# Is the following set inclusion true?

Let $f:D\rightarrow C$ be any injective function. Consider $D_1\subseteq D_2 \subseteq D$ and denote $C_1=\{f(d)|~ d \in D_1\}$, $C_2=\{f(d)|~ d \in D_2\}$. Is it true that $C_1 \subseteq C_2$?

• Yes.${}{}{}{}{}$ – user228113 Mar 3 '18 at 15:57
• I think it should be trivially true, but I'd like to check. – pulosky Mar 3 '18 at 15:57

The statement is true, and it would be true even if $f$ is not an injective function. It is easy to see that if $x \in C_1$, then there exists a $d \in D_1$ such that $x = f (d)$. Now since $D_1 \subseteq D_2$, we have $d \in D_2$ as well. Therefore $x = f (d) \in C_2$. Since $x \in C_1$ is arbitrary, we have proved that $D_1 \subseteq D_2$.

It is true. You can prove it like any other set inclusion: to prove that $C_1 \subseteq C_2$, take an element $c \in C_1$ and show that $c \in C_2$.

To see this, note that if $c \in C_1$, then $c=f(d)$ for some $d \in D_1$. Since $D_1 \subseteq D_2$, it follows that $d \in D_2$. But then $c=f(d)$ for some $d \in D_2$, which says precisely that $c \in C_2$.

Yes, $$x\in C_1=\{f(d)|~ d \in D_1\} \implies x=f(d)$$

 for some $d\in D_1$

Since $D_1 \subseteq D_2$ we also have $d\in D_2$

Thus $$x=f(d) \in C_2$$ Therefore $C_1 \subseteq C_2$