Set with $\|f(\vec{a})\|<1$ for $f$ Lipschitz

Question

Let $A\subseteq \mathbb{R}^n$ be open and $f:A\to\mathbb{R}^m$ be a Lipschitz function. Prove that $U:=\{\vec{a}\in A: \|f(\vec{a})\|<1\}$ is an open set.

My approach:

Since $f$ is Lipschitz, it is continuous on $A$, thus, $\forall\varepsilon >0$, $\exists \delta >0 $ such that $f(B(\vec{a}, \delta))\subseteq B(f(\vec{a}),\varepsilon)\iff B(\vec{a},\delta)\subseteq f^{-1}(B(f(\vec{a}), \varepsilon))$. Let $\|f(\vec{a})\|<1$, then ... actually I forgot how exactly I've approached this problem.

Would appreciate your advice.

Answer 1

Lipschitz is not needed, only continuity:

Let $V=\{ x \in \mathbb R^n\: ||x||<1\}$ and $g(a): =||f(a)||$ for $a \in A$. Then $g$ is continuous and $g^{-1}(V)=U$. Hence $U$ is open in $A$. Since $A$ is open, it follows that $U$ is open.