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).



Hausdorff metric on C[0,1]
Yes, this is true.


You need to show that for any $\varepsilon>0$ and any function $f$ in $C[0,1]$
there is $\delta>0$ such that $\varepsilon$-neighborhood in one metric contains $\delta$-neighborhood in the other metric and the other way around.


One way you know already; it follows since $\rho\le d\_{L^\infty}$.


The other way it follows since any continuous function $[0,1]$ is uniformly continuous. Indeed, it is easy to see that if $\omega$ is a modulus of continuity of $f$ and $\rho(f,g)<\varepsilon$ then $d\_{L^\infty}(f,g)<\omega(\varepsilon)+\varepsilon$.


