Mapping to Indiscrete Space is Continuous

Theorem
Let $T_1 = \left({X_1, \vartheta_1}\right)$ be any topological space.

Let $T_2 = \left({X_2, \vartheta_2}\right)$ be the indiscrete topological space on $X_2$.

Let $\phi: X_1 \to X_2$ be a mapping.

Then $\phi$ is continuous.

Proof
From the definition of continuous:
 * $U \in \vartheta_2 \implies \phi^{-1} \left({U}\right) \in \vartheta_1$

The only elements of $\vartheta_2$ are $X_2$ and $\varnothing$, from which:


 * $\phi^{-1} \left({X_2}\right) = X_1 \in \vartheta_1$


 * $\phi^{-1} \left({\varnothing}\right) = \varnothing \in \vartheta_1$