For any $y\neq 0$ in $Y$, one can define an element $v\in V$ by taking $0 such that: $v:=\frac{t}{\|y\|}y\in V$ Then $v\in V\subset \delta^{-1}\overline{A(U)}\quad (1)\:$ means that $v\in V$ is adherent to $\delta^{-1} A(U)$. Notice since $V$ is an open subset in a closed set, you can't get an immediat $x\in U$ s.t. $Ax=y$, but instead only that there exists a point $A(x)$ "$\varepsilon$"-close to $v$. By definition of $v$ being a limit point of $\delta A(U)$, any open ball centred at $v$ will intersect $\delta^{-1} A(U)$, so there exists an $x'\in U$ such that $A(x')\in B(v,\varepsilon)\cap\delta^{-1}A(U)$ for any $\varepsilon>0$. $\forall \varepsilon>0,\exists x'\in U\:\text{ s.t. } \;\| \delta^{-1}A(x')-v\|<\varepsilon$ Define $x:=\dfrac{\|y\|}{t\delta}x'$; then $\|x'\|<1 \Longleftrightarrow \|x\|<\dfrac{1}{t\delta}\|y\|$. So when $A$ is surjective - $(1)$ originates from this fact- we have found so far $c(A)=t/\delta>0$ such that: $\forall y\in Y,\,\forall \varepsilon'>0,\exists x\:\text{ s.t. } \|x\|< c\|y\|\:\text{ and }\;\| y-A(x)\|<\varepsilon'\quad (2)$ Compare this to the openness criteria (in this case, we can find exactly an $x\in U$ s.t. $A(x)=y$ since $A(U)$ is open): openess => surjectivity. $\forall y\in Y,\exists x\in X\;\text{ s.t. }\:\|x\| So how to prove surjectivity => openess or how to take $\varepsilon=0$ in $(2)$ ? you need to build a sequence of Ax's which approximates $y$ better and better and ensure that it converges to $y$. Take $y\in Y$ with $\|y\|<1/c$, and $\varepsilon>0$, then by $(2)$, $\exists x_0\in X$ such that $ \|x_0\|\leqslant c\|y\|<1$ and $\|y-Ax_0\|<\varepsilon$. Then by $(2)$ again $\exists x_1$ such that $\|x_1\|\leqslant c\|y-Ax_0\|<\varepsilon$ and $\|y-Ax_0-Ax_1\|<\varepsilon/2$. Then $\exists x_2$ s.t. $\|x_2\|\leqslant c \|y-A(x_0+x_1)\|<\varepsilon/2$ and $\|y-A(x_0+x_1-x_2)\|<\varepsilon/4$ so on by induction. Finally, $\|x_n\|<\varepsilon/2^{n-1}$ and $\|y-A(x_0+\cdots+x_n)\|<\varepsilon/2^n$ so $A(\sum x_n)\to Ax$ since it is absolutely convergent in a complete space, and by continuity, $A(\sum x_n)\to y$ so $\lim A(\sum x_n)=Ax=y$ Then, since $\sum x_n \leqslant 2\delta$, it follows that $y\in 2\delta A(U)$ Note: the finite codomain case is (Openness of linear mapping 2).