Linear Transformation between Normed Vector Spaces is Open iff Image of Open Unit Ball is Open

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.

Let $T : X \to Y$ be a linear transformation.

Let $B_X^O$ be the open unit ball of $X$.

Then $T$ is open $T \sqbrk {B_X^O}$ is open.

Necessary Condition
Suppose that $T$ is open.

From Open Ball is Open Set, $B_X^O$ is open in $\struct {X, \norm {\, \cdot \,}_X}$.

So $T \sqbrk {B_X^O}$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$.

Sufficient Condition
Suppose that $T \sqbrk {B_X^O}$ is open.

Let $U$ be an open set in $X$.

From the definition of an open set:
 * for each $x \in U$ there exists $\epsilon_x > 0$ such that:
 * $x + \epsilon_x B_X^O \subseteq U$

Then we have:
 * $\ds U = \bigcup_{x \in U} \paren {x + \epsilon_x B_X^O}$

Hence:

From Dilation of Open Set in Normed Vector Space is Open:
 * $\epsilon_x T \sqbrk {B_X^O}$ is open for each $x \in U$.

From Translation of Open Set in Normed Vector Space is Open
 * $T x + \epsilon_x T \sqbrk {B_X^O}$ is open for each $x \in U$.

Hence from the definition of a topology:
 * $\ds T \sqbrk U = \bigcup_{x \in U} \paren {T x + \epsilon_x T \sqbrk {B_X^O} }$ is open.

So $T \sqbrk U$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$ whenever $U$ is open in $\struct {X, \norm {\, \cdot \,}_X}$.

So $T$ is an open mapping.