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

## Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let the real number line $\R$ be considered as a topology under the usual (Euclidean) topology.

Let $f: T \to \R$ be a real-valued function on $T$.

Then:

- $f$ is continuous

- for all $a \in \R$: $f^{-1} \openint \gets a$ and $f^{-1} \openint a \to$ are open in $T$.

## Proof 1

### Sufficient Condition

Let $f$ be a continuous mapping.

From the corollary to Open Real Interval is Open Set, both $\openint \gets a$ and $\openint a \to$ are open in $\R$.

Then by definition of continuous mapping, $f^{-1} \openint \gets a$ and $f^{-1} \openint a \to$ are both open in $T$.

$\Box$

### Necessary Condition

Let $f$ be such that:

- for all $a \in \R$: $f^{-1} \openint \gets a$ and $f^{-1} \openint a \to$ are open in $T$.

Let $a, b \in \R$ be arbitrary.

From Sub-Basis for Real Number Line:

- $\set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$.

Hence:

- $\forall a, b \in \R: \openint \gets a \cap \openint b \to$ is open in $T$.

Let $A = \openint \gets a$ and $B = \openint b \to$.

From Preimage of Intersection under Mapping:

- $f^{-1} \sqbrk {A \cap B} = f^{-1} \sqbrk A \cap f^{-1} \sqbrk B$

We have been given that $T$ is a topological space.

By assertion:

- $f^{-1} \sqbrk A$ and $f^{-1} \sqbrk B$ are open in $T$.

Then:

- $f^{-1} \sqbrk A \cap f^{-1} \sqbrk B$ is open in $T$.

That is:

- $f^{-1} \sqbrk {A \cap B}$ is open in $T$.

But:

- $A \cap B = \openint a b$

Hence:

- $f^{-1} \openint a b$ is open in $T$.

$a$ and $b$ are arbitrary.

Hence the preimage of an open real interval is open in $T$.

Let $\family {U_i}_{i \mathop \in I}$ be a family of open intervals of $\R$.

Then from Preimage of Union under Mapping: Family of Sets:

- $\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i}$

where:

- $\ds \bigcup_{i \mathop \in I} U_i$ denotes the union of $\family {U_i}_{i \mathop \in I}$
- $f^{-1} \sqbrk {U_i}$ denotes the preimage of $U_i$ under $f$.

Each of $U_i$ is an open interval in $\R$.

Hence from above each of $f^{-1} \sqbrk {U_i}$ is open in $T$.

As $T$ is a topological space:

- $\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i}$ is open in $T$.

From Open Sets in Real Number Line, the open sets of $\R$ are:

It has been shown that all subsets of $\R$ of this form have preimages under $f$ which are open in $T$.

That is:

- $f$ is a continuous mapping.

$\blacksquare$

## Proof 2

From Sub-Basis for Real Number Line:

- $\set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$.

The result follows from Continuity Test using Sub-Basis.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 7$