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.

  • $\begingroup$ 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. $\endgroup$ – VARulle Jan 25 at 10:18

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

  • 1
    $\begingroup$ 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. $\endgroup$ – VARulle Jan 25 at 10:25

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