4
$\begingroup$

I'm reading a book of Ergodic Theory and in a proof of certain result I've found a fact I don't understand how to explain. The text assures that for any separable metric space there exists a countable basis of open sets $\{U_n\}$ whose diameters satisfy $\lim_{n\rightarrow \infty} \mbox{diam} (U_n)=0$. Any idea about how to assure this property for the diameters of the countable basis? Thanks.

$\endgroup$
  • $\begingroup$ There's an obvious way to do it if the metric space is $\sigma$-compact, or $\sigma$-totally bounded. Let $X=\bigcup_n X_n$ where each $X_n$ is totally bounded, and $X_1\subseteq X_2\subseteq\cdots.$ For each $n$ choose a finite cover of $X_n$ with open sets of diameter $\lt1/n$ and take the union. Don't see how to do it with just a separable metric space. $\endgroup$ – bof Sep 15 '17 at 22:53
1
$\begingroup$

Let $\{x_n: n ∈ ω\}$ be a dense set. Let $U_{n, m} := B(x_n, \frac{1}{m})$, i.e. the open ball of radius $\frac{1}{m}$. Clearly, $\{U_{n, m}: n, m ∈ ω\}$ is a countable base. Moreover, $\{U_{n, m}: m ≥ n ∈ ω\}$ is also a countable base, and this one has diameters converging to $0$ since for every $k$ only sets $U_{n, m}$ with $n ≤ m ≤ k$ may have diameter $> \frac{2}{k}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.