Neat way to prove that $\mathbb{Q}_p$ is complete via 'field of fractions of inverse limit' definition Define the ring $\mathbb{Z}_p$ of $p$-adic integers to be the inverse limit of  $\mathbb{Z}/p^k\mathbb{Z}$ for $k\geq 1$ and the $p$-adic field $\mathbb{Q}_p$ to be its field of fractions. I have shown, starting with this definition, that $\mathbb{Z}_p$ is a discrete valuation ring (see my earlier question) and that each $\alpha\in\mathbb{Q}_p$ can be expressed uniquely in the form $p^ku$ where $k\in\mathbb{Z}$ and $u\in\mathbb{Z}_p^\times=\mathbb{Z}_p\setminus p\mathbb{Z}_p$.
This then defines a valuation $\nu:\mathbb{Q}_p\to\mathbb{Z}\cup\{\infty\}$ given by $\nu(p^ku):=k$ and $\nu(0):=\infty$, which gives an absolute value $|\alpha|:=e^{-\nu(a)}$ (with the convention that $e^{-\infty}=0$) and a metric $d(\alpha,\beta):=|\alpha-\beta|$. So $\mathbb{Q}_p$ is a metric space and contains $\mathbb{Z}_p$ as its closed unit ball.
I want to show that $\mathbb{Q}_p$ is complete in as neat a way as possible (and also that $\mathbb{Z}_p$ is the closure of $\mathbb{Z}$ in $\mathbb{Q}_p$). The alternative definition of $\mathbb{Q}_p$ as the quotient of the set of Cauchy sequences in $\mathbb{Q}$ with respect to the $p$-adic metric mod the null sequences makes completeness clear, so ideally I would be able to find a homeomorphism from the topological definition to the algebraic one. How might I go about doing this?
EDIT: To prove that $\mathbb{Z}_p$ is complete:
If $a^{(n)}$ is a Cauchy sequence in $\mathbb{Z}_p$, then $b^{(n)}:=a^{(n+1)}-a^{(n)}\to 0$ and $b^{(n)}=(b_k^{(n)}+p^k\mathbb{Z})_{k=1}^\infty$ for some $b_k^{(n)}\in\mathbb{Z}$. Then $b^{(n)}$ has a subsequence $b^{(n_m)}$ for which $\nu((b_k^{(n_m)}+p^k\mathbb{Z}))\geq m$ for all $m$, and then $a_k^{(n_m+1)}+p^k\mathbb{Z}=a_k^{(n_m)}+p^k\mathbb{Z}$ for all $1\leq k\leq m$....?
 A: Let's first show that $\mathbb Z_p$ is complete.
For that, just note that under the isomorphism to (or the definition as) the projective limit $\mathbb Z_p = \projlim{\mathbb Z/ p^k}$, the valuation translates to $\nu ((x_{(k)})_k) = \min \{k: x_{(k)} \neq 0\}$. That means that a sequence $x_n$ in that projective limit (that is, each $x_n$ is an element $(x_{n \;(k)})_k \in \projlim{\mathbb Z/ p^k}$) is Cauchy if and only if for each $k$, the sequence of $k$-th components $x_{n \;(k)}$ is (for $n \to \infty$) eventually constant, i.e. goes to a certain $y_{(k)} \in \mathbb Z/ p^k$.
But then it's  almost immediate that the sequence $x_n$ converges to the element $(y_{(k)})_k$.

Completeness of $\mathbb Q_p$ follows like this:
Let $(a_n)_n$ be a Cauchy sequence in $\mathbb Q_p$. There is $N \in \mathbb N$ such that for all $k \ge N$, $\lvert a_k-a_N \rvert \le \lvert p \rvert$; in particular, the "tail end" $(a_n)_{n \ge N}$ of our sequence is contained in
$$a_N + \mathbb Z_p$$
which is a closed neighbourhood of $a_N$, and complete by $\mathbb Z_p$ being complete. So our sequence will have a limit in there.
(Thanks for pointing out one needs an extra argument here. Indeed locally complete metric spaces are not necessarily complete. But as just shown, this one is.)
