Closed Subgroups of P-adic Integers

Theorem
Let $\Z_p$ be the $p$-adic integers for some prime $p$.

Then the closed subgroups of $\Z_p$ are the principal ideals:
 * $\text a) \quad \set 0$
 * $\text b) \quad \forall k \in \N : p^k \Z_p$

Proof
From Metric Space is Hausdorff:
 * $\Z_p$ is a Hausdorff space

From Finite Subspace of Hausdorff Space is Closed:
 * $\set 0$ is closed

From Cosets Form Local Basis of P-adic Number:
 * $\forall k \in \N : p^k \Z_p = 0 + p^k \Z_p$ is closed

Hence the ideals:
 * $\text a) \quad \set 0$
 * $\text b) \quad \forall k \in \N : p^k \Z_p$

are closed subgroups.

It remains to show that an arbitrary closed subgroups is indeed one of these ideals.

Let $H$ be a closed subgroup of the additive group of $\Z_p$.

From Correspondence between Abelian Groups and Z-Modules:
 * $\Z H \subseteq H$

Let $a \in \Z_p$.

Let $h \in H$.

From Integers are Dense in P-adic Integers, there exists a sequence $\sequence {a_n}$:
 * $\forall n \in \N: a_n \in Z$
 * $\ds \lim_{n \mathop \to \infty} a_n = a$

From Multiple Rule for Sequences in Normed Division Ring:
 * $\ds \lim_{n \mathop \to \infty} a_n h = a h$

We have:
 * $a_n h \in \Z H \subseteq H$

From Subset of Metric Space contains Limits of Sequences iff Closed:
 * $a h = \ds \lim_{n \mathop \to \infty} a_n h \in H$

Since $a \in \Z_p$ and $h \in H$ were arbitrary, it follows that:
 * $\Z_p H \subseteq H$

By definition of ideal:
 * $H$ is an ideal of $\Z_p$

From Ideals of P-adic Integers, $H$ is one of the principal ideals:
 * $\text a) \quad \set 0$
 * $\text b) \quad \forall k \in \N : p^k \Z_p$