(x')-\varepsilon<u(x)<u(x')+\varepsilon$$

In particular, there exists a $\delta>0$ such that for all $x\in X$:

$$|x-x'|<\delta \implies u(x)>u(x')-\varepsilon=\frac{1}{2}[u(x')+u(x^*)]>u(x^*)$$

In particular, we know that if $x''=\alpha x'$ with: $$|\alpha x'-x'|<\delta \iff (1-\alpha)<\frac{\delta}{|x'|} \iff \alpha>1-\frac{\delta}{|x'|}$$

then $u(x'')>u(x^*)$.

  • $\begingroup$ Thanks for your answer. How do you guarantee that $\alpha\in(0,1)$? $\endgroup$
    – manifold
    Sep 8, 2023 at 15:30
  • 1
    $\begingroup$ Since $\delta>0$, it must be that $1-\frac{\delta}{|x'|}<1$, so there certainly exist $\alpha\in(0,1)$ such that $\alpha>1-\frac{\delta}{|x'|}$. $\endgroup$
    – smcc
    Sep 8, 2023 at 16:44

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