Lemma of open mapping theorem Kreyszig Revising the Proof of the Kreyszig Open Mapping Theorem, being more specific the proof of the lemma 4.12-3, the central idea is find a open ball $B(0,r)$ such that exists $B(0,\delta)\subseteq T(B(0,r))$.
In the book they prove it for the case $r = 1$,
however it is valid for any $r$.
I understand the proof correctly, however in the final part I have a question that may be a bit subtle.
The sequence $z_n$ converge to $x$, i.e
$$ \sum_{n=1}^{\infty}{x_n} = x $$
So
$$ \lim_{n\to \infty} \Vert z_n \Vert = \Vert x \Vert $$
Then how $\Vert x_n \Vert < 1/2^n$ for all $n$ the book says:
$$ \Vert x \Vert = \lim_{n\to \infty} \Vert z_n \Vert < \sum_{n=1}^{\infty} \frac{1}{2^n} = 1  $$
And conclude that $\Vert x \Vert < 1$. My question is simple, it shouldn't be
$$ \Vert x \Vert = \sum_{n=1}^{\infty}\Vert x_n \Vert \leq \sum_{n=1}^{\infty} \frac{1}{2^n} = 1  $$
that is $\Vert x \Vert \leq 1$.
As I said at the beginning, it doesn't matter if we prove for $r = 2$, it still doesn't affect the proof of the theorem. However, I think we cannot say that $\Vert x \Vert < 1$.


Suppose that $$\tag1\sum_k\|x_k\|=\sum_k2^{-k}.$$ Then
$$
0=\sum_k\tfrac1{2^k}-\sum_k\|x_k\|=\sum_k\big(\tfrac1{2^k}-\|x_k\|\big).
$$
This last series has non-negative terms; actually, all of its terms are positive. That is, the sequence of partial sums is positive and increasing, so its limit cannot be zero. The contradiction shows that the equality $(1)$ is impossible. It would have been enough for a single term to be nonzero, to guarantee the strict inequality.
On a separate note, you say that $\sum_k\|x_k\|=\|x\|$. That's rarely true. You have the inequality $\|x\|\leq\sum_k\|x_k\|$, though, so the proof has no issue.
