Restriction of Continuous Mapping is Continuous

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ 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 $\left \langle {z_n} \right \rangle$ be a sequence in $S$ such that $\displaystyle \lim_{n \to \infty} z_n = \alpha$.

Since $\left \langle {z_n} \right \rangle$ and $\alpha$ both lie in $S$, $\displaystyle \lim_{n \to \infty} f \left({z_n}\right) \to \alpha$.

But $\forall n \in \N: g \left({z_n}\right) = f \left({z_n}\right)$, and also $g \left({\alpha}\right) = f \left({\alpha}\right)$.

The result follows from Limit of Image of Sequence.