# 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.

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#### SlipEternal

MHF Helper
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?

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#### 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?

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