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

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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 if and only if $T \sqbrk {B_X^O}$ is open.


Proof

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 $\struct {Y, \norm {\, \cdot \,}_Y}$.

$\Box$

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:

\(\ds T \sqbrk U\) \(=\) \(\ds T \sqbrk {\bigcup_{x \in U} \paren {x + B_X^O} }\)
\(\ds \) \(=\) \(\ds \bigcup_{x \in U} T \sqbrk {x + \epsilon_x B_X^O}\) Image of Union under Mapping: General Result
\(\ds \) \(=\) \(\ds \bigcup_{x \in U} \paren {T x + T \sqbrk {\epsilon_x B_X^O} }\) Image of Translation of Set under Linear Transformation is Translation of Image
\(\ds \) \(=\) \(\ds \bigcup_{x \in U} \paren {T x + \epsilon_x T \sqbrk {B_X^O} }\) Image of Dilation of Set under Linear Transformation is Dilation of Image

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.

$\blacksquare$