# A question on the proof of Open mapping theorem

I was following the proof of the Open Mapping Theorem in functional analysis in Wikipedia, and I came across a line in the proof that did not make sense.

Some notations: $U,V$ are open unit balls in $X,Y$ respectively, and $A$ is a bounded linear transformation.

I understood to the part where if $v \in V$ then, $v \in \overline{A(\delta^{-1}U)}$. However, the article claims that this implies that for every $y \in Y$ and $\epsilon > 0$, there is some $x \in X$ such that $$\|x\| < \delta^{-1}\|y\|\text{ and }\|y - Ax\| < \epsilon.$$

I am having trouble with the aforementioned implication. Any help is greatly appreciated.

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Given $0\ne y\in Y$, let $v=\frac y{||y||}$. There exists $u\in U$ such that $A(\delta^{-1}u)\approx v$, e.g. we can enforce $||A(\delta^{-1}u)- v||<\frac{\varepsilon}{||y||}$. Letting $x=\delta^{-1}||y||u$, we find $$||A(x)-y||=||y||\cdot ||A(\delta^{-1}u)-v||<||y||\cdot\frac{\varepsilon}{||y||}=\varepsilon$$ and of course $||x||=\delta^{-1}||y||\cdot||u||<\delta^{-1}||y||$.
For any $y\neq 0$ in $Y$, one can define an element $v\in V$ by taking $0<t<1$ 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. $$\forall y\in Y,\exists x\in X\;\text{ s.t. }\:\|x\|<c\|y\|\:\text{and}\:A(x)=y\quad (3)$$ 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).
If $y\neq 0$, then there is $x_n\in \delta^{-1}U$ such that $\lVert \frac y{\lVert y\rVert+n^{—1}}-Ax_n\rVert<\frac{\varepsilon}{1+\lVert y\rVert}$. We have $$\lVert (\lVert y\rVert+n^{—1})x_n\rVert\leq \delta^{—1}(\lVert y\rVert+1/n),$$ so pick $x_n$ for a $n$ such that $\lVert y\rVert+1/n<1$.