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A group $G$ is conjugacy separable when any two nonconjugate elements remain nonconjugate in some finite quotient. Equivalently, every conjugacy class is closed in the profinite topology.

Let $I(L/K)$ be the inertia group and let $\phi$ denote Frobenius on the residue-field extension. The Relative Weil group isIts Weil-group topology makes $I(L/K)$ an open profinite group with its usual topology and gives $W(L/K)/I(L/K)$ the discrete topology. Thus every inertia coset is an open copy of $I(L/K)$.

Take $L=K^{\mathrm{nr}}$, the maximal unramified extension of $K$. ThenThe subgroup $\{0\}\subset \mathbb{Z}$ is open in the discrete Weil-group topology, but it is not open in the profinite topology inherited from $\hat{\mathbb{Z}}$.