> [!definition] > > Let $X, Y$ be [[Topological Space|topological spaces]]. A [[Function|function]] $f: X \to Y$ is **open** if $f(U)$ is [[Open Set|open]] in $Y$ whenever $U$ is open in $X$. > [!theorem] > > Let $X, Y$ be [[Metric Space|metric spaces]]. A function $f: X \to Y$ is open if and only if for any $B = B(x, r) \subset X$, $f(B) \supset B(f(x), r')$ for some $r' > 0$. > > *Proof*. If $f$ is open, then $f(B)$ is open and contains an open ball centred at $f(x)$. > > Suppose that $f(B(x, r))$ contains an open ball centred at $f(x)$ for any $x \in X$, $r > 0$. Let $U \in \topo_X$ be an open set and $f(x) \in f(U)$, then there exists a ball $B(x, \varepsilon_x) \subset U$. As the image of an open ball contains an open ball, > $ > B(f(x), \varepsilon_{f(x)}) \subset f(B(x, \varepsilon_x)) \subset F(U) > $ > therefore for any $f(x) \in f(U)$, there exists $\varepsilon_{f(x)} > 0$ such that $B(f(x), \varepsilon_{f(x)}) \subset f(U)$, and $f(U)$ is open. Let $\cx, \cy$ be [[Normed Vector Space|normed spaces]] and $f: \cx \to \cy$ be a [[Linear Transformation|linear map]], then $f$ is open if and only if $f(B)$ contains a ball centred at $0$ for any ball $B$ centred at $0$. *Proof*. Firstly, $f$ commutes under addition and dilation, meaning that for any $E \subset \cx$, $\lambda \in \complex$ and $y \in \cx$, $ \begin{align*} f(\lambda E + y) &= \bracs{f(\lambda x + y): x \in E} \\ &= \bracs{\lambda f(x) + f(y): x \in E} \\ &= \bracs{\lambda f(x): x \in E} + f(y) \\ &= \lambda f(E) + f(y) \end{align*} $