Real Line Continuity by Inverse of Mapping

Theorem
Let $f$ be a real function.

Let the domain of $f$ be open.

Let $f^{-1}$ be the inverse of $f$.

Then $f$ is continuous :
 * for every open real set $O$ that overlaps with the image of $f$, the preimage $f^{-1} \sqbrk O$ is open.

Necessary Condition
Let $\Dom f$ be the domain of $f$.

Let $\Img f$ be the image of $f$.

Let $f^{-1} \sqbrk O$ be the preimage of $O$ under $f$.

Thus by definition:
 * $\Img f$ is the set of points $q$ in the codomain of $f$ satisfying $q = \map f p$ for a point $p$ in $\Dom f$.
 * $f^{-1} \sqbrk O$ is the set of points $p$ in $\Dom f$ such that $\map f p \in O$.

Let $f$ be continuous.

Let $O$ be an open real set that overlaps with $\Img f$.

We need to show that $f^{-1} \sqbrk O$ is open.

$(1): \quad$ It is shown that $f^{-1} \sqbrk O$ is non-empty.

A point $q_1$ exists in $O \cap \Img f$ as $O$ and $\Img f$ overlap.

In particular, $q_1 \in \Img f$.

Therefore, by definition of $\Img f$, a point $p_1$ in $\Dom f$ exists satisfying $\map f {p_1} = q_1$.

Also, $q_1 \in O$, which gives:

Accordingly, $f^{-1} \sqbrk O$ is non-empty.

$(2): \quad$ It is shown that the function $f$ maps $\openint {x - \delta} {x + \delta}$ into $O$.

Let $x$ be a point in $f^{-1} \sqbrk O$.

This means that $x \in \Dom f$ and $\map f x \in O$.

We know that $f$ is continuous.

Accordingly, $f$ is continuous at $x$ as $x \in \Dom f$.

Let an $\epsilon > 0$ be given.

That $f$ is continuous at $x$, means that:
 * a $\delta > 0$ exists such that $\map f y \in \openint {\map f x - \epsilon} {\map f x + \epsilon}$ whenever $y \in \openint {x - \delta} {x + \delta} \cap \Dom f$.

We know that $\map f x \in O$.

Also, $O$ is open.

This allows us to choose $\epsilon \in \R_{>0}$ small enough such that:


 * $\openint {\map f x - \epsilon} {\map f x + \epsilon} \subseteq O$

We know that $x \in \Dom f$.

Also, $\Dom f$ is open.

This allows us to choose $\delta \in \R_{>0}$ small enough such that:


 * $\openint {x - \delta} {x + \delta} \subseteq \Dom f$

Having chosen $\epsilon$ and $\delta$ in this way, we have, where $f \sqbrk {\openint {x - \delta} {x + \delta} }$ is the image of $\openint {x - \delta} {x + \delta}$ by $f$:


 * $f \sqbrk {\openint {x - \delta} {x + \delta} } \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$ as $\map f y \in \openint {\map f x - \epsilon} {\map f x + \epsilon}$ whenever $y \in \openint {x - \delta} {x + \delta}$

which implies:


 * $f \sqbrk {\openint {x - \delta} {x + \delta} } \subseteq O$ as $\openint {\map f x - \epsilon} {\map f x + \epsilon} \subseteq O$.

$(3): \quad$ It is shown that the interval $\openint {x - \delta} {x + \delta}$ is a subset of $f^{-1} \sqbrk O$.

Keep in mind that by Subset of Domain is Subset of Preimage of Image:
 * $\openint {x - \delta} {x + \delta} \subseteq f^{-1} \sqbrk {f \sqbrk {\openint {x - \delta} {x + \delta} } } $

Continuing by elaborating on $f \sqbrk {\openint {x - \delta} {x + \delta} } \subseteq O$:

Because:
 * $f \sqbrk {\openint {x - \delta} {x + \delta} } \subseteq O$

it follows that:
 * $\openint {x - \delta} {x + \delta} \subseteq f^{-1} \sqbrk O$

Since $x$ is an arbitrary point in $f^{-1} \sqbrk O$, it follows by definition of open set that $f^{-1} \sqbrk O$ is open.

Sufficient Condition
Let $\Dom f$ be the domain of $f$.

Let $\Img f$ be the image of $f$.

Let $f^{-1} \sqbrk O$ be the preimage of $O$ under $f$.

Thus by definition:
 * $\Img f$ is the set of points $q$ in the codomain of $f$ satisfying $q = \map f p$ for a point $p$ in $\Dom f$.
 * $f^{-1} \sqbrk O$ is the set of points $p$ in $\Dom f$ such that $\map f p \in O$.

Let $f^{-1} \sqbrk O$ be open for every open real set $O$ that overlaps with $\Img f$.

We need to show that $f$ is continuous.

Let $O$ be an open real set that overlaps with $\Img f$.

$(1): \quad$ It is shown that $\Dom f$ is non-empty.

A point $q_1$ exists in $O \cap \Img f$ as $O$ and $\Img f$ overlap.

In particular, $q_1 \in \Img f$.

Therefore, by definition of $\Img f$, a point $p_1$ in $\Dom f$ exists satisfying $q_1 = \map f {p_1}$.

Accordingly, $\Dom f$ is non-empty.

$(2): \quad$ It is shown that the set $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$ is open.

Let $x$ be a point in $\Dom f$.

Let $\epsilon \in \R_{>0}$ be given.

The open interval $\openint {\map f x - \epsilon} {\map f x + \epsilon}$ overlaps with $\Img f$ as $\map f x \in \Img f$.

In other words, $\openint {\map f x - \epsilon} {\map f x + \epsilon}$ is an open real set that overlaps with $\Img f$.

Accordingly, $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$ is open by assumption.

$(3): \quad$ It is shown that the interval $\openint {x - \delta} {x + \delta}$ is a subset of $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$.

By definition of preimage of $\openint {\map f x - \epsilon} {\map f x + \epsilon}$ under $f$:


 * $x \in f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$

as:
 * $x \in \Dom f$ and $\map f x \in \openint {\map f x - \epsilon} {\map f x + \epsilon}$

Since $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$ is open, a $\delta \in \R_{>0}$ exists such that:


 * $\openint {x - \delta} {x + \delta} \subseteq f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$

which implies:


 * $\openint {x - \delta} {x + \delta} \subseteq \Dom f$ as $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} } \subseteq \Dom f$ by definition of $f^{-1}$

$(4): \quad$ It is shown that the set $f \sqbrk {\openint {x - \delta} {x + \delta} }$ is a subset of $\openint {\map f x - \epsilon} {\map f x + \epsilon}$.

Keep in mind that because:
 * $\openint {x - \delta} {x + \delta} \subseteq \Dom f$

it follows that:
 * $f \sqbrk {\openint {x - \delta} {x + \delta} }$ is defined.

Because:
 * $f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} } \subseteq \Dom f$

it follows that:
 * $f \sqbrk {f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} } }$ is defined.

By Subset of Codomain is Superset of Image of Preimage:
 * $f \sqbrk {f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} } } \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$

We continue by elaborating on $\openint {x - \delta} {x + \delta} \subseteq f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$:

Because:
 * $\openint {x - \delta} {x + \delta} \subseteq f^{-1} \sqbrk {\openint {\map f x - \epsilon} {\map f x + \epsilon} }$

it follows that:
 * $f \sqbrk {\openint {x - \delta} {x + \delta} } \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$

In other words, a point in $f \sqbrk {\openint {x - \delta} {x + \delta} }$ is also a point in $\openint {\map f x - \epsilon} {\map f x + \epsilon}$.

Accordingly, let $y \in \openint {x - \delta} {x + \delta}$.

Because:
 * $\map f y \in f \sqbrk {\openint {x - \delta} {x + \delta} }$

it follows that:
 * $\map f y \in \openint {\map f x - \epsilon} {\map f x + \epsilon}$

Therefore, by definition of continuity, $f$ is continuous at $x$.

Since $x$ is an arbitrary point in the domain of $f$, $f$ is continuous.

Also see

 * Metric Space Continuity by Inverse of Mapping between Open Balls