In the problem set 2 of Rubinsteins Microeconomics (btw is there a comparably nice written book on macroeconomics?) there is the following question: Let $\succ_n$ be the preference relations defined on $\mathbb{R}^2_+$ by the utility $x_1^n + x_2^n$. Let the preference relations $\succ$ be defined by the utility $\max\{x_1, x_2\}$. Show that $\succ_n$ converge to $\succ$.

Preference relations are said to converge if for $a \succ b$ we have that $a \succ_m b$ for sufficiently large $m$. I am indeed able to prove that.

Now w.l.o.g. $x_1 = \max\{x_1, x_2\}$ and $y_1 = \max\{y_1, y_2\}$. Assuming that $x_1 = y_1$ we have $x \sim y$. But the only case when $x \sim_m y$ is when also $x_2 = y_2$. In other cases we would have either $x \succ_m y$ or $y \succ_m x$ for all $m$ (depending on $x_2$ and $y_2$, the "smaller component". These are actually lexicographic preferences!)

Is there a reason that the convergence of preference relations is done in this way to ignore such subtilities? In my opinion we should have that $\succeq = \lim_{n \to \infty} \succeq_n$ iff $a \succeq b = \lim_{n \to \infty} a \succeq_n b$.

  • $\begingroup$ “the only case when $x\sim_m y$ is when $x_2=y_2$”. Did you mean "$x_1=y_1$"? $\endgroup$
    – Herr K.
    Commented Apr 11, 2019 at 23:23
  • $\begingroup$ I wanted to assume $x_1 = y_1$ w.l.o.g. but forgot to note it down. I now edited the question to make that part more clear, thank you. $\endgroup$
    – Lochend
    Commented Apr 12, 2019 at 6:38

2 Answers 2


A source of confusion comes from the distinction between strict and weak preferences. Define $\succ_n$ strict preferences and $\succsim_n$ weak preferences.

1) The sequence of preference relations $\succ_n$ does not converge to preference relation $\succ$, precisely because of the indifference case. If we take your example with $x_1=y_1$, $x_1=\max(x_1,x_2)$ and $y_1=\max(y_1,y_2)$. Assume also that $x_2>y_2$. Then, for any $n>0$, $x\succ_n y$ but we do not have $x\succ y$.

2) However, one can show that the sequence $\succsim_n$ converges to $\succsim$, which is what you suggest at the end of your question. We can check that the counterexample above does not apply here: $x\succsim_n y$ for $n>0$ and $x\succsim y$.

3) Is there an error in Rubinstein's textbook? No. The problem set (2018 update) actually frames the question in terms of weak preferences $\succsim$ (note there is still a mix of notations that is puzzling).


I am assuming you're familiar with the concept of monotonic transformations.

Let $f(z)=z^n$ be a monotonic transformation. $f(z)$ is an order preserving transformation for all positive $n$. Let $v:\mathbb{R_+^2} \to\mathbb{R}$ be an utility function, where $v(x_1,x_2)=(x_1^n + x_2^n)^{1/n}$. The function $f(v(x_1,x_2))=((x_1^n + x_2^n)^{1/n})^n=x_1^n + x_2^n$ will represent the same preferences as $v(x_1,x_2)$. Therefore, $(x_1^n + x_2^n)^{1/n}$ should also converge to $\max\{x_1,x_2\}$.

To prove the same, consider $v(x_1,x_2)=\Big(1+\cfrac{x_1^n }{x_2^n}\Big)^{1/n}x_2$. Assume $x_2 = \max\{x_1,x_2\}. $Now, as $n \to \infty$, two cases arise.

1) If $x_2>x_1$, then $\cfrac{x_1^n }{x_2^n} \to 0$ as $n \to \infty$. In that case, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.

2) If $x_2=x_1$, the limit can be calculated by taking log on both sides as follows- \begin{align*} \ln(v(x_1,x_2)) = \cfrac{\ln(1+\cfrac{x_1^n }{x_2^n})}{n} +\ln(x_2) \,\,\,\,-(1) \end{align*} Since $x_1/x_2=1$,we have $(x_1/x_2)^n=1^n$. Using this in equation $(1)$, we get \begin{align*} \ln(v(x_1,x_2)) = \cfrac{\ln(1+1^n)}{n} +\ln(x_2) \end{align*} As $n\to \infty$, the term $\cfrac{\ln(1+1^n)}{n}$ becomes $\infty/\infty$. This limit can be solved using L'Hospital's rule. The value of the limit(you may check it for yourself) comes out to be $0$. Now we have $\ln(v(x_1,x_2)) = 0 +\ln(x_2)$, or, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.

  • $\begingroup$ The part "If $x_2>x_1$, then $\cfrac{x_1^n }{x_2^n} \to 0$ as $n \to \infty$. In that case, $v(x_1,x_2)=x_2$" is shaky. By the same reasoning $$ \lim_{n \to \infty} \left( \frac{x_1^n}{x_2^n}\right)^{1/n} = 0 $$ if $x_1 < x_2$, but this is of course not true. $\endgroup$
    – Giskard
    Commented Apr 12, 2019 at 4:55
  • 1
    $\begingroup$ Also I am not sure if your calculations really answer the OP's question, which seems to be about "why", not "how". $\endgroup$
    – Giskard
    Commented Apr 12, 2019 at 4:58
  • $\begingroup$ Yes, thank you for the calculation, but the question is why we define convergence in this way (only requesting that $\lim a \succ_n b = a \succ b$ if $a \succ b$ and not $a \succeq b = \lim_n a \succeq_n b$ for every $a, b$.) $\endgroup$
    – Lochend
    Commented Apr 12, 2019 at 6:35

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