Given two nonempty closed sets $C$ and $D$, the set $C-D$ is defined by $$ C-D=\{x-y|x \in C, y \in D\}. $$ Is the set $C-D$ closed? Here is my proof:
Take $\{x_i\} \subseteq C$ satisfy $x_i \to x$, and $\{y_i\} \subseteq D$ satisfy $y_i \to y$, then we have $z_i=x_i-y_i \in C-D$ and $z_i \to x-y \in C-D$, thus, we can conclude that $C-D$ is closed. I know the conclusion is not right, but I do not know where I make some mistakes?
In order to argue that $C - D$ is (sequentially) closed, you would need to start with some sequence $(z_n)_{n \in \mathbb{N}}$ converging to some $z$ such that $\forall n \in \mathbb{N} \ z_n \in (C - D)$, and then try to conclude that $z \in (C - D)$. You can then indeed write each $z_n$ as $z_n = x_n - y_n$ such that $x_n \in C$ and $y_n \in D$.
If $(x_n)_{n \in \mathbb{N}}$ would converge, then so would $(y_n)_{n \in \mathbb{N}}$, the respective limits would belong to $C$ and $D$ respectively, the difference of the limits would be $z$, which then belongs to $C - D$.
However, it is possible that neither $(x_n)_{n \in \mathbb{N}}$ nor $(y_n)_{n \in \mathbb{N}}$ converges; and that is where we get stuck.
