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 $\infty > [G:U] > [G:N(U)]$.