Restriction of Continuous Mapping is Continuous/Metric Spaces

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $S \subseteq M_1$ be a subset of $M_1$.

Let $f: M_1 \to M_2$ be a mapping which is continuous at a point $\alpha \in S$.

Let $f \restriction_S = g: S \to M_2$ be the restriction of $f$ to $S$.

Then $g$ is continuous at $\alpha$.

Proof
Let $\sequence {z_n}$ be a sequence in $S$ such that:
 * $\ds \lim_{n \mathop \to \infty} z_n = \alpha$

Since $\sequence {z_n}$ and $\alpha$ both lie in $S$:
 * $\ds \lim_{n \mathop \to \infty} \map f {z_n} \to \alpha$

But:
 * $\forall n \in \N: \map g {z_n} = \map f {z_n}$

and also:
 * $\map g \alpha = \map f \alpha$

The result follows from Limit of Function by Convergent Sequences.