# Removing finite points from closed set.

I have the following homework problem:

"Let $$S$$ de 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 it's 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 subset of itself then it holds 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?

• $\emptyset$ is closed. – Lord Shark the Unknown Oct 30 '19 at 19:15
• This depends on the topology. – Bernard Oct 30 '19 at 19:19
• On which topology are you working? – mathcounterexamples.net Oct 30 '19 at 19:19

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.

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.

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.

Recall that in order for a set to be closed, it must contain all of it's accumulation points.

Well, a point $$x$$ of a set $$A$$ is said to be an accumulation point if for any neighborhood of $$x$$, the neighborhood contains at least one other point in the set $$A$$.

We know that any finite set is clearly not infinite. So, you can generate an extremely small neighborhood about any point in a finite set and get infinitely many points not in the original set.

Your question doesn't specify if this set is infinite or finite. Assuming finite $$S$$, then yes, any $$S\setminus S_1$$ would contain no accumulation points and be closed.

If $$S$$ is infinite, closed, and you remove any part it, you're removing accumulation points and thus ensuring the set is no longer closed. However, this is only true unless the infinite set has no accumulation points to begin with.

Another problem arises with the topology being used here. It's not specified, but in a general case this is the structure.

• There are infinite closed subsets of $\mathbb R$ with no accumulation points, such as the set of integers, where you can remove any finite subset and have it remain closed. – Matt Samuel Oct 31 '19 at 2:03
• It also need not be true for finite sets if points are not necessarily closed. – Matt Samuel Oct 31 '19 at 2:04
• @Matt Samuel Completely slipped my mind. That's entirely true. – help Oct 31 '19 at 3:21