In this paper, we show that the protocol complex of a Byzantine synchronous system can remain $(k - 1)$-connected for up to $\lceil t/k \rceil$ rounds, where $t$ is the maximum number of Byzantine processes, and $t \ge k \ge 1$. This topological property implies that $\lceil t/k \rceil + 1$ rounds are necessary to solve $k$-set agreement in Byzantine synchronous systems, compared to $\lfloor t/k \rfloor + 1$ rounds in synchronous crash-failure systems. We also show that our connectivity bound is tight as we indicate solutions to Byzantine $k$-set agreement in exactly $\lceil t/k \rceil + 1$ synchronous rounds, at least when $n$ is suitably large compared to $t$. In conclusion, we see how Byzantine failures can potentially require one extra round to solve $k$-set agreement, and, for $n$ suitably large compared to $t$, at most that.