Does limit of "approximate zero set" converge to the zero set?

[Vulture]

Let \(\displaystyle f:\mathbb{R}^m\rightarrow\mathbb{R}^m\).
Define the zero set by \(\displaystyle \mathcal{Z}\triangleq\{x\in\mathbb{R}^m | f(x)=\mathbf{0}\}\) and an \(\displaystyle \epsilon\)-approximation of this set by \(\displaystyle \mathcal{Z}_\epsilon\triangleq\{x\in\mathbb{R}^m|~||f(x)||\leq\epsilon\}\) for some \(\displaystyle \epsilon>0\). Clearly \(\displaystyle \mathcal{Z}\subseteq \mathcal{Z}_\epsilon\). Can one assume any condition on the function \(\displaystyle f\) so that
\(\displaystyle
\lim_{\epsilon\rightarrow 0}~\max_{x\in \mathcal{Z}_\epsilon}~\text{dist}(x, \mathcal{Z})=0,
\)
holds?


I know in general this doesn't hold by this example (function of a scalar variable):
\(\displaystyle
f(x)=\left\{\begin{align}
0,\quad{x\leq 0};
\\
1/x,\quad x>0.
\end{align}
\right.
\)

I really appreciate any help or hint.
Thank you.

 

[SlipEternal]

If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion.

 

[Vulture]

Thank you very much for your answer.
Actually \(\displaystyle f\) here is the gradient of a non-convex function \(\displaystyle g\), i.e. \(\displaystyle f=\nabla g\) which is not monotone, and the zero set is the set of critical points. However, I assume \(\displaystyle g\) is \(\displaystyle \mathcal{C}^\infty\).
Do you have any thought how to approach this?

 

[Vulture]

If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion.

Thank you very much for your answer.
Actually \(\displaystyle f\) here is the gradient of a non-convex function \(\displaystyle g\), i.e. \(\displaystyle f=\nabla g\) which is not monotone, and the zero set is the set of critical points. However, I assume \(\displaystyle g\) is \(\displaystyle \mathcal{C}^\infty\).
Do you have any thought how to approach this?