# Image of Open Set under Continuous Mapping in Metric Space may not be Open

## Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a $\left({d_1, d_2}\right)$-continuous mapping from $A_1$ to $A_2$.

Let $U \subseteq A_1$ be an open set of $M_1$.

Then it is not necessarily the case that $f \left({U}\right)$ is an open set of $M_2$.

## Proof

Consider the constant mapping $f_0: \R^2 \to \R$ defined as:

$\forall \left({x, y}\right) \in \R^2: f \left({x, y}\right) = 0$

Then by Constant Mapping is Continuous $f$ is a continuous mapping.

But consider any open set $U \subseteq A_1$ of $M_1$ such that $U \ne \varnothing$.

Then $f \left({U}\right) = \left\{{0}\right\} = \left[{0 \,.\,.\, 0}\right]$ which is a closed interval of $\R$.

The result follows from Closed Real Interval is not Open Set.

$\blacksquare$