Continuous Mapping on Union of Open Sets

Theorem
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.

Let $I$ be an indexing set.

Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.

Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.

Then $f$ is continuous on $C = \ds \bigcup_{i \mathop \in I} C_i$, that is, $f \restriction_C$ is continuous.

Proof
Let $V$ be an open set of $S$.

By assumption, we have that, for all $i \in I$, $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also open in $T$.

From the definition of a restriction, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.

Therefore, we can compute:

That is, $U = \paren {f \restriction_C}^{-1} \sqbrk V$ is a union of open sets.

Therefore, $U$ is itself open by definition of a topology.

It follows that $f \restriction_C$ is also continuous by the definition of continuity.

Also see

 * Continuous Mapping on Finite Union of Closed Sets for an analogous statement for closed sets.