# Prove that any finite set is closed

## Homework Statement

As the title says

## Homework Equations

Definitions of "open" and "closed"

## The Attempt at a Solution

Suppose a finite set S is not closed. Then Sc is not open, and there exists an element x of Sc, so that for all µ > 0, either x + u, or x - u, is an element of S.

Hmmm ... Where do I go from here? I need a springboard.

Your problem is not well posed. To be able to say that a set is closed you need to specify a topology. Every finite subset of a general topological space?

Of course one can argue that it does not really matter, that it will be closed in any "reasonable" topology, but still this should have been precised.

So, what was the exact statement of your problem? What was the context? The proof will depend on this precise statement.

P.S. The statement may not be true in a "trivial topology", when the only open (an therefore also closed) sets are the space itself and the empty set.

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Your problem is not well posed. To be able to say that a set is closed you need to specify a topology. Every finite subset of a general topological space?

Of course one can argue that it does not really matter, that it will be closed in any "reasonable" topology, but still this should have been precised.

So, what was the exact statement of your problem? What was the context? The proof will depend on this precise statement.

P.S. The statement may not be true in a "trivial topology", when the only open (an therefore also closed) sets are the space itself and the empty set.

That is exactly what the problem says. It's Problem 4 from Page 516 of Taylor and Mann's Advanced Calculus

But then the context is that it is a set of points on the real line, with the standard epsilon-neighborhoods. I suggest you first prove that a one-point set is closed.

But then the context is that it is a set of points on the real line, with the standard epsilon-neighborhoods. I suggest you first prove that a one-point set is closed.

Hmmmm ...

Let S={a}. Then Sc={b: ba}.

Because ab, there must exist µ such that b + µ = a.

b + µ is an element of S, but since there is only one of S, b - µ is an element of Sc.

Let ß = µ/2, and it is obvious that both b + ß and b - ß are elements of Sc.

(Horrible, I know.)

Now where do I go from here?

What about a simpler proof, like this one:

Take b in the complement of S. Then b is different from a. Let d=|a-b| be the distance from a to b. Then d/2 neighborhood of b is evidently contained in the complement of S.

Now, if you understand the above, you should have no difficulty proving that the set of two points is also closed.

But then you will see how it works for n-points as well.

Moral: Don't overcomplicate the problem! Most problems in math are a lot easier than they look, especially when they come from textbooks, so it is advisable to think in different ways rather than follow one set path.

I think you should have given a chance for the OP to find the solution. You have taken out all the fun and happiness of a discovery.

What about a simpler proof, like this one:

Take b in the complement of S. Then b is different from a. Let d=|a-b| be the distance from a to b. Then d/2 neighborhood of b is evidently contained in the complement of S.

Now, if you understand the above, you should have no difficulty proving that the set of two points is also closed.

But then you will see how it works for n-points as well.

I see, I see. Is this sort of the idea?

Let S={a}. If y is an element of Sc, then ya, and there exists a single µ0 such that y+ µ = a. But y + [I ]µ/2[/I] ≠ a; so y is not an accumulation point of S, and all S's accumulation points must be contained in the set itself. Thus S is finite. In fact, any finite set is composed of a number of single-element sets like S, and therefore has no accumulation points outside itself, and therefore is closed.

Probably you proof would fly, but it seems not to be the optimal one. Why? Because it can be made simpler. You know the definition of the open set and of the closed set. That is all you need. Take S to be a finite set: S={a1,...,an}. Take any point a that is not in S. Let {d1,...,dn} be the set of distances |a-an|. They are all positive since a is different from each of the points a1,...,an. Let d be the smallest of these n numbers. Then the set

a-d<x<a+d

is also in the complement of S. Thus the complement of S is open, thus S is closed. You do not even need d/2 neighborhood, d is enough. You do not need to bring in the concept of accumulation points, which comes later. What you need are just definitions.
Drawing a picture usually helps in such proofs.

HallsofIvy
Homework Helper
What about a simpler proof, like this one:

Take b in the complement of S. Then b is different from a. Let d=|a-b| be the distance from a to b. Then d/2 neighborhood of b is evidently contained in the complement of S.

Now, if you understand the above, you should have no difficulty proving that the set of two points is also closed.

But then you will see how it works for n-points as well.
You are assuming a metric space- and the OP has not said that we are working in a metric space.

Jamin2112, you started by asserting that "Closed" and "Open" are relevant definitions but did NOT give those definitions! You are aware, are you not, that there are a number of different definitions depending upon exactly what topology you are working with- which is what arakajad asked in his first response- and you did not answer.

OP explicitly quoted "Advanced Calculus" by Taylor and Mann, giving even page number there. So I adapted my answer to what has learned so far and to what notation is being used there. I think this is the best way of helping. So I decided that there is no point of diverging into the subject of different types of topological spaces when we know the exact textbook and the exact page that the student has the problem with. This particular textbook introduces topology through what it calls "point sets" of the real line and the standard distance function there.

Probably you proof would fly, but it seems not to be the optimal one. Why? Because it can be made simpler. You know the definition of the open set and of the closed set. That is all you need. Take S to be a finite set: S={a1,...,an}. Take any point a that is not in S. Let {d1,...,dn} be the set of distances |a-an|. They are all positive since a is different from each of the points a1,...,an. Let d be the smallest of these n numbers. Then the set

a-d<x<a+d

is also in the complement of S. Thus the complement of S is open, thus S is closed. You do not even need d/2 neighborhood, d is enough. You do not need to bring in the concept of accumulation points, which comes later. What you need are just definitions.
Drawing a picture usually helps in such proofs.

Tell me what you think.

Let be S={a1, a2, ..., an} be a set of distinct elements ordered so that a1 < a2 < ... < an. Sc, the complement of S, will consist of all the real numbers greater than an, all the real numbers less than a1, and, if there at least two elements S, the spaces between ai and ai+1 for each 1 < i < n.

If y > a, then y + µ > an for all µ > 0. If we choose a small enough µ, then also y - µ > a. Thus the set {y: y > an}, a subset of Sc, is open. Using similar reasoning, the set {y: y < a1} is open.

If S is a one-element set, then we have shown it is closed. Otherwise, we need also consider the spaces between ai and ai+1. We know there exists an infinite number of real numbers between ai and ai+1 (See Homework 2, Exercise 4). Let ∂ be such a number. then ai < ∂ < ai+1. If we choose a small enough positive µ, then ai < ∂ - µ < ∂ < ∂ + µ < ai+1. Thus the set of all numbers between the elements of S is open, and S is closed.

Would it be simpler like this:
Let S = {a1, a2, a3, ....., an} be a finite subset of R. Then, R-S (i.e. complement of S) can be represented as a union of finite number of open intervals: (-infinity, a1), (a1, a2),
(a2, a3), ..., (an, +infinity) such that R-S is open, since open intervals in R are open. Hence, S is closed.

Indeed, it is simpler.

Deveno