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Because the vertex set is finite, at least one vertex occurs infinitely often, so $\mathrm{Inf}(\rho )$ is nonempty. Let $u,v$ belong to this set. After any chosen visit to $u$, a later visit to $v$ exists; the intervening edges give a directed walk from $u$ to $v$. After that visit to $v$, a later visit to $u$ also exists, giving a walk in the other direction. Removing repeated portions gives directed paths if desired.

Thus every pair in $\mathrm{Inf}(\rho )$ is mutually reachable and belongs to the same strongly connected component. This proves the claimed inclusion, which need not be equality: an infinite walk can repeatedly follow one cycle while omitting other vertices in its component. This is the recurrence-set part of the infinite walk criterion in a finite directed graph.

An infinite trace exists exactly when a directed cycle is reachable from $s$, by the infinite walk criterion in a finite directed graph. Compute the reachable set with depth-first search, compute its strongly connected components, and return true if any component has at least two vertices or contains a self-loop. A singleton without a self-loop does not qualify.

If the test succeeds, follow a path from $s$ to the cycle and traverse the cycle forever. Conversely, an infinite trace in a finite graph repeats a vertex, producing a reachable nonempty closed walk and hence a directed cycle. Reachability, component construction and the self-loop check all scan adjacency lists only a constant number of times. The worst-case time is $O(|V|+|E|)$, with $O(|V|+|E|)$ auxiliary space for a straightforward transpose-based implementation.