这是一份warwick华威大学MA222-10 的成功案例

By the LUB axiom, there exists $\ell \in \mathbb{R}$ which is sup $s$. We clamm that $\lim x_{n}=\ell$. Let $\varepsilon>0$ be given. As $\ell$ is an upper bound for $S$ and $x_{N}-\varepsilon / 2 \in S$ (by (i)) we infer that $x_{N}-\varepsilon / 2 \leq \ell$. Since $\ell$ is the least upper bound for $S$ and $x_{N}+\varepsilon / 2$ is an upper bound for $S$ (from (ii)) we see that $\ell \leq x_{N}+\varepsilon / 2$. Thus we have $x_{N}-\varepsilon / 2 \leq \ell \leq x_{N}+\varepsilon / 2$ or
$$
\left|x_{N}-\ell\right| \leq \varepsilon / 2
$$
For $n \geq N$ we have
$$
\begin{aligned}
\left|x_{n}-\ell\right| & \leq\left|x_{n}-x_{N}\right|+\left|x_{N}-\ell\right| \
&<\varepsilon / 2+\varepsilon / 2=\varepsilon
\end{aligned}
$$
We have thus shown that $\lim {n \rightarrow \infty} x{n}=\ell$.

MA222-10  COURSE NOTES ：

Let $A$ be a nonempty subset of a metric space $(X, d)$. Define
$$
d_{A}(x):=\inf {d(x, a): a \in A}, \quad x \in X .
$$
Then $d_{A}$ is continuous. (Geometrically, we think of $d_{A}(x)$ as the distance of $x$ to $A$.)

We give a proof even though it is easy, because of the importance of this result. Let $x, y \in X$ and $a \in A$ be arbitrary. We have, from the triangle inequality $d(a, x) \leq d(a, y)+d(y, x)$,
$$
\begin{aligned}
&d(a, y) \geq d(a, x)-d(y, x) \
&d(a, y) \geq d_{A}(x)-d(y, x)
\end{aligned}
$$
since $d(a, x) \geq \inf \left{d\left(a^{\prime}, x\right): a^{\prime} \in A\right}:=d_{A}(x)$. The inequality says that $d_{A}(x)-d(y, x)$ is a lower bound for the set ${d(a, y): a \in A}$. Hence the greatest lower bound of this set, namely, $\inf {d(a, y)=a \in \backslash A} n d$ is greater than or equal to this lower bound, that is,
$$
d_{A}(y) \geq d_{A}(x)-d(y, x)
$$