Given some ultrametric space $X$, is its completed metric $\hat{X}$ necessarily an ultrametric? I have a lot of difficulty understanding completed metrics. In fact, I don't think I understand them at all!
How could I start showing this? I'm nearly certain that the first two properties of a metric would hold, but the third property is the one I'm not too sure about $\forall x,y,z \in X$ $$d(x,y) \leq max\{d(x,y),d(y,z)\}$$
It almost seems intuitive that it would need to be an ultrametric, just because $i(X)$ is dense in $\hat{X}$, its completed metric with ($i$ an isometry), but I still can't see it...
Any clues? Much appreciated! 
 A: Suppose that $x=\langle x_n:n\in\Bbb N\rangle$, $y=\langle y_n:n\in\Bbb N\rangle$, and $z=\langle z_n:n\in\Bbb N\rangle$ are Cauchy sequences in $X$. Let $\hat x,\hat y$, and $\hat z$ be the equivalence classes of these sequences, interpreted as point of $\widehat X$. We’d like to show that $\hat d(\hat x,\hat y)\le\max\{\hat d(\hat x,\hat z),\hat d(\hat z,\hat y)\}$. For each $n\in\Bbb N$ we know that $$d(x_n,y_n)\le\max\{d(x_n,z_n),d(z_n,y_n)\}\;,$$ so
$$\hat d(\hat x,\hat y)=\lim_{n\to\infty}d(x_n,y_n)\le\lim_{n\to\infty}\max\{d(x_n,z_n),d(z_n,y_n)\}\;.$$
If $d(\hat x,\hat z)<\hat d(\hat z,\hat y)$, then there is an $m\in\Bbb N$ such that $d(x_n,z_n)<d(z_n,y_n)$ for all $n\ge m$, in which case
$$\hat d(\hat x,\hat y)\le\lim_{n\to\infty}\max\{d(x_n,z_n),d(z_n,y_n)\}=\lim_{n\to\infty}d(z_n,y_n)=\hat d(\hat z,\hat y)\;.$$
Similarly, $\hat d(\hat x,\hat y)\le\hat d(\hat x,\hat z)$ if $\hat d(\hat z,\hat y)<\hat d(\hat x,\hat z)$. The only remaining possibility is that $d(\hat x,\hat z)=\hat d(\hat z,\hat y)$, in which case
$$\lim_{n\to\infty}\max\{d(x_n,z_n),d(z_n,y_n)\}=d(\hat x,\hat z)=\hat d(\hat z,\hat y)\;.$$
In all cases, therefore, $$\hat d(\hat x,\hat y)\le\max\{\hat d(\hat x,\hat z),\hat d(\hat z,\hat y)\}\;,$$ and $\hat d$ is an ultrametric.
