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 $$\lim_{n \to \infty} z_n = \alpha$$.

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

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.