Definition: Given $x,$ $y\in X$ and $\alpha>0,$ $x$ is $\alpha$ -related to $y$ (written $x\stackrel{\alpha}{\sim} y$ ) if there are $\alpha$ -pseudo-orbits of $f$ such that $x_{0}=x,$ $x_{1},$ $\cdots,$ $x_{k}=y$ and $y_{0}=y,$ $y_{1},$ $\cdots,$ $y_{l}=x$ . If $x\stackrel{\alpha}{\sim} y$ for every $\alpha>0$ , then $x$ is related to $y$ (written $x\sim y$ ). The chain recurrent set of $f,$ $R$ is $\{x\in X:x\sim x\}$.
My question: If $x\sim x$, then for each $\alpha>0$ there exists a sequence of points $x_0,x_1,\cdots,x_k=x$ such that $d(f(x_i),x_{i+1})<\alpha.$ Hence, $d(f(x),x)<\alpha$. Since, $\alpha>0$ was arbitrary we have $d(f(x),x)=0.$ Thus, $f(x)=x.$ So isn't the chain recurrent set of $f$ just the set of fixed points?
When you wrote out the meaning of $x \sim x$ under the heading of "My question", you made two mistakes which, if fixed, should clear this up.
First, the sequence of points should be written $x=x_0,x_1,...,x_k=x$ (you left out the initial $x=x_0$).
Second, the distance condition is that $d(f(x_i),x_{i+1}) < \alpha$ for each $i=0,...,k-1$ (you left out the quantifier "for each...").
So we have $$d(f(x),x_1) = d(f(x_0),x_1) < \alpha $$ and so on as $i$ increases, up to $$d(f(x_{k-1}),x) = d(f(x_{k-1}),x_k) < \alpha $$ but none of this implies $d(f(x),x) < \alpha$.