> [!NOTE] Theorem (Continuity Via Open and Closed Sets) > The following statements are equivalent: > $(1)$ $f:\mathbb{R}^n \to \mathbb{R}^k$ is [[Continuous maps|continuous]] at all points in $\mathbb{R}^n$; > $(2)$ for all [[Open subsets of metric spaces|open]] subsets $V$ of $\mathbb{R}^k$, $f^{-1}(V)$ is open; > $(3)$ for all [[Closed Sets|closed]] subsets $\mathcal{F}$ of $\mathbb{R}^k$. $f^{-1}(V)$ is closed. ###### Proof of Continuity Via Open and Closed Sets Note that the $\varepsilon-\delta$ definition of continuity at $p$ may be written as for all $\varepsilon>0$, there exists $\delta>0$ such that $f(\mathbb{B}(p,\delta)\cap U) \subset \mathbb{B}(f(p),\varepsilon)$ (where $\mathbb{B}(a,r)$ denotes the [[Open Euclidean Ball|open Euclidean ball]] of radius $r$ centred at $a$) or equivalently $\mathbb{B}(p,\delta)\cap U \subset f^{-1}(\mathbb{B}(f(p),\varepsilon)).\tag{*}$ $(1)\implies(2)$ Suppose $f$ is continuous at all points in $\mathbb{R}^n$ and let $V \subset \mathbb{R}^k$ be open. Then for each $p\in f^{-1}(V)$, there exists $\varepsilon>0$ such that $\mathbb{B}(f(p),\varepsilon)\subset V$. By continuity of $f$ at $p$ as stated in $(*)$, there exists $\delta>0$ such that $\mathbb{B}(p, \delta)\subset f^{-1}(\mathbb{B}(f(p),\varepsilon))\subset f^{-1}(V)$, which shows that $f^{-1}(V)$ is open. $(2) \implies (1)$ Conversely, given $p\in\mathbb{R}^n$ and $\varepsilon>0$, the ball $\mathbb{B}(f(p),\varepsilon)$ is open and therefore, it follows (by assumption) that $f^{-1}(\mathbb{B}(f(p),\varepsilon))$ is open. In particular, $\exists\delta>0$ such that $\mathbb{B}(p,\delta)\subset f^{-1}(\mathbb{B}(f(p),\varepsilon))$, which is precisely the statement of the $\varepsilon$-$\delta$ definition of continuity as expressed by $(*)$. $(2)\iff (3)$ follows from $f^{-1}(V^c)=(f^{-1}(V))^c$ and the facts that [[Characterization of Open and Closed Subsets of Euclidean Space by Set Complement|a set is closed if, and only if its complement is open]].