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;
 * $$\lim_{x \to a} f \left({x}\right) = f \left({a}\right)$$.

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 $$f: \R^2 \to \reals$$ 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 $$M_2$$.