Final Topology for Preference Spaces

Proof.

Let D be the basis for U generated by products of open intervals. That is, each D ∈ D is of the form D = Π _x I _x where each I _x is an open interval and I x ≠ R for a finite subset of elements x ∈ X. We show B = F ( D ) . Because F is open and D is a basis for U , then B is a basis for P .

B ⊂ F ( D ) : Let B ( ≻ , A ) ∈ B , where ≻ ∈ P and A ∈ α ( ≻ ) . Let u be any representation of ≻. Let A ∈ A , so that A is ≻-strictly ranked. Let ɛ = min{|u(x) − u(y)|: x, y ∈ A, x ≠ y} > 0, where positivity is guaranteed because A is finite and ≻-strictly ranked. For each x ∈ A, let I x = ( u ( x ) − ε 2 , u ( x ) + ε 2 ) ⊂ R . For each x ∉ A, let I x = R . Notice that I _x ∩ I _y = ∅ for all x, y ∈ A, x ≠ y. Let D = Π x I x ∈ D . Then, B ( ≻ , A ) = F ( D ) ∈ F ( D ) for each A ∈ A . Therefore B ( ≻ , A ) = ∩ A ∈ A B ( ≻ , A ) ∈ F ( D ) so B ⊂ F ( D ) .^[10]

B ⊃ F ( D ) : Pick any F(D) where D ∈ D . Let A = { x : I x ≠ R } , which is finite by definition. First, assume ∩ _x∈A I _x ≠ ∅. Then, ∩ _x∈A I _x = (a, b) for some a , b ∈ R . Thus, F ( D ) = P ∈ B . Second, assume ∩ _x∈A I _x = ∅. Then there exists A ₀ ⊂ A such that ∩ x ∈ A 0 I x = ∅ . Let A = { A 0 ⊂ A : ∩ x ∈ A 0 I x = ∅ } Therefore, if v, u ∈ D and x, y ∈ A ₀ then u(x) ≠ u(y), v(x) ≠ v(y) and u(x) > u(y) ⇔ v(x) > v(y). That is, any utility function in D ranks the alternatives in each A 0 ∈ A the same way and no two alternatives receive equal utility. Then F ( D ) = ∩ A 0 ∈ A B ( ≻ , A 0 ) for all ≻ ∈ F(D). Therefore, F ( D ) ∈ B for all D ∈ D and so F ( D ) ⊂ B .□

Theorem 2.

The space P s is Hausdorff and path-disconnected.

Proof.

We first show P s is Hausdorff. Let ≻ 0 , ≻ 1 ∈ P s such that ≻₁ ≠ ≻₀. Then there exists x, y ∈ X such that x ≻ ₀ y and y ≻ ₁ x. Consider the sets B ₀ = B(≻₀, {x, y}) and B ₁ = B(≻₁, {x, y}). By Theorem 1, B ₀ and B ₁ are neighbordoods of ≻₀ and ≻₁ respectively. Furthermore, B ₀ ∩ B ₁ = ∅. This shows P s is Hausdorff.

We now argue that P s is totally path-disconnected. It suffices to show that U s is totally path-disconnected. Take u , v ∈ U s to be distinct functions. Assume there is a continuous path t joining u and v. That is, t : [ 0,1 ] → U s with t(0) = u, t(1) = v and t continuous. Since u ≠ v and u , v ∈ U s , there are points x, y ∈ X such that u(x) > u(y) and v(y) > v(x). Let Δ(s) = t(s)(x) − t(s)(y) for s ∈ [0, 1]. Then, Δ(0) > 0 > Δ(1). Thus, there is a point s* such that Δ(s*) = 0. Then, t(s*)(x) = t(s*)(y), contradicting that t ( s * ) ∈ U s . Thus, no two distinct functions are connected via a continuous path.□

Proposition 1.

Let P 0 be Hausdorff and P s ⊂ P 0 . Then, P 0 = P s .

Proof.

We proceed in two steps. Recall that ≃ denotes the total indifference preference.

Step 1: if ≃ ∈ P 0 then P 0 is not Hausdorff. By Theorem 1, the only open neighborhood of ≃ is P 0 . Therefore, ≃ is topologically indistinguishable from any other preference, and thus P 0 is not Hausdorff.

We may alternatively proceed using limit arguments. Let 0 denote constant function 0. Clearly F(0) = ≃. Let ≻ ∈ P s be any preference and let u be any representation of ≻. Then u n = 1 n u represents ≻. Since u _n → 0, F(u _n ) = ≻ for all n, F(0) = ≃, then ≃ ∈ { ≻ } ̄ , so P 0 cannot be Hausdorff.

Step 2: if ≃ ∉ P 0 then P ⁰ is not Hausdorff. Take ≻ ∈ P 0 \ P s . For each x ∈ X, let < x > be the indifference class of x according to ≻. For each < x > , let f < x > : < x > → R be any function such that f(a) ≠ f(b) for all a, b ∈ < x >.^[11] Construct ≻ 0 ∈ P s as follows: for any pair x, y ∈ X, if x ≻ y then x ≻ ₀ y; if x ∼ y then x ≻ ₀ y if and only if f _<x>(x) > f _<x>(y). Notice that ≻ 0 ∈ P s and that ≻₀|A = ≻|A for any finite set A that is ≻-strictly ranked. Let G be any open set such that ≻ ∈ G. Then there is a finite collection A and a preference ≻ 1 ∈ P 0 such that ≻ ∈ B ( ≻ 1 , A ) . Without loss of generality B ( ≻ 1 , A ) = B ( ≻ , A ) .^[12] By construction, ≻ 0 ∈ B ( ≻ , A ) . Thus ≻ and ≻₀ are topologically indistinguishable, so P 0 is not Hausdorff.□

Lemma 4.

Let ≻ ∈ P c i . Then, ≻ is locally strict in ( X , T X ( ≻ ) ) .

Proof.

Let (x, y) ∈ X × X be such that x ≿ y. If x ≻ y, then the locally strict property holds vacuously. Assume then that x ∼ y. Then, let N be a neighborhood of (x, y). Then, there is an open set G = G ₁ × G ₂ ⊂ X × X such that (x, y) ∈ G ⊂ N. Assume that for all (z, z′) ∈ G ₁ × G ₂ we had z ∼ z′. Because x ∈ G ₁, for any z′ ∈ G ₂ we get x ∼ z′. Thus, G ₂ ⊂ < x >. By analogous reasoning, G ₁ ⊂ < y >. Then, G ₁ ⊂ < y > and G ₂ ⊂ < x >, contradicting that G is open. Hence, there must exist (z, z′) ∈ G such that z ≻ z′, completing the proof.□

Proposition 2.

If X is finite, then P c i = ∅ . If X is infinite, then P c i is not Hausdorff in the final topology.

Proof.

Part 1: If X is finite, then P c i = ∅ .

Let ≻ be a preference that is locally strict. We start with two observations about ≻: first, ≻ ≠ ≃ and, second ≻ must contain an infinite number of distinct indifference classes. Indeed, assume that ≻ only contains finitely many distinct indifference classes (say, N indifference classes). Then we can pick a representative from each indifference class and label them in an increasing way: x ₁ ≻ x ₂ ≻ (…) ≻ x _N . Then < x 1 > is open as it is the upper contour set of x ₂, thus ≻ is not locally strict. We conclude that if ≻ is locally strict, then it has infinitely many (distinct) indifference classes.^[13] Consequently, if ≻ is locally strict, then X must have infinitely many members. Alternatively, if X is finite, then P c i = ∅ .

Part 2: If X is infinite, then P c i is not Hausdorff in the final topology. We start with an observation about the topology T X ( ≻ ) . The basis for such a topology is given by sets of the form I(z, y) = {x: y ≻ x ≻ z}, where y, z ∈ X ∪ {∞, − ∞}, and we use the convention I(−∞, y) ≡ Lower(y) and I(z, ∞) ≡ Upper(z).^[14] Furthermore, we say a point x is isolated if the following is true: there are points a and b in X such that (1) a ≻ x ≻ b, (2) I(x, a) = I(b, x) = ∅. Next, we make two observations about locally strict preferences. First, a locally strict preference cannot have isolated points. If x was an isolated point then < x > = I ( a , b ) for the points a and b that make x isolated. This is a contradiction because it would imply < x > is open. Second, if ≻ is locally strict, ≻ must have infinitely many distinct indifference classes. This is a consequence of the arguments in Part 1. Consequently, there is a point x ∈ X such that x is neither maximal nor minimal (i.e., there exist a, b ∈ X such that a ≻ x ≻ b) because otherwise every point would be either maximal or minimal and there would only be two indifference classes, contradicting that ≻ is locally strict. Putting these observations together we conclude that any locally strict preference must have a point that is neither maximal nor minimal, and this point cannot be isolated. We now use these observations to prove that P c i is not Hausdorff. Pick any triplet x, y, z ∈ X and choose any preference ≻ ∈ P c i that satisfies z ≻ x ≻ y. Such a preference exists because every preference must have an x that is neither maximal nor minimal. Because x cannot be isolated then at least one of the following must hold: either I(x, z′) ≠ ∅ for all z′ ∈ Upper(x), or I(y′, x) ≠ ∅ for all y′ ∈ Lower(x). We will now construct a preference ≻₁ such that ≻₁ ≠ ≻ and ≻ belongs to every neighborhood of ≻₁, thus showing P c i is not Haursdorff. Without loss of generality, assume I(x, z′) ≠ ∅ for all z′ ∈ Upper(x); we can carry out an analogous construction for the other case. For each pair a, b ∈ X define ≻₁ as follows: if a ∈ Upper(x) and b ∈ Upper(x) then ≻|{a, b} = ≻₁|{a, b}, a ∈ Upper(x) and b ∉ Upper(x) then a ≻₁ b, and if a ∉ Upper(x), b ∉ Upper(x) then a ∼ ₁ b. We make three observations about ≻₁. First, ≻₁ ≠ ≻ because x ∼ ₁ y and x ≻ y. Second, ≻₁ is locally strict. To see this it suffices to show that no subset of the indifference class of x under ≻₁ is open in T X ( ≻ 1 ) . Let G ⊂ {w ∈ X: w ∼ ₁ x} be an open set. Because G is open there must be a non-empty set I(a, b) such that I(a, b) ⊂ G.^[15] Because every point in G is minimal by construction, a = −∞. Then, there must be an alternative b ∈ X such that I(−∞, b) = Lower(b) ⊂ G. Because I(x, z′) ≠ ∅ for all z′ ∈ Upper(x), such a b ∈ X cannot exist: if b ∈ {w: w ∼ ₁ x} then Lower(b) = ∅, and if b ≻ ₁ x then there exists b′ ∈ Lower(b) ∩ Upper(x) so Lower(b) ⊄ G. Thus, G cannot be open. Lastly, ≻₁ satisfies the following property: a ≻₁ b ⇒ a ≻ b for all a, b ∈ X. This last observation implies that if N is a neighborhood of ≻₁ then ≻ ∈ N. Therefore, P c i is not Hausdorff.□

Theorem 3.

The space P clsg is Hausdorff.

Proof.

Let ≻ , ≻ ̂ ∈ P clsg , ≻ ≠ ≻ ̂ . We will show that there are neighborhoods of ≻ and ≻ ̂ that are disjoint. Because ≻ ≠ ≻ ̂ , there must be x, y ∈ X such that x ≻ y and x ⊁ ̂ y .

Case 1: y ≻ ̂ x . Then B(≻, {x, y}) and B ( ≻ ̂ , { x , y } ) are disjoint neighborhoods of ≻ and ≻ ̂ respectively, and this concludes the proof.

Case 2: y ∼ ̂ x . Because ≻ has no gaps, then {w: x ≻ w ≻ y} ≠ ∅. Furthermore, {w: x ≻ w ≻ y} is open in T X by continuity of ≻. First, observe that there must be z ∈ {w: x ≻ w ≻ y} such that z ≁ ̂ x , otherwise {w: x ≻ w ≻ y} would be an open set such that z ∼ ̂ z ′ for all z, z′ ∈ {w: x ≻ w ≻ y}, contradicting that ≻ ̂ is locally strict. Take z ∈ {w: x ≻ w ≻ y} such that z ≁ ̂ x . If x ≻ ̂ z then we get y ∼ ̂ x ≻ ̂ z . Thus, y ≻ ̂ z and z ≻ y, implying B(≻, {z, y}) and B ( ≻ ̂ , { z , y } ) are disjoint neighborhoods of ≻ and ≻ ̂ thereby concluding the proof. If z ≻ ̂ x then we get z ≻ ̂ x and x ≻ z, implying B(≻, {z, x}) and B ( ≻ ̂ , { z , x } ) are disjoint neighborhoods of ≻ and ≻ ̂ thereby concluding the proof.

Therefore, we can always find neighborhoods of ≻ and ≻ ̂ that are disjoint, and this concludes the proof.□