Transitivity of chained inclusion

Statement

Let $A$, $B$, $C$, $D$ be sets such that $A \subseteq B$, $B \subseteq C$ and $C \subseteq D$. Prove that $A \subseteq D$.

Solution

Let $x \in A$ be arbitrary.

• From $A \subseteq B$: $x \in B$.
• From $B \subseteq C$: $x \in C$.
• From $C \subseteq D$: $x \in D$.

Hence every element of $A$ belongs to $D$, i.e. $A \subseteq D$. (We used transitivity of inclusion repeatedly.) $\blacksquare$