A question on the proof of Open mapping theorem

Question

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.

Answer 1

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 $||x||=\delta^{-1}||y||\cdot||u||<\delta^{-1}||y||$.

Answer 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 $\lVert y\rVert+1/n<1$.