Union of contractions is a contraction

Question

Can someone give me a proof of the following statement:

Let $(X,d)$ be a metric space. If $w_{i} : D_{i} \rightarrow X$ , $D_{i} \subset X$, are contraction mappings with corresponding contractivity factors $s_{i}, i =1,...,n$, then $W= \cup_{i=1}^n w_{i}$ is a contraction mapping with contractivity factor $s= \max\limits_{i=1,...,n}s_{i}$.
( $W(A) = \cup_{i=1}^n w_{i}(A)$, $A \in X$)

Thanks in advance.

Answer 1

As stated, the theorem is nonsensical. If $w_i = w_j$ in the intersection of the domains, the union of functions makes sense. But without this hypothesis, if the definition is $$W(A) = \cup_{i=1}^n w_{i}(A)$$ for $A\subset X$ (not $A\in X$), the domain isn't $\bigcup_{i=1}^n D_i$.

EDIT: the original source (Fractals Everywhere) abominably quote in the linked paper: