Clarify if $A⊆\mathcal R^{-1}\big[\mathcal R[A]\big]$ and if $\mathcal R\big[\mathcal R^{-1}[B]\big]=B\cap\text{ran }\mathcal R$ for any relation

Question

Definition

A set $\mathcal R$ is a binary relation if any its element is an ordered pair. In particular a function $\mathcal f$ is a binary relation such that $$ (x,y_1),(x,y_2)\in\mathcal f\Leftrightarrow y_1=y_2 $$ for any $x$.

Definition

Given a relation $\mathcal R$ the subset $\text{dom }\mathcal R$ of $\bigcup\big(\bigcup\mathcal R\big)$ whose elements are the first coordinates of $\mathcal R$ is called the domain of $\mathcal R$.

Definition

Given a relation $\mathcal R$ the subset $\text{ran}\;\mathcal R$ of $\bigcup\big(\bigcup\mathcal R\big)$ whose elements are the second coordinates of $\mathcal R$ is called the range of $\mathcal R$.

So with this definitions is not hard to show that $$ \mathcal R\subseteq\text{dom }\mathcal R\times\text{ran }\mathcal R $$ so that we can give the following definition.

Definition

If $\mathcal R$ is a relation then the image $\mathcal R[A]$ of $A$ and the preimage $\mathcal R^{-1}[B]$ of $B$ under $\mathcal R$ are the sets $$ \mathcal R[A]:=\{y\in\text{ran }\mathcal R:x\mathcal Ry\,\text{for some }x\in A\}\\ \mathcal R^{-1}[B]:=\{x\in\text{dom }\mathcal R:x\mathcal Ry\,\text{for some }y\in B\} $$

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Now if $\mathcal f$ is a function then it is a well-known result that $$ A\subseteq\mathcal f^{-1}\big[\mathcal f[A]\big]\,\,\,\text{and}\,\,\,\mathcal f\big[\mathcal f^{-1}[B]\big]=B\cap\text{ran }\mathcal f $$ for any $A\subseteq\text{dom }\mathcal f$ and for any other set $B$ so that I ask myself if this these results hold replacing $\mathcal f$ with a general binary relation $\mathcal R$.

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1. So first of all I observed that if $A\subseteq\text{dom }\mathcal R$ then there exist $b\in\bigcup\big(\bigcup\mathcal R\big)$ such that $a\mathcal R b$ and necessarly $b\in\mathcal R[A]$ because $b$ is surely an element of $\text{ran }\mathcal R$. So I concluded that the inclusion $$ A\subseteq\mathcal R^{-1}\big[\mathcal R[A]\big] $$ holds for any $A\subseteq\text{dom }\mathcal R$, provided that the argumentations I gave are correct.
2. Now if $y$ is an element of $B\cap\text{ran }\mathcal f$ then there exists $x\in\text{dom }\mathcal R$ such that $x\mathcal R y$ and surely this such $x$ is an element of $\mathcal R^{-1}[B]$ by definition; on the other hand for this such $x$ there exist an element (precisely $y$) of $\text{ran }f$ related with $x$ through $\mathcal R$ so that $y$ is an element of $\mathcal R\big[\mathcal R^{-1}[B]\big]$ but we initially assume that $y$ is an element of $B\cap\text{ran }\mathcal R$ so that we can conclude that the inclusion $$ B\cap\text{ran }\mathcal R\subseteq\mathcal R\big[\mathcal R^{-1}[B]\big] $$ holds.

Now unfortunately I was not able to prove that the inverse inclusion holds: indeed conversely if $y$ is an element of $\mathcal R\big[\mathcal R^{-1}[B]\big]$ then there exist $x\in\mathcal R^{-1}[B]$ such that $x\mathcal R y$ and this last element $x$ is such that $x\mathcal Ry'$ for any $y'\in B$ so that if the relation $\mathcal R$ was a function then surely $y'=y$ and so we could conclude that $$ \mathcal R\big[\mathcal R^{-1}[B]\big]\subseteq B\cap\text{ran }R $$ since obviously $y'$ is an element of $\text{ran }R$. So by the last argumentation I suspect that the identity $$ \mathcal R\big[\mathcal R^{-1}[B]\big]=B\cap\text{ran }\mathcal R $$ does not hold when the relation $\mathcal R$ is not a function but unfortunately I was not able to elaborate a counterexample so that I thought to pay a specific question where I ask to clarify this and where I ask also if the argumentations I gave to prove that $$ A\subseteq\mathcal R^{-1}\big[\mathcal R[A]\big] $$ are correct.

So could someone help me, please?

Answer 1

The reasoning for

$$A \subseteq R^{-1}\left[R[A]\right] \text{ if } A \subseteq \operatorname{dom}(R)$$

is quite correct: recap let $x \in A$. That $A \subseteq \operatorname{dom}(R)$ gives us some $y$ so that $(x,y) \in R$. By definition $y \in R[A]$ and again by definition $x \in R^{-1}\left[R[A]\right]$. QED.

(notational side-remark : I always use $R$ for a relation, capitals like $A,B$ for subsets, and $x,y$ for elements; It's easier on the typing and my eyes. When I see $\mathcal R$ in this context I go up a type and think that it's a set of relations, not just one. That's my mathematical upbringing, I suppose; even in high school they do it like that)

Also, $$B \cap \operatorname{ran}(R) \subseteq R\left[R^{-1}[B]\right]$$ also holds quite easily:

Let $y \in B \cap \operatorname{ran}(R)$. So $y \in B$ and for some $x$ we have $(x,y) \in R$. So by definition $x \in R^{-1}[B]$ (as witnessed by $y \in B$) and again by definition $y \in R\left[R^{-1}[B]\right]$, as now witnessed by $x$. QED.

The reverse does not hold as you suspected. Trivial almost minimal example:

Let $R=\{(1,1),(1,2),(2,1),(2,2)\}$ then $\operatorname{ran}(R)=\{1,2\}$ and taking $B=\{2\}$ we see that $R^{-1}[B]=\{1,2\}$ and $R[R^{-1}[B]]=\{1,2\}$ and we have inequality of the latter set and $B \cap \operatorname{ran}(R)= \{2\}$...