这是一份Sydney悉尼大学MATH1921的成功案例

Let $\varepsilon>0$ be given and let $n_{1}$ be large enough that if $n \geq n_{1}$,
$$
\left|a_{n}-a\right|<\varepsilon / 2 \text { and }\left|b_{n}-a\right|<\varepsilon / 2 \text {. }
$$
Then for such $n$,
$$
\left|c_{n}-a\right| \leq\left|a_{n}-a\right|+\left|b_{n}-a\right|<\varepsilon
$$
The reason for this is that if $c_{n} \geq a$, then
$$
\left|c_{n}-a\right|=c_{n}-a \leq b_{n}-a \leq\left|a_{n}-a\right|+\left|b_{n}-a\right|
$$
because $b_{n} \geq c_{n}$. On the other hand, if $c_{n} \leq a$, then
$$
\left|c_{n}-a\right|=a-c_{n} \leq a-a_{n} \leq\left|a-a_{n}\right|+\left|b-b_{n}\right|
$$
This proves the theorem.
As an example, consider the following.

MATH1921 COURSE NOTES ：

Here is why. Let $x \in D(z, r)$. Then $|x-z| \leq r$ and so
$$
|x| \leq|x-z|+|z| \leq r+|z|<2 r+|z| . $$ It remains to verify it is closed. Suppose then that $y \notin D(z, r)$. This means $|y-z|>r$. Consider the open ball $B(y,|y-z|-r)$. If $x \in B(y,|y-z|-r)$, then
$$
|x-y|<|y-z|-r $$ and so by the triangle inequality, $$ |z-x| \geq|z-y|-|y-x|>|x-y|+r-|x-y|=r
$$
Thus the complement of $D(z, r)$ is open and so $D(z, r)$ is closed.
For the second type of set, if $x+i y \in[a, b]+i[c, d]$, then
$$
|x+i y|^{2}=x^{2}+y^{2} \leq(|a|+|b|)^{2}+(|c|+|d|)^{2} \equiv r / 2 .
$$