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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.

A Hausdorff, noncompact topology. The condition says precisely that each point of $U$ lies in a residue class $n+k\mathbb{Z}$ contained in $U$. These classes form a basis: the intersection of two classes containing a common point contains the class through that point with modulus equal to the least common multiple of their moduli. Thus arbitrary unions obey the condition, and so do finite intersections. Also $\varnothing$ and $\mathbb{Z}$ obey it. This is the profinite topology on $\mathbb{Z}$.

If $a\ne b$, choose $k>|a-b|$. Then $a+k\mathbb{Z}$ and $b+k\mathbb{Z}$ are disjoint open neighbourhoods, proving the Hausdorff property. To test compactness, defineEvery $F_n$ is nonempty and both open and closed: its complement is a finite union of other residue classes. Since $a_{n+1}-a_n=4^n$, $F_{n+1}\subseteq F_n$. If $x$ belonged to their intersection, then $3x+1$ would be divisible by $4^n$ for every $n$, forcing $3x+1=0$, impossible for an integer. Therefore $U_n=\mathbb{Z}\setminus F_n$ is an increasing open cover of $\mathbb{Z}$. Any finite subcollection has union $U_N$ for its largest index and misses the nonempty $F_N$. This proves that the profinite topology on the integers is not compact.

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 $\hat{\mathbb{Z}}$.

The profinite topology on $\mathbb{Z}$ has basis the residue classes $a+k\mathbb{Z}$, $k\ge 1$. These classes are both open and closed, since their complements are finite unions of other residue classes. Set $F_n=(4^n-1)/3+4^n\mathbb{Z}$ for $n\ge 1$. Then $F_{n+1}\subseteq F_n$ and every $F_n$ is nonempty. An integer $x$ in their intersection would make $3x+1$ divisible by every $4^n$, forcing $3x+1=0$, impossible. Thus the open complements $\mathbb{Z}\setminus F_n$ form an increasing cover with no finite subcover. The topology is Hausdorff because distinct integers are distinct modulo a sufficiently large integer, but is not compact.