Linear order cardinality.

Question

Could you please help me with this question from elementary set theory exam.

Suppose R is linear order on some unfinite set A.

Prove that $|R|=|A|$.

My attempt:

$R\subseteq A\times A$, so $|R|\le|A|$.

How to prove that $|R|\ge|A|$ ?

I suppose that there exists some bijectional function from R onto A, but how to construct one ?

Edit: R is non-reflexive,transitive and complete order on A, $|A|\ge\aleph_0$.

Thank you.

Answer 1

Fix $a \in A$. As $R$ is a linear order, it contains either $(x, a)$ or $(a,x)$ for all $x \in A$.