# Strange Property of Ultrametric Spaces and Metric Completion

The following property of ultrametric spaces seems quite strange:

(No new values of the metric after completion) Let $x_1, x_2, \ldots$ be a sequence in $X$ converging to $x \in X$. Suppose $a \in X, a \ne x$. Then $d(x_n, a) = d(x,a)$ for large $n$. [...] The metric completion of $X$ is again an ultrametric space, denoted by $X'$, with the metric on $X'$ also denoted by $d$. By the above remark we have $d(X\times X) = d(X' \times X')$.

What does this mean, "no new values after completion" could be interpreted that the space equals it's own completion, but that I am sure is not the case. So what does this property mean, and does it has any applications?

Remark: This is taken from page 4 of this book.

The distance between any two $x,y\in\Bbb Q$ is a value in $\Bbb Q$. If we take the metric completion of $\Bbb Q$ we obtain $\Bbb R$. The distance between two $x,y\in\Bbb R$ can be any real number, not necessarily rational.

So distances between points in $\Bbb R$ can be values that distances between points in $\Bbb Q$ can't be.

What the text is saying is that in an ultrametric space, when you take the completion, no new values will be taken by the distance function. Since $d(X\times X)=\{d(a,b):(a,b)\in X\times X\}$ is the set of all values taken by $d$ on $X\times X$ (i.e. between points of $X$), one way to say this is that the value sets of $X$ and $X'$ are the same, i.e. $d(X\times X)=d(X'\times X')$.

I think "no new values of the metric completion" is just the statement $d(X \times X) = d (X' \times X')$. And I think that $d(X,X)$ is just the set of all distances between pairs of points in $X$, i.e. $d(X \times X) = \{ d(p,q): p,q \in X \}$. This statement is certainly not true in general about any old metric space (e.g. take the open unit interval with Euclidean metric).

It's not the case that any ultra-metric space is complete (i.e. equal to its own completion).

• The $p$-adic integers are metrically complete. The algebraic closure of the $p$-adic numbers are however not metrically complete. – blue Mar 20 '14 at 19:45
• Oh, thanks, sea turtles. I've never really studied $p$-adics, and I thought I had recalled hearing in some brief overview that $\mathbb{Q}_p$ was a metric completion of $\mathbb{Z}_p$, but clearly that's not actually the case! I've edited my original answer accordingly. – Abram Lipman Mar 21 '14 at 21:37
• You are probably thinking of the following two facts: $\Bbb Z_p$ is the metric completion of $\Bbb Z$ and $\Bbb Q_p$ the metric completion of $\Bbb Q$ with respect to the $p$-adic metric. (In particular, this means both $\Bbb Z_p$ and $\Bbb Q_p$ are complete.) – blue Mar 21 '14 at 21:54
• Oh, yes, those two facts each independently make sense. Great! – Abram Lipman Mar 21 '14 at 22:21