Equivalence of Definitions of Continuity on Metric Spaces

Theorem
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 $$A_1$$.

The definitions for continuity of $$f$$ from $$A_1$$ to $$A_2$$ are equivalent.

That is, the following definitions say exactly the same thing:

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

Open Set Definition
$$f$$ is continuous (with respect to the metrics $$d_1$$ and $$d_2$$) iff:


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

Proof

 * Suppose that $$f$$ is defined to be continuous (in the sense of the definition sing limits) for all $$a \in A_1$$.

Let $$U \subseteq M_2$$ be open in $$M_2$$.

Let $$x \in f^{-1} \left({U}\right)$$.

Since $$U$$ is open in $$M_2$$, $$\exists \epsilon > 0: \forall N_\epsilon \left({f \left({x}\right)}\right) \subseteq U$$.

By hypothesis, $$\exists \delta > 0: f \left({N_\delta \left({x}\right)}\right) \subseteq N_\epsilon \left({f \left({x}\right)}\right)$$.

So $$f \left({N_\delta \left({x}\right)}\right) \subseteq U$$ and so $$N_\delta \left({x}\right) \subseteq f^{-1} \left({U}\right)$$.

Thus $$\subseteq f^{-1} \left({U}\right)$$ is open in $$M_1$$.


 * Now suppose $$f$$ is defined to be continuous (in the sense of the Open Set Definition).

Let $$\epsilon > 0$$.

Then by Neighborhood of Point Inside Neighborhood, $$N_\epsilon \left({f \left({x}\right)}\right)$$ is open in $$M_2$$.

So by hypothesis, $$f^{-1} \left({N_\epsilon \left({f \left({x}\right)}\right)}\right)$$ is open in $$M_1$$.

As $$f \left({x}\right) \in N_\epsilon \left({f \left({x}\right)}\right)$$, and therefore $$x \in f^{-1} \left({N_\epsilon \left({f \left({x}\right)}\right)}\right)$$.

So by hypothesis $$\exists \delta \ge 0: N_\delta \left({x}\right) \subseteq f^{-1} \left({N_\epsilon \left({f \left({x}\right)}\right)}\right)$$.

Then $$f \left({N_\delta \left({x}\right)}\right) \subseteq N_\epsilon \left({f \left({x}\right)}\right)$$.

Thus $$f$$ is continuous at $$x$$ in the sense of the Epsilon-Neighborhood Definition.