Mapping from Discrete Space is Continuous

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ be the discrete topological space on $S_1$.

Let $T_2 = \struct {S_2, \tau_2}$ be any other topological space.

Let $\phi: S_1 \to S_2$ be a mapping.

Then $\phi$ is continuous.

Proof
From the definition of continuous:
 * $U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$

But as $\phi^{-1} \sqbrk U \subseteq S_1$ it follows from the definition of discrete space that $\phi^{-1} \sqbrk U \in \tau_1$.