<p>The preference relation is not transitive. As a counterexample take <span class="math-container">$x=(0,0), y=(0,1), z=(1,1)$</span>.</p>
<p>For monotonicity let <span class="math-container">$x=(x_1,x_2)=(y_1+1, y_2)$</span> so that the <span class="math-container">$x$</span> bundle contains more of some commodity but no less of any, than <span class="math-container">$y=(y_1,y_2)$</span>. Then
<span class="math-container">$$ (y_1+1+y_2-y_1-y_2) = 1 \leq 1 $$</span></p>
<p>so that the preference relation is not monotonic.</p>
<p>so that the preference relation is strictly monotonic.</p>
<p>The preference relation is continuous since
<span class="math-container">$$ \{x| x\succeq y\} = \{x_1, x_2 | x_1 + x_2 - y_1 - y_2 \geq -1\} $$</span>
is closed and
<span class="math-container">$$ \{x| x\preceq y\} = \{x_1, x_2 | x_1 + x_2 - y_1 - y_2 \leq 1\} $$</span>
is also closed</p>
<p>Finally, it is not locally non satiated. Take any <span class="math-container">$\epsilon\leq 1$</span> as an example.</p>
<p>EDIT: It is not strictly monotonic, see the comment below</p>

CC BY-SA 4.0
2021-01-25T10:25:03.897Z
Let $x=(x_1,x_2)=(y_1+0.1, y_2+0.1)$ so that the $x$ bundle contains more of all commodities than $y=(y_1,y_2)$. Then $ x_1+x_2-y_1-y_2 = 0.2 < 1 $, so it is *not* strictly monotonic.
VARulle
CC BY-SA 4.0
2021-01-25T00:36:48.783Z
2021-01-25T13:53:13.673Z
2021-01-25T13:53:13.673Z
Pekisch
<p>This question is from Harvard seminar problem set (Q-3 part b)  <a href="https://www.studocu.com/en-us/document/harvard-university/economics/mandatory-assignments/econ2020a-14-ps01-please-give-as-much-additional-information-as-possible/3513583/view" rel="nofollow noreferrer">https://www.studocu.com/en-us/document/harvard-university/economics/mandatory-assignments/econ2020a-14-ps01-please-give-as-much-additional-information-as-possible/3513583/view</a></p>
<p>Show that it holds continuous, monotonic, LNS and transversality.
<span class="math-container">$$|x_1-x_2 -y_1-y_2| \le 1$$</span> if x~y</p>
<p>I can just show transversality condition</p>
<p>first, for x~y <span class="math-container">$$-1\le x_1-x_2-y_1-y_2\le 1 $$</span>
for y~z <span class="math-container">$$-1\le y_1-y_2-z_1-z_2\le 1 $$</span></p>
<p>When I sum these in equalities, I will obtain</p>
<p><span class="math-container">$$-2\le x_1-x_2-z_1-z_2\le 2 $$</span></p>
<p>So, At the same time, I can say that</p>
<p><span class="math-container">$$-1\le x_1-x_2-z_1-z_2\le 1$$</span></p>
<p>This implies that x~z. Thus, transversality holds. But I cannot prove for others even though I know their definitions. Please help me to do this.</p>

CC BY-SA 4.0
2021-01-25T10:18:55.433Z
There's a typo, it should be $|x_1+x_2 -y_1-y_2|$. Then it should be "transitivity" instead of "transversality". And your proof is wrong: $-2\le a \le 2$ it doesn't imply $-1\le a \le 1$, as $a=1.5$ shows.
VARulle
<microeconomics><preferences><self-study>
Assumptions on preference relation
studentp
# Assumptions on preference relation

(Asked by: username_0 on 2021-01-24)

This question is from Harvard seminar problem set (Q-3 part b) <https://www.studocu.com/en-us/document/harvard-university/economics/mandatory-assignments/econ2020a-14-ps01-please-give-as-much-additional-information-as-possible/3513583/view>

Show that it holds continuous, monotonic, LNS and transversality.
$$|x\_1-x\_2 -y\_1-y\_2| \le 1$$ if x~y

I can just show transversality condition

first, for x~y $$-1\le x\_1-x\_2-y\_1-y\_2\le 1 $$
for y~z $$-1\le y\_1-y\_2-z\_1-z\_2\le 1 $$

When I sum these in equalities, I will obtain

$$-2\le x\_1-x\_2-z\_1-z\_2\le 2 $$

So, At the same time, I can say that

$$-1\le x\_1-x\_2-z\_1-z\_2\le 1$$

This implies that x~z. Thus, transversality holds. But I cannot prove for others even though I know their definitions. Please help me to do this.

**Tags:** microeconomics, preferences, self-study
> **Comments:**
> **username_1**: There's a typo, it should be $|x_1+x_2 -y_1-y_2|$. Then it should be "transitivity" instead of "transversality". And your proof is wrong: $-2\le a \le 2$ it doesn't imply $-1\le a \le 1$, as $a=1.5$ shows.  

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## 1 answers

### Answer by username_2 on 2021-01-25. Score: 1 | Accepted answer

The preference relation is not transitive. As a counterexample take $x=(0,0), y=(0,1), z=(1,1)$.

For monotonicity let $x=(x\_1,x\_2)=(y\_1+1, y\_2)$ so that the $x$ bundle contains more of some commodity but no less of any, than $y=(y\_1,y\_2)$. Then
$$ (y\_1+1+y\_2-y\_1-y\_2) = 1 \leq 1 $$

so that the preference relation is not monotonic.

so that the preference relation is strictly monotonic.

The preference relation is continuous since
$$ \{x| x\succeq y\} = \{x\_1, x\_2 | x\_1 + x\_2 - y\_1 - y\_2 \geq -1\} $$
is closed and
$$ \{x| x\preceq y\} = \{x\_1, x\_2 | x\_1 + x\_2 - y\_1 - y\_2 \leq 1\} $$
is also closed

Finally, it is not locally non satiated. Take any $\epsilon\leq 1$ as an example.

EDIT: It is not strictly monotonic, see the comment below

> **Comments:**
> **username_1**: Let $x=(x_1,x_2)=(y_1+0.1, y_2+0.1)$ so that the $x$ bundle contains more of all commodities than $y=(y_1,y_2)$. Then $ x_1+x_2-y_1-y_2 = 0.2 < 1 $, so it is *not* strictly monotonic.  

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