Open Mapping Theorem, Brezis, step 2

Question

I'm trying to understand the second step of Open Mapping Theorem's proof by Brezis ( Functional Analysis, Sobolev Spaces and Partial Differential Equations ).

The statement of the theorem is the following.

Let $E$ and $F$ be two Banach spaces and let $T$ be a continuous linear operator from $E$ into $F$ that is surjective. Then there exists a constant $c>0$ such that: $T(B_{E}(0,1))\supset B_{F}(0,c)$.

With the first step it is proved that: $$\overline{T(B(0,1))} \supset (B(0,2c))\tag{1}$$ I can't understand why ( and this is the beginning of the second step of the proof ) if we choose any $y \in F$ with $\Vert y \Vert<c$ then by $(1)$ we know that: $\forall \varepsilon>0\exists z\in E$ with $\Vert z \Vert<\frac{1}{2}$ and $\Vert y-Tz \Vert < \varepsilon$. Could anyone help me?

There appear to be some typos in what you have written. As written, the claim is definitely false. — Eric Wofsey, Aug 29 '18 at 17:45
Your last sentence doesn't make sense, there is a syntax problem or a typo when you write with $|y|<\frac12$ — Arnaud Mortier, Aug 29 '18 at 17:50
Please correct your post: replace $|y|<\frac{1}{2}$ by $|z|<\frac{1}{2}$, and maybe then you will understand what's going on. — uniquesolution, Aug 29 '18 at 18:06
Thank you for your answers. I'm sorry, it was a typo. I still can't understand why it should be true. — F. Buseghin, Aug 29 '18 at 18:32
It still isn't true; there must be at least one more typo. It would be true if the condition were $|z|<1$ instead... — Eric Wofsey, Aug 29 '18 at 18:35

score 2 · Accepted Answer · answered Aug 29 '18 at 18:44

Since $2y\in B_F(0,2c)\subseteq \overline{T(B_E(0,1))}$, there exists $x\in T(B_E(0,1))$ such that $\|2y-x\|<\epsilon$. We can write $x=Tw$ for some $w\in B_E(0,1)$. Then for $z=w/2$, we have $\|z\|<1/2$ and $\|y-Tz\|=\|\frac{1}{2}(2y-x)\|=\frac{1}{2}\|2y-x\|<\epsilon$.

Open Mapping Theorem, Brezis, step 2

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