In the following picture two preference relations are defined -

Ist preference relation defines lexicographic relation. Though we know that lexicographic preferences doesn't have utility function. They are defined at a point so we cannot define a utility function - is discontinuous. But how can we show it in terms of upper and lower contour set being closed?

Why second one is continuous?

Note - some terminologies -

A binary relation >~ on the metric space X is continuous if, for all x in X, the upper and lower contour sets, { y in X : y > ~ x} and {y in X : x >~ y}, respectively, are closed.

The upper contour set of x is >~ (x) = {y in X : y >~ x}

The lower contour set of x is >~ (x) = {y in X : x>~ y}

enter image description here

up vote 3 down vote accepted

Take $y_1=y_2=1$. As the figure below shows, the lower left segment of the boundary of the upper contour set of the lexicographic preference is open.

enter image description here

The continuity axiom can be interpreted as follows: Let $(\textbf{x}^n)$ and $(\textbf{y}^n)$ be two sequence of goods, such that $(\textbf{x}^n)$ converges to $\textbf{x}$ and $(\textbf{y}^n)$ to $\textbf{x}$. Then the continuity axiom says that if $\forall n$ $(\textbf{x}^n)$$\succsim$$(\textbf{y}^n)$, then $\textbf{x}$$\succsim$$\textbf{y}$ also holds true(here, $(\textbf{x}^n)$,$(\textbf{y}^n),\textbf{x}, \textbf{y} \in\mathbb{R_+^n} $). This statement is equivalent to the fact that the upper contour set and lower contour set is closed (think of $(\textbf{x}^n)$ and $(\textbf{y}^n)$ to be convergent sequences of the upper and lower contour sets respectively,i.e;$(\textbf{x}^n)\in (upper contour)$, and $(\textbf{y}^n)\in (lower contour)$, $\forall n$. Then, the sequences should converge within their respective sets).

The above statement, however, does not hold true for lexicographic preferences. To see that, consider any two bundles of $\mathbb{R_+^2} $represented by $((1/2)^n,0) \& (0,k)$ where, $k>0$. It is routine to check that the bundle $((1/2)^n,0)$ converges to $(0,0)$. Now, according to the concept of lexicographic preference(I hope you're familiar with it), $((1/2)^n,0)\succsim(0,k)$. However, the preferences change as the bundle $((1/2)^n,0)$ converges to $(0,0)$(i.e; now the bundle $(0,k)\succsim(0,0)$,as n tends to an infinite value). Thus, lexicographic preferences violate the continuity axiom.

  • Thank you for great explanation! It couldn't have been made more clear. – Elina Gilbert Sep 13 at 16:12

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