Hausdorff distance and $C^0$-distance

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

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

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 $$g(y)\in g(X)$$

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

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

So

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

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

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