Proof of continuity via open sets in metric spaces

Proof

We prove each implication separately.

1. Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and let $f \colon X \to Y$.
2. Write $B_X(a, \delta)$ and $B_Y(b, \varepsilon)$ for the open balls of $X$ and $Y$. By the symmetry of a metric, $x \in B_X(a, \delta)$ if and only if $d_X(x, a) < \delta$.
3. Assume condition 1. Let $V$ be an open set of $Y$, and let $a$ be in the preimage $f^{-1}(V)$.
4. Then $f(a) \in V$. By the description of the open sets of a metric space, there is $\varepsilon > 0$ with $B_Y(f(a), \varepsilon) \subseteq V$.
5. Condition 1 gives $\delta > 0$ such that $d_X(x, a) < \delta$ implies $d_Y(f(x), f(a)) < \varepsilon$.
6. Let $x \in B_X(a, \delta)$. By step 2 and step 5, $d_Y(f(x), f(a)) < \varepsilon$.
7. By symmetry, $f(x) \in B_Y(f(a), \varepsilon) \subseteq V$, so $x \in f^{-1}(V)$. Thus $B_X(a, \delta) \subseteq f^{-1}(V)$.
8. Each point of $f^{-1}(V)$ has a ball about it inside $f^{-1}(V)$. By the description of open sets, $f^{-1}(V)$ is open. This is condition 2.
9. Now assume condition 2. Let $a \in X$ and $\varepsilon > 0$. Put $V = B_Y(f(a), \varepsilon)$.
10. By open balls are open, $V$ is open. By condition 2, $f^{-1}(V)$ is open in $X$.
11. Since $d_Y(f(a), f(a)) = 0 < \varepsilon$, we have $f(a) \in V$, so $a \in f^{-1}(V)$.
12. By the description of open sets, there is $\delta > 0$ with $B_X(a, \delta) \subseteq f^{-1}(V)$.
13. Let $x \in X$ with $d_X(x, a) < \delta$. By step 2, $x \in B_X(a, \delta)$, so $f(x) \in V$.
14. So $d_Y(f(a), f(x)) < \varepsilon$, and by symmetry $d_Y(f(x), f(a)) < \varepsilon$. This is condition 1.