<p>A source of confusion comes from the distinction between strict and weak preferences.
Define <span class="math-container">$\succ_n$</span> strict preferences and <span class="math-container">$\succsim_n$</span> weak preferences. </p>

<p>1) The sequence of preference relations <span class="math-container">$\succ_n$</span> does <strong>not</strong> converge to preference relation <span class="math-container">$\succ$</span>, precisely because of the indifference case.
If we take your example with <span class="math-container">$x_1=y_1$</span>, <span class="math-container">$x_1=\max(x_1,x_2)$</span> and <span class="math-container">$y_1=\max(y_1,y_2)$</span>. Assume also that <span class="math-container">$x_2&gt;y_2$</span>.
Then, for any <span class="math-container">$n&gt;0$</span>, <span class="math-container">$x\succ_n y$</span> but we do not have <span class="math-container">$x\succ y$</span>. </p>

<p>2) However, one can show that the sequence <span class="math-container">$\succsim_n$</span> converges to <span class="math-container">$\succsim$</span>, which is what you suggest at the end of your question. We can check that the counterexample above does not apply here: <span class="math-container">$x\succsim_n y$</span> for <span class="math-container">$n&gt;0$</span> and <span class="math-container">$x\succsim y$</span>.</p>

<p>3) Is there an error in Rubinstein's textbook? No. The problem set (2018 update) actually frames the question in terms of weak preferences <span class="math-container">$\succsim$</span> (note there is still a mix of notations that is puzzling).</p>

CC BY-SA 4.0
2019-04-21T10:28:36.610Z
2019-04-21T10:28:36.610Z
GuiWil
<p>I am assuming you're familiar with the concept of monotonic transformations.</p>

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

<p>To prove the same, consider <span class="math-container">$v(x_1,x_2)=\Big(1+\cfrac{x_1^n }{x_2^n}\Big)^{1/n}x_2$</span>. Assume <span class="math-container">$x_2 = \max\{x_1,x_2\}. $</span>Now, as <span class="math-container">$n \to \infty$</span>, two cases arise. </p>

<p>1) If <span class="math-container">$x_2&gt;x_1$</span>, then <span class="math-container">$\cfrac{x_1^n }{x_2^n} \to 0$</span> as <span class="math-container">$n \to \infty$</span>. In that case, <span class="math-container">$v(x_1,x_2)=x_2=\max\{x_1,x_2\}$</span>.</p>

<p>2) If <span class="math-container">$x_2=x_1$</span>, the limit can be calculated by taking log on both sides as follows-
<span class="math-container">\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*}</span>
Since <span class="math-container">$x_1/x_2=1$</span>,we have <span class="math-container">$(x_1/x_2)^n=1^n$</span>. Using this in equation <span class="math-container">$(1)$</span>, we get
<span class="math-container">\begin{align*}
\ln(v(x_1,x_2)) = \cfrac{\ln(1+1^n)}{n} +\ln(x_2) 
\end{align*}</span>
As <span class="math-container">$n\to \infty$</span>, the term <span class="math-container">$\cfrac{\ln(1+1^n)}{n}$</span> becomes <span class="math-container">$\infty/\infty$</span>. 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 <span class="math-container">$0$</span>.
Now we have <span class="math-container">$\ln(v(x_1,x_2)) = 0 +\ln(x_2)$</span>, or, <span class="math-container">$v(x_1,x_2)=x_2=\max\{x_1,x_2\}$</span>.</p>

CC BY-SA 4.0
2019-04-12T04:55:57.007Z
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.
Giskard
CC BY-SA 4.0
2019-04-12T04:58:42.953Z
Also I am not sure if your calculations really answer the OP's question, which seems to be about "why", not "how".
Giskard
CC BY-SA 4.0
2019-04-12T06:35:08.010Z
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 $a, b$.)
Lochend
CC BY-SA 4.0
2019-04-11T23:09:23.670Z
2019-04-11T23:09:23.670Z
superhulk
<p>In the problem set 2 of Rubinsteins Microeconomics (btw is there a comparably nice written book on macroeconomics?) there is the following question:
Let <span class="math-container">$\succ_n$</span> be the preference relations defined on <span class="math-container">$\mathbb{R}^2_+$</span> by the utility <span class="math-container">$x_1^n + x_2^n$</span>. Let the preference relations <span class="math-container">$\succ$</span> be defined by the utility <span class="math-container">$\max\{x_1, x_2\}$</span>. Show that <span class="math-container">$\succ_n$</span> converge to <span class="math-container">$\succ$</span>.</p>

<p>Preference relations are said to converge if for <span class="math-container">$a \succ b$</span> we have that <span class="math-container">$a \succ_m b$</span> for sufficiently large <span class="math-container">$m$</span>. I am indeed able to prove that.</p>

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

<p>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 <span class="math-container">$\succeq = \lim_{n \to \infty} \succeq_n$</span> iff <span class="math-container">$a \succeq b = \lim_{n \to \infty} a \succeq_n b$</span>. </p>

CC BY-SA 4.0
2019-04-11T23:23:58.423Z
“the only case when $x\sim_m y$ is when $x_2=y_2$”. Did you mean "$x_1=y_1$"?
Herr K.
CC BY-SA 4.0
2019-04-12T06:38:46.013Z
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.
Lochend
<utility><preferences><definition>
Preference relations defined by $x_1^n + x_2^n$ converge to $\max\{x_1, x_2\}$
Lochend
# Preference relations defined by $x_1^n + x_2^n$ converge to $\max\{x_1, x_2\}$

(Asked by: username_0 on 2019-04-11)

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 $a \succeq b = \lim\_{n \to \infty} a \succeq\_n b$.

**Tags:** utility, preferences, definition

> **Comments:**
> **username_1**: “the only case when $x\sim_m y$ is when $x_2=y_2$”. Did you mean "$x_1=y_1$"?  
> **username_0**: 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.  

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## 2 answers

### Answer by username_2 on 2019-04-21. Score: 0 | Accepted answer

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).


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### Answer by username_3 on 2019-04-11. Score: 1

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, $v(x\_1,x\_2)=x\_2=\max\{x\_1,x\_2\}$.

> **Comments:**
> **username_4**: 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.  
> **username_4**: Also I am not sure if your calculations really answer the OP's question, which seems to be about "why", not "how".  
> **username_0**: 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 $a, b$.)  

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