A relation $R$ has a "chain" that connects $a$ to $b$ if there exists some sort of
$$(a, x_0),(x_0, x_1),\cdots,(x_{n-1}, x_n),(x_n,b)$$
made out of the elements in $R$. Why doesn't there exist a relation with an infinitely long chain?
My first idea to create a counterexample started with $(0,1)$, and repeatedly replacing $(0,a)$ with $(0,a/2),(a/2,a)$, doing so infinitely many times. Then there exists an infinitely long chain from $0$ to $1$, but then you can no longer tell what $0$ is related to so it breaks down.
The question arose after seeing my book's definition for transitive closure, which appears to imply that such chains are always finite.
For any $2$ binary relations $R$ and $R'$ on a set $S$ we define the composition of $R$ and $R'$ as the binary relation $R\circ R'$ on $S$ such that for any $x,y\in S$ we have
$$(x,y)\in R\circ R'\Leftrightarrow\exists z\in S. (x,z)\in R'\land(z,y)\in R$$
Let $R^1=R$ and define the $m^{th}$ iterate of $R$ as $R^m=R\circ R^{m-1}$ for any integer $m>1$. Furthermore, define the transitive closure of R by
$$Tran(R)=\bigcup_{m\in\mathbb{N}}R^m$$