1. Assume that the preference relation ≿ on a metric space X is continuous. Show that this implies that the sets > (y) = {x : x>y} and < (y) = {x : x2. Conversely, assume that the sets > (y) = {x : x>y} and < (y) = {x: xz0.
a. Suppose there exists some y ∈ X such that x0>y>z0. Show that there exist neighborhoods S…(x0) and S (z0) of x0 and z0 such that x>z for every x ∈ S(x0) and z ∈ S(z0).
b. Now suppose that there is no such y with x0>y>z0. Show that
i. > (z0) is an open neighborhood of x0
ii. > (z0) = ≿ (x0)
iii. x>z0 for every x ∈ > (z0)†
iv. There exist neighborhoods S(x0) and S (z0) of x0 and z0 such that
x>z for every x ∈ S…(x0) and z ∈ S(y0)
This establishes that a preference relation ≿ on a metric space X is continuous
if and only if the sets >(y) = {x : x>y} and <(y) = (x : x3. Show that a preference relation 7 on a metric space X is continuous if and only if the sets ≿(y) = (x : x≿y} and ≾(y) = {x : x≾} are closed for every y in X.