Product of a sequence of completely metrizable spaces is completely metrizable (p13 (C))

The product of a sequence of completely metrizable spaces is completely metrizable.
Proof.
Let (X_n) be a sequence of completely metrizable spaces.  For each X_n, let d_n denote a compatible metric.

Let X = \prod_{n=0}^{\infty} X_n have the product topology, which is given by the metric d(x,y) = \sum_{n=0}^{\infty}2^{-n-1}d_n'(x_n,y_n), where d_n' = \frac{d_n}{1 + d_n}.  I will show that this metric d is complete.

Let (x_n) be a Cauchy sequence in X.  Let \epsilon > 0.  There exists N, such that m,n \ge N \Rightarrow d(x_m, x_n) = \sum_{k=0}^{\infty} 2^{-k-1} d_k'(x_m(k),x_n(k)) < \epsilon. Hence, for each k, d_k'(x_m(k), x_n(k)) < 2^{k + 1}\epsilon.  Thence, (x_n(k))_{n=0}^{\infty} is Cauchy with respect to d_k'.  By p13(D), d_k' is complete since d_k is, so x_n(k) \to x(k) (with respect to both d_k, d_k').  This defines x \in X.  Now, is it true that (x_n) \to x \in (X,d)?

Consider (\mathbf{N}, \mu), the measure space of the natural numbers with counting measure.  Let m \in \mathbf{N}, and f_n(k) = 2^{-k-1}d_k'(x_m(k),x_n(k)).  Then \lim_{n \to \infty} f_n(k) = 2^{-k-1}d_k'(x_m(k),x(k)) := f(k), by the continuity of d_k' (see p13(E)).  Also, |f_n(k)| \le 2^{-k-1} := g(k) for all k,n and \int_{\mathbf{N}} g \text{ d} \mu = 1.

Let \epsilon > 0. Since (x_n) is Cauchy, there exists N, such that m, n \ge N \Rightarrow \int_{\mathbf{N}} f_n \text{ d} \mu < \epsilon.
By Lebesgue’s Dominated Convergence Theorem, if m \ge N,
\epsilon \ge \lim_{n \to \infty} \int_{\mathbf{N}} f_n d \mu = \int_{\mathbf{N}} f d \mu = \sum_{k = 0}^{\infty} 2^{-k-1}d_k'(x_m(k),x(k)) = d(x_m,x).

Hence (x_n) \to x, as desired.  So the metric d completely metrizes X.

Advertisements
This entry was posted in Descriptive Set Theory, Kechris, Logic. Bookmark the permalink.

2 Responses to Product of a sequence of completely metrizable spaces is completely metrizable (p13 (C))

  1. Jindrich Zapletal says:

    I prefer to cut off the n-th metric at 1 instead of considering d’. This is to say, consider d”(x,y)=min( d(x,y), 1). Comparisons between different metrics for the same space can be interesting. Say two metrics are equivalent if one is between two constant multiples of the other. Can you produce many inequivalent metrics for the same given topology?

  2. alanmath says:

    Sorry it has taken me a while to respond. I have been on break for a week, and went down some wrong paths before that. But I was making my formulas overcomplicated; it appears that, given a metric space X with an unbounded metric d, \{ d_n : d_n(x , y ) = (d(x,y))^{1/n} \text{ for all } x,y \in X, n \in \mathbb{N} \} is an infinite family of inequivalent metrics that give the same topology on X.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s