# 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 if and only if:

for every open real set $O$ that overlaps with the image of $f$, the preimage $f^{-1} \left [{O}\right]$ is open.

## Proof

### Necessary Condition

Let $\operatorname{Dom} \left( {f} \right)$ be the domain of $f$.

Let $\operatorname{Im} \left( {f} \right)$ be the image of $f$.

Let $f^{-1} \left [{O}\right]$ be the preimage of $O$ under $f$.

Thus by definition:

$\operatorname{Im} \left( {f} \right)$ is the set of points $q$ in the codomain of $f$ satisfying $q = f \left( {p} \right)$ for a point $p$ in $\operatorname{Dom} \left( {f} \right)$.
$f^{-1} \left [{O}\right]$ is the set of points $p$ in $\operatorname{Dom} \left( {f} \right)$ such that $f \left( {p} \right) \in O$.

Let $f$ be continuous.

Let $O$ be an open real set that overlaps with $\operatorname{Im} \left( {f} \right)$.

We need to show that $f^{-1} \left [{O}\right]$ is open.

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

A point $q_1$ exists in $O \cap \operatorname{Im} \left( {f} \right)$ as $O$ and $\operatorname{Im} \left( {f} \right)$ overlap.

In particular, $q_1 \in \operatorname{Im} \left( {f} \right)$.

Therefore, by the definition of $\operatorname{Im} \left( {f} \right)$, a point $p_1$ in $\operatorname{Dom} \left( {f} \right)$ exists satisfying $f \left( {p_1} \right) = q_1$.

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

 $\displaystyle q_1 \in O$ $\implies$ $\displaystyle f \left( {p_1} \right) \in O$ as $f \left( {p_1} \right) = q_1$ $\displaystyle$ $\implies$ $\displaystyle p_1 \in f^{-1} \left[ {O} \right]$ by the definition of $f^{-1} \left[ {O} \right]$

Accordingly, $f^{-1} \left [{O}\right]$ is non-empty.

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

Let $x$ be a point in $f^{-1} \left [{O}\right]$.

This means that $x \in \operatorname{Dom} \left( {f} \right)$ and $f \left( {x} \right) \in O$.

We know that $f$ is continuous.

Accordingly, $f$ is continuous at $x$ as $x \in \operatorname{Dom} \left( {f} \right)$.

Let an $\epsilon > 0$ be given.

That $f$ is continuous at $x$, means that:

a $\delta > 0$ exists such that $f \left( {y} \right) \in \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ whenever $y \in \left( {x - \delta \,.\,.\, x + \delta} \right) \cap \operatorname{Dom} \left( {f} \right)$.

We know that $f \left( {x} \right) \in O$.

Also, $O$ is open.

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

$\left({f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right) \subseteq O$

We know that $x \in \operatorname{Dom} \left( {f} \right)$.

Also, $\operatorname{Dom} \left( {f} \right)$ is open.

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

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq \operatorname{Dom} \left( {f} \right)$

Having chosen $\epsilon$ and $\delta$ in this way, we have, where $f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ is the image of $\left( {x - \delta \,.\,.\, x + \delta} \right)$ by $f$:

$f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ as $f \left( {y} \right) \in \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ whenever $y \in \left( {x - \delta \,.\,.\, x + \delta} \right)$

which implies:

$f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq O$ as $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right) \subseteq O$.

$(3): \quad$ It is shown that the interval $\left( {x - \delta \,.\,.\, x + \delta} \right)$ is a subset of $f^{-1} \left [{O}\right]$.

Keep in mind that by Subset of Domain is Subset of Preimage of Image:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left[ {f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]} \right]$

Continuing by elaborating on $f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq O$:

 $\displaystyle f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq O$ $\implies$ $\displaystyle f^{-1} \left[ {f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]} \right] \subseteq f^{-1} \left [{O}\right]$ by Image of Subset is Subset of Image $\displaystyle$ $\implies$ $\displaystyle \left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left [{O}\right]$ as $\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left[ {f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]} \right]$

Because:

$f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq O$

it follows that:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left [{O}\right]$

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

$\Box$

### Sufficient Condition

Let $\operatorname{Dom} \left( {f} \right)$ be the domain of $f$.

Let $\operatorname{Im} \left( {f} \right)$ be the image of $f$.

Let $f^{-1} \left [{O}\right]$ be the preimage of $O$ under $f$.

Thus by definition:

$\operatorname{Im} \left( {f} \right)$ is the set of points $q$ in the codomain of $f$ satisfying $q = f \left( {p} \right)$ for a point $p$ in $\operatorname{Dom} \left( {f} \right)$.
$f^{-1} \left [{O}\right]$ is the set of points $p$ in $\operatorname{Dom} \left( {f} \right)$ such that $f \left( {p} \right) \in O$.

Let $f^{-1} \left [{O}\right]$ be open for every open real set $O$ that overlaps with $\operatorname{Im} \left( {f} \right)$.

We need to show that $f$ is continuous.

Let $O$ be an open real set that overlaps with $\operatorname{Im} \left( {f} \right)$.

$(1): \quad$ It is shown that $\operatorname{Dom} \left( {f} \right)$ is non-empty.

A point $q_1$ exists in $O \cap \operatorname{Im} \left( {f} \right)$ as $O$ and $\operatorname{Im} \left( {f} \right)$ overlap.

In particular, $q_1 \in \operatorname{Im} \left( {f} \right)$.

Therefore, by the definition of $\operatorname{Im} \left( {f} \right)$, a point $p_1$ in $\operatorname{Dom} \left( {f} \right)$ exists satisfying $q_1 = f \left( {p_1} \right)$.

Accordingly, $\operatorname{Dom} \left( {f} \right)$ is non-empty.

$(2): \quad$ It is shown that the set $f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$ is open.

Let $x$ be a point in $\operatorname{Dom} \left( {f} \right)$.

Let $\epsilon > 0$ be given.

The open interval $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ overlaps with $\operatorname{Im} \left( {f} \right)$ as $f \left( {x} \right) \in \operatorname{Im} \left( {f} \right)$.

In other words, $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ is an open real set that overlaps with $\operatorname{Im} \left( {f} \right)$.

Accordingly, $f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$ is open by assumption.

$(3): \quad$ It is shown that the interval $\left( {x - \delta \,.\,.\, x + \delta} \right)$ is a subset of $f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$.

By the definition of preimage of $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ under $f$:

$x \in f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$

as:

$x \in \operatorname{Dom} \left( {f} \right)$ and $f \left( {x} \right) \in \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$

Since $f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$ is open, a $\delta \in \R_{>0}$ exists such that:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$

which implies:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq \operatorname{Dom} \left( {f} \right)$ as $f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right] \subseteq \operatorname{Dom} \left( {f} \right)$ by the definition of $f^{-1}$

$(4): \quad$ It is shown that the set $f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ is a subset of $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$.

Keep in mind that:

Because:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq \operatorname{Dom} \left( {f} \right)$

it follows that:

$f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ is defined.

Because:

$f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right] \subseteq \operatorname{Dom} \left( {f} \right)$

it follows that:

$f \left[ {f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]} \right]$ is defined.
$f \left[ {f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]} \right] \subseteq \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$

We continue by elaborating on $\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$:

 $\displaystyle \left( {x - \delta \,.\,.\, x + \delta} \right)$ $\subseteq$ $\displaystyle f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$ $\displaystyle \implies \ \$ $\displaystyle f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ $\subseteq$ $\displaystyle f \left[ {f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]} \right]$ by Image of Subset is Subset of Image $\displaystyle \implies \ \$ $\displaystyle f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ $\subseteq$ $\displaystyle \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$ as $f \left[ {f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]} \right] \subseteq \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$

Because:

$\left( {x - \delta \,.\,.\, x + \delta} \right) \subseteq f^{-1} \left[ {\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)} \right]$

it follows that:

$f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right] \subseteq \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$

In other words, a point in $f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$ is also a point in $\left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$.

Accordingly, let $y \in \left( {x - \delta \,.\,.\, x + \delta} \right)$.

Because:

$f \left( {y} \right) \in f \left[ {\left( {x - \delta \,.\,.\, x + \delta} \right)} \right]$

it follows that:

$f \left( {y} \right) \in \left( {f \left( {x} \right) - \epsilon \,.\,.\, f \left( {x} \right) + \epsilon} \right)$

Therefore, by the definition of continuity, $f$ is [Definition:Continuous Real Function at Point|continuous]] at $x$.

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

$\blacksquare$