1522 CHAPTER 47. THE YANKOV VON NEUMANN AUMANN THEOREM

∏ki=1 Xi and for k large enough the remaining component’s contribution to the metric, ρ is

very small. Therefore, obtaining d ∈ D close to x ∈ X may be accomplished by findingd ∈ D such that d is close to x in the first k components for k large enough. Note that wedo not use ∏

∞k=1 Dk!

Definition 47.0.3 A complete separable metric space is called a polish space.

Theorem 47.0.4 Let X be a polish space. Then there exists f : NN→ X which is onto andcontinuous. Here NN ≡∏

∞i=1N and a metric is given according to the above theorem. Thus

for n,m ∈ NN,

ρ (n,m)≡∞

∑i=1

2−i |ni−mi|1+ |ni−mi|

.

Proof: Since X is polish, there exists a countable covering of X by closed sets havingdiameters no larger than 2−1,{B(i)}∞

i=1 . Each of these closed sets is also a polish spaceand so there exists a countable covering of B(i) by a countable collection of closed sets,{B(i, j)}∞

j=1 each having diameter no larger than 2−2 where B(i, j) ⊆ B(i) ̸= /0 for all j.Continue this way. Thus

B(n1,n2, · · · ,nm) = ∪∞i=1B(n1,n2, · · · ,nm, i)

and each of B(n1,n2, · · · ,nm, i) is a closed set contained in B(n1,n2, · · · ,nm) whose diam-eter is at most half of the diameter of B(n1,n2, · · · ,nm) . Now we define our mapping fromNN to X . If n ={nk}∞

k=1 ∈ NN, we let f (n)≡ ∩∞m=1B(n1,n2, · · · ,nm) . Since the diameters

of these sets converge to 0, there exists a unique point in this countable intersection andthis is f (n) .

We need to verify f is continuous. Let n ∈ NN be given and suppose m is very closeto n. The only way this can occur is for nk to coincide with mk for many k. Therefore,both f (n) and f (m) must be contained in B(n1,n2, · · · ,nm) for some fairly large m. Thisimplies, from the above construction that f (m) is as close to f (n) as 2−m, proving fis continuous. To see that f is onto, note that from the construction, if x ∈ X , then x ∈B(n1,n2, · · · ,nm) for some choice of n1, · · · ,nm for each m. Note nothing is said about fbeing one to one. It probably is not one to one.

Definition 47.0.5 We call a topological space X a Suslin space if X is a Hausdorff spaceand there exists a polish space, Z and a continuous function f which maps Z onto X .

Zf onto→

continuousX

These Suslin spaces are also called analytic sets in some contexts but we will use the termSuslin space in referring to them.

Corollary 47.0.6 X is a Suslin space, if and only if there exists a continuous mapping fromNN onto X .

Proof: We know there exists a polish space Z and a continuous function, h : Z → Xwhich is onto. By the above theorem there exists a continuous map, g : NN → Z whichis onto. Then h ◦ g is a continuous map from NN onto X . The “if” part of this theorem isaccomplished by noting that NN is a polish space.