Function on the set of relations: $f(R)=RK$ where $K$ is a fixed relation

Question

$\ M$ is the set of all relations on $\ A = \{1,2,3\}$

$\ K$ is the following relation on A

$\ K=\{(1,1),(2,1),(3,1)\}$

let there be $\ f :M\rightarrow M$

$\ f(R) = RK$

1. prove that R$\subseteq$K$\rightarrow$f(R)=R

2. prove that for every $\ R\in M$

   $\ f(R)\subseteq K$

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1. I wrote

   (a,b)$\in$R $\rightarrow$ 1=b$\in$Range(K)={1} (because R$\subseteq$K) $\rightarrow$ (a,1)$\in$R , (1,1)$\in$K $\rightarrow$(a,1)$\in$f(R)=RK$\rightarrow$(a,b)$\in$f(R)

   but I don't know if it's good enough and also I'm struggling to show it the other way as well (which is needed)

2. I'm thinking about something like

   R$\subseteq$AXA

   f(AXA)$\subseteq$K

   I can't put the pieces together but I think there is a way that I can say that

   R$\subseteq$K

can you please direct me to the correct answers? I'm stuck and have been thinking about this over a week

Answer 1

For the first part, you have the good intuition :

Let $(a,b) \in R$, we know that $b=1$ since $R \subset K$, and since $\{ (b,1) \mid b \in A \} \subset K$ then $(b,1) \in K$ and by definition $(a,b) \in RK = f(R)$. So $R \subset f(R)$.

Conversly, if $(a,c)$ is in $f(R)$, then $\exists b, (b,c) \in K,(a,b) \in R$ and it implies $b=c=1$, and so $(a,c)=(a,b)=(a,1) \in R$, hence $f(R) \subset R$ and finally $f(R)=R$.

2) This is not true in general that $R \subset K$, since $R$ can be any relation on $M$. But it is true for its image $f(R)$, since by definition $(a,c) \in f(R) \iff \exists b,(a,b) \in R,(b,c) \in K$. But this imposes $c=1$. So all the couple of $f(R)$ have the form $(a,1)$ and are hence (since you can easily see as I previously said that $\{ (a,1) \mid a \in A \} \subset K$) also in $K$, id est $f(R) \subset K$