Prove that if $A \subset B$ then $R[A] \subset R[B]$ I have a question saying:

Prove that if $A \subseteq B$, then $R[A] \subseteq R[B]$

I am having some trouble getting a rigorous proof to me. Intuitively I can see how this is true, but I am struggling to prove this rigorously. So far I have:
$\forall x \in A$ it is true that $x \in B$. This implies that $(\forall y\in R[A]) x \in B$, as $(\forall y \in R[x] |x \in A)$ it is also true that $x\in B$. Therefore, $R[A] \subseteq R[B]$
I am aware there are probably many gaps in my logic but I am not sure where. Any hints or help would be greatly appreciated.
EDIT:
Thanks for your help. Is this now correct, or is there still more to do:
$\forall y \in R[A]$ there exist an $x$ such that $_xR_y$. This implies that $\forall y \in R[A]$ there exists an $x \in B$, as $\forall x\in A$ it is true that $x\in B$. Therefore, $y \in B$, and so $R[A] \subseteq R[B]$
 A: Equivalent are:


*

*$y\in R[A]$

*$\langle x,y\rangle\in R$ for some $x\in A$


Since $A\subseteq B$ the second bullet implies that $\langle x,y\rangle\in R$ for some $x\in B$ or equivalently $y\in R[B]$.
Proved is not that $y\in R[A]\implies y\in R[B]$ or equivalently $R[A]\subseteq R[B]$.
A: I think you're misguided by all those $\forall$ symbols.


*

*$X$ is a set, $R$ is a relation on $X$;

*if $C$ is a subset of $X$, then $R[C]=\{y\in X: x\mathrel{R}y\text{, for some } x\in C\}$

*$A$ and $B$ are subsets of $X$.



Prove that, if $A\subseteq B$, then $R[A]\subseteq R[B]$.

Suppose $y\in R[A]$. Then there is $x\in A$ such that $x\mathrel{R}y$. Since $A\subseteq B$, we can say that $x\in B$. Therefore, by definition, $y\in R[B]$, because $x\mathrel{R}y$ and $x\in B$.
A: First off, to do this proof, you need to know and use what $\;y \in R[X]\;$, how it is defined.$%
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\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
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%$ Why is that?  Because of the definition of $\;\subseteq\;$, which tells us that $$
\tag{0}
R[A] \subseteq R[B] \;\equiv\; \langle \forall y :: y \in R[A] \;\then\; y \in R[B] \rangle$$so you need to know what $\;y \in R[X]\;$ means.
Since you say in a comment that $\;R[A]\;$ is "the image of the set $\;A\;$ under the relation $\;R\;$", presumably your definition is something like
$$
\tag{1}
y \in R[A] \;\equiv\; \langle \exists x :: x \in A \land (x,y) \in R \rangle
$$
for every $\;y,R,A\;$.
Now we need to prove the right hand side of $\Ref{0}$, assuming $\;A \subseteq B\;$. Starting on the left of $\;\then\;$ and working towards the other side, we have for every $\;y,R,A,B\;$
$$\calc
    y \in R[A]
\op\equiv\hint{definition $\Ref{1}$}
    \langle \exists x :: x \in A \;\land\; (x,y) \in R \rangle
\op\then\hints{$\;x \in A \then x \in B\;$ by our assumption -- the only}\hint{way to use the only thing we know about $\;A\;$}
    \langle \exists x :: x \in B \;\land\; (x,y) \in R \rangle
\op\equiv\hint{definition $\Ref{1}$}
    y \in R[B]
\endcalc$$
And that completes the proof.
$%
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%$
