Any finite index subgroup of $\mathbb Z_p$ is open I'm trying to show that every finite index subgroup $H$ of $\mathbb Z_p$ is open.  Since $H$ has finite index, it is equivalent (and perhaps easier) to show that it is closed.  But I've tried showing that, and I can't work out how to use the finite index condition at any point.
Another thing I tried was showing that any non-closed subgroup $K$ of $\mathbb Z_p$ must have infinite index, by using non-convergent sequences in $K$ to construct infinitely many cosets.  But I don't think it's obvious even that $\mathbb Z$ has infinite index in $\mathbb Z_p$ (apart from using a cardinality argument that might not generalize to general non-closed subgroups).  
As far as I'm aware, this is a property that is not shared by general profinite groups.  So any proof must use some other part of the structure of $\mathbb Z_p$ in an essential way.  
 A: Let $H_n=H+p^n\mathbb Z_p$ be the subgroup of $\mathbb Z_p$ genarated by $H$ together with $p^n$. Let $a_n=[\mathbb Z_p:H_n]$. Then from $H\subseteq H_{n+1}\subseteq H_n$ we see that the seequence $(a_n)_{n\in\mathbb N}$ is nondecreasing and bounded from above by $[\mathbb Z_p:H]$, hence it is eventually constant, say $a_n=a_N$ for all $n\ge N$.
But then $[\mathbb Z_p:H_{n}]=[\mathbb Z_p:H_N][H_N:H_{n}]$  implies that also $H_n=H_N$ for $n\ge N$. Therefore
$$H=\bigcap_{n\in\mathbb N}H_n = H_0\cap H_1\cap \ldots\cap H_N = H_N, $$
which immediately gives us $p^N\mathbb Z_p\subseteq H$.
A: I've voted to close the question, since it's answered here.  
For completeness, the answer is:
Let $n=[\mathbb Z_p:H]$.  We claim that $n\mathbb Z_p\subset H$.  But this is just Lagrange's theorem applied to the quotient group $\mathbb Z_p/H$.  
Now it can be shown that $[\mathbb Z_p:n\mathbb Z_p]=|n|_p^{-1}$.  So if $p^k$ is the largest power of $p$ dividing $m$ then $[\mathbb Z_p:n\mathbb Z_p]=p^{k}=[\mathbb Z_p:p^k\mathbb Z_p]$.  Therefore, $n\mathbb Z_p=p^k\mathbb Z_p\subset H$, so $H$ is open.
