Lexicographic Indifference Curve
relation that is not monotone in !R . For example, x » y but y > x.
3.C.1 Let v be a lexicographic ordering. To prove the completeness, suppose
that we do not have x > y. Then "y^ 2: x^ and "Xj * y^ 0r y^ > Hence either uy^ > x^' or "y £ x^ and y^ > Thus y > x.
To prove the transitivity, suppose that x > y and y >- z. Then x & y
and y £ z., Hence x. If x. > z,, then x >- z. If x. ~ z then x, -
yT = z.. Thus X- 2: y and y_ ^ z Hence £ z_. Thus x > z.
To show that the strong monotonicity, suppose that x 2: y and y * x. This implies either that x^ > y^ and x^ £ y^, or that x^ - y^ and x^ > y^. In either case x > y.
To show the strict convexity, suppose that y > x, z x, y * z, and cc e (0,1). Without loss of generality, assume that x * y. By the definition of the lexicographic ordering, we have either "y^ > or "y^ = ^^ >'2 > *
On the other hand, since z >- x, we have either "z, > x," or "z, = x, and x^ 2:
y^." Hence, we have either May^ + (1 - cciz^ > x^" or "ay^ + (1 - a)z - x^ and cxy^ + (1 - cOz^ > x^." Thus ay + (I - cc)z >- x.
3,C.2 Take a sequence of pairs {(x? yn)}m- t such that xn > yn for all n, xn n=l ~
n n - n x, and y y. Then u(x') £ u(y ) for all nT and the continuity of uM
implies that u(x) £ u(y). Hence x >- y. Thus > is continuous.
3.C,3 One wav to prove the assertion is to assume that > is monotone and notice that the proof actually make use only of the closedness of upper and lower contour sets. Then the proposition is applicable to >-, implying that it has a continuous utility function. Thus, by Exercise 3.C.2, y is continuous.
A mere direct proof (without assuming monotonicity or using a utility function) goes as follows. Suppose that there exist two sequences {xn} and
{yn) in X such that xn > yn for every n, xn x e X, yD y € X, and y >- x. Since {z: y >- 2> is openr there exists a positive integer N^ such that y > xn for every n > Ny Since {z: 2 >- x) is open there exists a positive integer N
p such that y ' > x for every n > N^. Conceivably, there are two cases on the
Case 1: There exists a positive integer N such that y >- y for every n > N .
Case 2: There exists a subsequence such that y v for every n.
If Case 1 applies, then, by Proposition I.B.l(iii), we have yn y xn for every n > Max <N This is a contradiction. If Case 2 applies, then there exists a positive integer m such that k(m) > N^. Since {z: z >- is open, there exists a positive integer N^ such that yn > for every n >
By xn > yn and Proposition LB.l(iii), xn >- for every n > N . ■
Since {z: z > is closed, x > But, since Jc(m) > N' this is a
contradiction.
3.C.4 We provide two examples. The first one is simpler, but the second one satisfies monotonicity, which the first does not.
Example I. Let X = and define uM: !R -> IR by letting u(x) = 0 for x < 1, u(x) = 1 for x > 1, and nil} be any number in [0,1 j. Denote by > the preference relation represented by u(-). We shall now prove that > is not continuous. In fact, if u(l) > 0, then consider a sequence {x11} with = 1 -1/n for every n. Although - 0 for every n and xn 1, we have 1 > 0. If ii(I) < I, then consider a a sequence {xn} with xn = 1 + 1/n for every n. Although xn - 2 for every n and xn 1, we have 2 >- 1. Note that if ii(x) = 0, then all lower contour sets are closed. If u(l) = 1, then all upper contour sets are closed.
2 2 Example 2. Take X = and define a utility function uM: R R by the following rule:
Case 1. If x^ + x^ ^ 2 and x * (1,1), then u(x) = x^ + x . Case 2. If miiKx^x^) £ 1 and x * (1,1), then u(x) = minCx^x^} + 2. Case 3. If x^ ♦ x0 > 2, minix^x^} < 1, and x^ > x^, Then u(x) = 3 - (1 - x )/ix - 1). Case 4. If x^ + x^ > 2, minix^x^} < 1, and x^ < x^> then u(x) = 3 - (1 - x )/(x - a
The indifference curves of the preference relation > represented by *) are described in the following picture:
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