Composite Relations - reflexive, antisymmetric, and transitive

Question

I think I may not be understanding the composition of a relation, especially when it is composed with itself. Here is an example I am struggling with and my reasoning...

Prove or give a counterexample to each of the following statements:

1. If $R$ is a reflexive relation on $A$, then $ R \circ R$ is a reflexive relation on A.

   Let $a∈A$ and $R = f(a)$
   Since R is reflexive we know that $\forall a∈A \,\,\,,\,\, \exists (a,a)∈R$ then $f(a)=(a,a)$
   Since $R \circ R = f(f(a))$ and $f(a)=(a,a)\implies f(f(a))=f((a,a))$

This is where I get confused with this proof. I can't seem to wrap my head around this composed relation. I know there must be something I am not understanding.

Answer 1

Yes, that is an awkward "proof".

Is this clearer?:

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For any relations $S\subseteq X\times Y, T\subseteq Y\times Z$, the definition of composition is that: $$(x,z)\in T\circ S ~\iff ~\exists y\in Y~((x,y)\in S\wedge (y,z)\in T)$$

So clearly we have that for any $a\in A$ we have $(a,a)\in R\circ R$, if $(a,a)\in R$.

Now, $R$ is reflexive on $A$, iff $\forall a\in A: (a,a)\in R$.

Apply universal modus ponens.

$$\begin{align}&\forall a\in A:((a,a)\in R) && \text{assuming $R$ is reflexive}\\& \forall a\in A:((a,a)\in R\to (a,a)\in R\circ R) && \text{from the definition of composition}\\ \hline &\forall a\in A:((a,a)\in R\circ R) && \text{conclude $R\circ R$ is reflexive}\end{align}$$

So therefore $R\circ R$ is reflexive on $A$, if $R$ is reflexive on $A$.

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[Remark: note the swap of order in the definition of composition.   Well, it is not relavant for this, but worth remembering.]

Answer 2

Because $R$ is reflexive then $aRa$, for all $a \in A$. By definition $$R \circ R=\{(x,z) \in A: \exists y \in A\: | \: xRy \: \wedge \: yRz\}.$$

Let $x \in A$, then since $xRx$ then $xRx \wedge xRx$ is a true statement. Therefore $x(R \circ R)x$. Hence $R \circ R$ is reflexive.