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

[Vulture1991]

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

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

I really appreciate any help or hint.
Thank you.

 

[caffeinemachine]

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

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

I really appreciate any help or hint.
Thank you.

If one assumes that $f$ is a proper map then the conclusion holds. By proper I mean $f^{-1}(K)$ is compact whenever $K\subseteq \mathbf R^n$ is compact. Let's prove this.

Assume by way of contradiction that there is a proper map $f:\mathbf R^n\to \mathbf R^m$ such that
$$\lim_{\epsilon\to 0} \max_{x\in \mathcal Z_\epsilon} \text{dist}(x, \mathcal Z)$$
is greater than $0$. Call this limit $\alpha$.

So for each positive natural number $n$ we can find $x_n\in \mathcal Z_{1/n}$ such that $\text{dist}(x_n, \mathcal Z)\geq \alpha$. But each $x_n$ lies is $\mathcal Z_1$, which by assumption of properness is a compact set. Thus the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})_{k=1}^\infty$ which, say, converges to $x_0$. It follows that $\text{dist}(x_0, \mathcal Z)\geq \alpha$.

But for each $N>0$, the points in the sequence $(x_{n_k})$ eventually lie in $\mathcal Z_{1/N}$ which is a closed set, since it is compact. Thus $x_0\in Z_{1/N}$ for each $N$. Therefore $f(x_0)\leq 1/N$ or all $N$ which shows that $f(x_0)=0$. But then $\text{dist}(x_0, \mathcal Z)=0$ and we have a contradiction.