Mapping is Continuous iff Inverse Images of Open Sets are Open

Theorem

Let $X$ and $Y$ be normed vector spaces.

Let $f : X \to Y$ be a mapping.

Then:

$f$ is continuous on $X$

if and only if:

for every $V$ open in $Y$, $\map {f^{-1}} V$ is open in $X$.

Corollary

Let $X$ and $Y$ be normed vector spaces.

Let $f : X \to Y$ be a mapping.

Then:

$f$ is continuous on $X$

if and only if:

for every $F$ closed in $Y$, $\map {f^{-1}} F$ is closed in $X$.

Proof

Sufficient Condition

Let $c \in X$.

Let $\epsilon \in \R_{\mathop > 0}$.

Let $V := \map {B_\epsilon} {\map f c}$ be an open ball in $Y$.

By Open Ball is Open Set in Normed Vector Space, $V$ is an open set in $Y$.

Let $\map {f^{-1}} V = \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$ be an open set in $X$.

Note that $\map f c \in \map {B_\epsilon} {\map f c}$.

Hence, $c \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$.

By definition of open set:

$\exists \delta > 0 : \map {B_\delta} c \subseteq \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$

In other words:

$\forall x \in X : \norm {x - c} < \delta \implies x \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$

or

$\map f x \in \map {B_\epsilon} {\map f c}$

By definition of open ball:

$\norm {\map f x - \map f c} < \epsilon$

But $\epsilon$ was arbitrary.

Hence:

$\forall \epsilon \in \R_{\mathop > 0} : \exists \delta > 0 : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$

Therefore, $f$ is continuous at $c$.

However, $c \in X$ was arbitrary.

Hence, $f$ is continuous on $X$.

$\Box$

Necessary Condition

Let $f$ be continuous on $X$.

Let $V$ be an open subset of $Y$.

Let $c \in \map {f^{-1}} V$.

Hence $\map f c \in V$.

By definition of an open set:

$\exists \epsilon \in \R_{\mathop > 0} : \map {B_\epsilon} {\map f c} \subseteq V$

By continuity of $f$ at $c$:

$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$

In other words:

$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \map f x \in \map {B_\epsilon} {\map f c} \subseteq V$

or

$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \map f x \in V$

$\map f x \in V$ is equivalent to $x \in \map {f^{-1}} V$.

Suppose $\norm {x - c} < \delta$.

By definition of open ball:

$x \in \map {B_\delta} c$

But then:

$x \in \map {B_\delta} c \implies x \in \map {f^{-1}} V$

Hence:

$\map {B_\delta} c \subseteq \map {f^{-1}} V$

But $\delta$ exists for any $c \in \map {f^{-1}} V$.

By definition, $\map {f^{-1}} V$ is open in $X$.

$\blacksquare$

Also see

• Real-Valued Mapping is Continuous if Inverse Images of Unbounded Open Intervals are Open

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.2$: Continuous and linear maps. Continuous maps