06.04.2013 Views

Chapter 7 Infinite product spaces

Chapter 7 Infinite product spaces

Chapter 7 Infinite product spaces

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

7.2. PROOF OF BOREL ISOMORPHISM THEOREM 5<br />

1<br />

and noting that, since an = b x c, if x is irrational two distinct sequences will<br />

an 1<br />

lead to two distinct numbers. The fourth injection is a diagonal map: Write an<br />

element x =(x1, x2,...) 2 [0, 1] N by means of a matrix am,n 2{0, 1} such that<br />

Now consider the diagonal map<br />

xm = Â n 1<br />

k 1<br />

f(x) =Â Â<br />

k 1 m=1<br />

am,n<br />

, m 1. (7.13)<br />

2n a m,m k<br />

. (7.14)<br />

21+···+k 2+m<br />

It is easy to check that this map is continuous, one-to-one and the inverse is continuous.<br />

By Lemma 7.4, the images of these maps are Borel and so all maps are Borel<br />

measurable. By Theorem 7.3, all of these <strong>spaces</strong> are Borel isomorphic.<br />

Recall the well-known fact—which enables the definition of dimension—that [0, 1]<br />

and [0, 1] 2 are not homeomorphic. However, by the arguments in the above lemma,<br />

as standard Borel <strong>spaces</strong> they are indistinguishable.<br />

Having proved the Borel-equivalence of examples 2-4 in our list above, let us now<br />

address the embedding of a general Polish space. Here it is convenient to work<br />

with Hilbert cube:<br />

Proposition 7.6 [Universality of Hilbert cube] Every Polish space is homeomorphic<br />

to a Gd-subset of [0, 1] N .<br />

Proof. Let (X, d) be a Polish space and let us assume without loss of generality<br />

that d 1 (otherwise maps (X, d) ! (X, d<br />

1+d ) by identity map). Let (xn) be a<br />

countable dense set in X. Define the map f : X ! [0, 1] N by<br />

x 2 X 7! f (x) = d(x, xn) n 1 2 [0, 1] N . (7.15)<br />

As is easy to check, f is continuous and one-to-one and so we have f 1 : f (X) ! X.<br />

We claim that f 1 is also continuous. Indeed, suppose that (zn) is a sequence such<br />

that f (zn) ! f (z) in f (X). Let e > 0 and find n 1 such that d(xn, z) < e. Then<br />

d(zm, z) apple d(zm, xn)+d(xn, z)<br />

= f (zm)n + d(z, xn) !<br />

n!• f (z)n + d(z, xn) < 2e. (7.16)<br />

Hence zm ! z or, in explicit terms, f 1 ( f (zm)) ! f 1 ( f (z)), and so f 1 is continuous<br />

on f (X). Thus f : X ! f (X) is a homeomorphism and so, by Lemma 7.4, f (X)<br />

is a Gd subset of [0, 1] N .<br />

The previous proposition gives the desired embedding of X into the Hilbert cube<br />

and, by Lemma 7.5, a Borel isomorphism into the Cantor space. It remains to construct<br />

an embedding of the Cantor space into a general (uncountable) Polish space.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!