For some reason I cannot figure out the following.
Suppose $G$ is a compact, Hausdorff and totally disconnected (but I think compact should suffice) group and $U$ an open subgroup. Why is the normalizer of $U$ open?
This was mentioned in a book I read some time ago but I don't see why.
Edit: As said below $U \subseteq N(U)$ and therefore $\infty > [G:U] > [G:N(U)]$.