and also,
$$
x_{n}=y_{n}-\frac{1}{n}=\alpha+\frac{m_{n}-1}{n} \leq x \quad \text { for some } x \in A
$$
Hence,
$$
x_{m} \leq y_{n} \quad \text { for all } m, n \in \mathbf{N} .
$$
It follows that, for any $m, n \in \mathbf{N}$, we have
$$
x_{n}-x_{m} \leq y_{m}-x_{m}=\frac{1}{m} \text { and } x_{m}-x_{n} \leq y_{n}-x_{n}=\frac{1}{n} .
$$
MATH3961 COURSE NOTES ：
$x \in 5$
If $f$ and $g$ belong to $\mathcal{B}(S)$, there exist $M>0$ and $N>0$ such that
$$
\sup {x \in S}|f(x)| \leq M \text { and } \sup {x \in S}|g(x)| \leq N .
$$
It follows that $\sup {x \in S}|f(x)-g(x)|<\infty$. Indeed, $$ |f(x)-g(x)| \leq|f(x)|+|g(x)| \leq \sup {x \in S}|f(x)|+\sup {x \in S}|g(x)|, $$ and so $$ 0 \leq \sup {x \in S}|f(x)-g(x)| \leq M+N .
$$
Define
$$
d(f, g)=\sup _{x \in S}|f(x)-g(x)|, \quad f, g \in \mathcal{B}(S) .
$$