Vtyjkyuk

I'm reading the proof of the Open Mapping Theorem from "Analysis Now", by Pedersen.

Theorem: Let $X,Y$ be Banach spaces and $Tin B(X,Y)$ with $T(X)=Y$, then $T$ is an open map.

The proof goes like this: We can write $Y=cupbar T(B(0,n))$, where $B(0,n)$ is the closed ball of radius $n$ around $0$. Now, by the Baire category theorem we can say that there exists $n$ such that there is $B(y,epsilon) subset bar T(B(0,n))$. This means that $T(B(0,1))$ is dense in $B(y,n^-1epsilon)$, and therefore ... (it continues but we arrived at the point that I don't understand).

How do we know that $T(B(0,1))$ is contained in $B(y,n^-1epsilon)$? I know that, because $B(y,epsilon) subset bar T(B(0,n))$ then $B(n^-1y,n^-1epsilon) subset bar T(B(0,1))$.

EDIT: It can be worth noting that in the proof, the author invokes a Lemma (2.2.3. in the book):

Lemma 2.2.3. If $T in B(X,Y)$ and the image of the unit ball in $X$ is dense in some $B(0,r) subset Y, r>0$ then $B(0,(1-epsilon)r) subset T(B(0,1))$, for every $epsilon>0$

I suspect this might be a case of poor wording. The usual proof of Lemma 2.2.3 in fact proves a little bit more. We have

Lemma: Let $X$ be a Banach space and $Y$ be a normed space with $T in B(X,Y)$. Suppose that there exist $varepsilon in (0,1)$ and $r > 0$ such that for any $y in B(0,r)$, $operatornamedist(y, T(B(0,1))) < varepsilon$. Then $B(0,r(1-varepsilon)) subseteq T(B(0,1))$.

The usual proof of the OMT then proceeds by noting that since $T(B(0,1))$ is convex and symmetric about $0$, $B(n^-1y, n^-1 r) subseteq overlineT(B(0,1))$ implies that $B(0, r) subseteq overlineT(B(0,1))$.

In particular, since $B(0,r) subseteq overlineT(B(0,1))$, the lemma gives us that $B(0,r(1-varepsilon)) subseteq T(B(0,1))$ for every $varepsilon in (0,1)$ and so
$$B(0,r) = bigcup_varepsilon in (0,1) B(0,r(1-varepsilon)) subseteq T(B(0,1))$$
which implies that $T$ is an open mapping.