Definition:Continuous Mapping (Metric Space)

Definition
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 mapping from $A_1$ to $A_2$.

Let $a \in A_1$ be a point in $A_1$.

Definition using Limit
$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) iff:
 * The limit of $f \left({x}\right)$ as $x \to a$ exists
 * $\displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$.

Epsilon-Delta Definition
$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) iff:


 * $\forall \epsilon > 0: \exists \delta > 0: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon$

Epsilon-Neighborhood Definition
$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) iff:
 * $\forall N_\epsilon \left({f \left({a}\right)}\right): \exists N_\delta \left({a}\right): f \left({ N_\delta \left({a}\right)}\right) \subseteq N_\epsilon \left({f \left({a}\right)}\right)$

where $N_\epsilon \left({a}\right)$ is the $\epsilon$-neighborhood of $a$ in $M_1$.

That is, for every $\epsilon$-neighborhood of $f \left({a}\right)$ in $M_2$, there exists a $\delta$-neighborhood of $a$ in $M_1$ whose image is a subset of that $\epsilon$-neighborhood.

Continuous on a Space
$f$ is continuous from $\left({A_1, d_1}\right)$ to $\left({A_2, d_2}\right)$ iff it is continuous at every point $x \in A_1$.

Open Set Definition
$f$ is continuous iff:
 * for every set $U \subseteq M_2$ which is open in $M_2$, $f^{-1} \left({U}\right)$ is open in $M_1$.

Equivalence of Definitions
All these statements are equivalent by Equivalence of Metric Space Continuity Definitions.

If necessary, for clarity, we can say that $f$ is $\left({d_1, d_2}\right)$-continuous.

If $f$ is continuous in this sense for all $a \in A_1$, then $f$ is $\left({d_1, d_2}\right)$-continuous on $A_1$.

Warning
When $f: M_1 \to M_2$ is continuous, it does not necessarily follow that if $U$ is open in $M_1$ then $f \left({U}\right)$ is open in $M_2$.

For example, let $f: \R^2 \to \R$ such that $\forall x \in \R^2: f \left({x}\right) = 0$.

Then $f$ is continuous but for any non-empty open set $U \in M_1$, $f \left({U}\right) = \left\{{0}\right\}$ which is not open in $M_2$.