Does removing finite points from closed set give a closed set? I have the following homework problem:
"Let $S$ be a closed set and let
$$
S_1 = \{x_1, x_2,..., x_n\} \subset S
$$
Is $S \smallsetminus S_1$ also a closed set? If true, prove the statement. If false, give a counterexample".
I think I have a counterexample to the statement, but I'm not completely sure of its validity:
Let $S =\{p\} $ a set composed of a single point. Since $S$ contains only one point, $S$ is closed. Let $S_1 =S$. Since any set is a subset of itself, then it holds true that $S_1 \subset S$. We then have:
$$
S \smallsetminus S_1 = S \smallsetminus S = \emptyset
$$
Since $ \emptyset$ is an open and closed set, in particular, it is open. Hence the statement is false.
My question is if it's valid to just assume that the empty set is in particular open and "ignoring" that it's also closed. Does this counterexample hold?
 A: No, your example doesn't work because the empty set is closed. Being open does not prevent a set from being closed, as they are not mutually exclusive conditions. However you could just remove $0$ from the interval $[0,1]$ to obtain a counterexample.
This works if you want to show it's not true in general. More generally, if there exists a one point set in the space that is not open, then the whole space minus that point is a counterexample. Thus the discrete topology is actually the only situation where the statement holds. 
A: The answer to the question depends on the considered topology.
For example, the answer is positive for the discrete topology.
It may be negative for example with the coarse topology, if $S$ is equal to the space and $\{x_1, \dots, x_n\}$ is a proper subset.
A: In $\Bbb R$ take $S = [0,1]$, which is a closed set. Then take any non-empty finite $S_1 \subseteq S$, then $S\setminus S_1$ is not closed, because all points of $S$ are limit points of $S$ and so also limit points of $S$ minus any finite set. And so $S \setminus S_1$ has limit points (the points of $S_1$!) that are not in the set, and the set is not closed.
Your example fails because $\emptyset$ is always closed.
