While reading the section 7G of Loomis' Introduction to Abstract Harmonic Analysis, I have found the following content of proof very puzzling.
Lemma 1. Let $T$ be a bounded linear transformation of a Banach space $X$ into a Banach space $Y$. If the Image under $T$ of the unit sphere $S_{1}=S(0,1)$ in $X$ is dense in some sphere $U_{r}=S(0,r)$ about the origin of $Y$, then it includes $U_{r}$.
Proof. The set $A=U_{r}\cap T(S_{1})$ is dense in $U_{r}$ by hypothesis. Let $\bar{y}$ be any point of $U_{r}$. Given any $\delta>0$ and taking $y_{0}=0$, we choose inductively a sequence $y_{n}\in Y$ such that $y_{n+1}-y_{n}\in\delta^n A$ and $\lVert y_{n+1}-\bar{y}\rVert<\delta^{n+1}r$ for all $n\geq 0$. ... ... ... (and so on) ... ... ... Thus $U_{r(1-\delta)}\subset T(S_{1})$ for every $\delta$, and hence $U_{r}\subset T(S_{1})$.
The part I'm having troubles with is where we inductively choose the sequence $y_{n}$. Choosing $y_{1}$ is fine since $A$ dense in $U_{r}$ implies the existence of such $y_{1}$. But when choosing $y_{2}$, we have to choose carefully so that $y_{2}-y_{1}\in\delta A$ and $\lVert y_{2}-\bar{y}\rVert<\delta^{2}r$. I just don't see how one can do this in a consistent way for all $n\geq 0$. Please enlighten me!
The author is guilty of obfuscating the reasoning here. A couple of extra lines of text would have made the argument immediately obvious.
$A$ is dense in $U_r\implies\delta A$ is dense in $\delta U_r$. We already know that $(\bar y-y_1)\in B(0,\delta r)=\delta U_r$. Therefore we can find $y_2^*\in\delta A: ||y_2^*-(\bar y-y_1)||<\delta^2r$. Defining $y_2=y_2^*+y_1$ we see that $y_2$ satisfies $y_2-y_1\in\delta A$ and $||y_2-\bar y||<\delta^2r$. Once we have $y_2$ we note that $\delta A$ is dense in $\delta U_r\implies\delta^2 A$ is dense in $\delta^2 U_r$ and our construction continues.