Let $(X, d)$ be a compact metric space and $A, B\subseteq X$. The Hausdorff distance between $A$ and $B$ is defined by \begin{equation} d_H(A, B)= \max \{\sup_{a\in A}d(a, B), \sup_{b\in B}d(A, b)\}. \end{equation} Also $C^0$-distance between the maps $f:X\to X$ and $g:X\to X$ of the same metric space $(X, d)$ defined by \begin{equation} d_{C^0}(f, g)= \sup_{x\in X}(d(f(x), g(x)). \end{equation} Question. Let $d_{C^0}(f, g)<\delta$. Can we say that $d_H(f(X), g(X))<\delta$?

Please help me to know it.

  • $\begingroup$ @ Adam Chalumeau , yes. In my research, $X$ is compact metric space. $\endgroup$ Commented Jul 1, 2019 at 18:09

2 Answers 2


Fixe $\delta^\prime>0$ such that $d_{C^0}(f,g)<\delta^\prime<\delta$. Let $a=f(x)\in f(X)$. Because $g(x)\in g(X)$ you have $$d(a,g(X))=d(f(x),g(X))\leq d(f(x),g(x))<\delta^\prime$$ hence $\sup_{a\in f(X)}d(a,g(X))\leq\delta^\prime$. Similarly $\sup_{b\in g(X)}d(f(X),b)\leq\delta^\prime$. Finally you get $$d_H(f(X),g(X))\leq \delta^\prime<\delta.$$


I don’t know if your assertion is true but I think that you can get the following result:

For each fixed $f(x)\in f(X)$ you have that for each $g(y)\in g(X)$

$d(f(x),g(y))\leq d(f(x),g(x))+d(g(x),g(y))$

$\leq d_{C^0}(f,g)+d_{g(X)}$


$d(f(x),g(X))\leq d_{C^0}(f,g)+d_{g(X)}$

where $d_{g(X)}$ is the diameter of $g(X)$.

For the same reason you have that for each fixed $g(x)\in g(X)$

$d(f(X),g(x))\leq d_{C^0}(f,g)+d_{f(X)}$

So you have that

$d_H(f(X),g(X))\leq d_{C^0}(f,g)+max(d_{f(X)},d_{g(X)})$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .