Reflexivity of a relation $R$ transfered to other relations

Question

Let $ R $ be a relation on $ E $. Demonstrate that:

• $ R $ is reflexive then $R \subseteq R.R$ and $R$.$R$ are reflexive too;

Here is my work though confused about few steps:

$R$.$R$ = {$(x,y) | \exists z \in E | (x,z) \in R , (z,y) \in R$}

$R$ is reflexive $\Rightarrow$ {$(x,x) | x \in E$ } If we have $x \in E, \exists z | (x,z) \in R$ and $(z, x) \in R $ Then $(x,x) \in R.R$

which implies $R \subset R.R$

and as $R$ is relfexive and $R \subset R.R$

then $R.R$ is reflexive too

Answer 1

To prove that $R \circ R$ is reflexive, we have to prove that $(x,x) \in R \circ R$, for every $x$.

But, for every $x$, we have $(x,x) \in R$, because $R$ is reflexive.

Thus, $(x,x) \in R \text { and } (x,x) \in R$, that means that:

$\exists z [(x,z) \in R \text { and } (z,x) \in R]$,

and this is: $(x,x) \in R \circ R$.

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To show that $R \subseteq R \circ R$, we have to consider $(x,y) \in R$ and use reflexivity of $R$ to have:

$(x,y) \in R \text { and } (y,y) \in R$.