Let us consider $C[0,1]$, the space of continuous functions $f\colon [0,1] \to \mathbb{R}$. It comes usually with the metric of the maximum, or of the supremum, $d_{L^{\infty}}$. Each element $f$ in $C[0,1]$ corresponds uniquely to a compact subset of $\mathbb{R}^2$, by taking the graph of $f$. Thus, we can consider the Hausdorff metric $\rho$ induced on $C[0,1]$.

My question is: "Is the Hausdorff metric $\rho$ on $C[0,1]$ equivalent to the (usual) $L^{\infty}$-metric?". (equivalent metrics for me means the underlying topologies are the same).

I understand that $\rho \leq d_{L^{\infty}}$. Also, the two metrics are not uniformly equivalent, indeed there exists no universal constant $C$ such that $d_{L^{\infty}} \leq C \cdot \rho$. To see this, just take the sequence of functions $f_n(x)= nx$ and $g_n(x)= nx+1$. The infinity distance of $f_n$ to $g_n$ is constantly equal to $1$, but the Hausdorff distance equals $1/n$.

(I am sorry if the question is too trivial/elementary, feel free to cancel it).