Show that $-y + \delta B_X \subseteq \overline{T(B_X)}.$

Question

The following theorem and proof are taken from 'An Introductory Course in Functional Analysis' by Bowers and Kalton, page $74.$

Theorem 4.29: (Open Mapping Theorem) Suppose $X$ and $Y$ are Banach spaces. If $T:X \rightarrow Y$ is a bounded surjective operator, then $T$ is an open map.

Proof:Observe that $Y = T(X) = \cup_{n=1}^{\infty}\overline{T(B_X)}.$ Therefore, by theorem $4.7,$ then set $\overline{T(B_X)}$ has non-empty interior. Consequently, there exists an element $y \in Y$ and a number $\delta>0$ such that $$y + \delta B_Y \subseteq \overline{T(B_X)}.$$ A simple calculation reveals that $-y + \delta B_X \subseteq \overline{T(B_X).}$

Question: How to obtain $-y + \delta B_X \subseteq \overline{T(B_X)}$?

I do now know how simple is the calculation. But it seems that I couldn't get it.

Any hint would be appreciated.

Answer 1

Let $a \in -y + \delta B_Y$, then $a=-y+ \delta z$ for some $z \in B_Y$.

Hence $-a=y+\delta(-z) \in y + \delta B_Y$. Thus $-a \in \overline{T(B_X)}$.

Thus there is a sequence $(x_n)$ in $T(B_X)$ with $x_n \to -a$. Therefore $-x_n \to a$.

Since $-x_n \in T(B_X)$, we get $a \in \overline{T(B_X)}$.