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 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.
$\blacksquare$