# Accumulation points of sets – Serlo

## Motivation[Bearbeiten]

in the article accumulation point of a sequence we already mentioned that the accumulation point of a set is not to be confused with the accumulation point of a sequence. This article concerns the accumulation point of a set. Whenever we refer to an "accumulation point" within this article, we mean an "accumulation point of a set".

The name "accumulation point" of a set suggests, that elements of a set "accumulate" there. Let us make this intuition mathematically precise.

What does it mean that elements of "accumulate" around ? When we put a small interval around , there should still be an element of the set inside it. I this was not the case, we could choose some , such that each is outside the interval , i.e. .

. So there are no points inside the interval and the set elements intuitively do not "accumulate" around .

So we could use the following (preliminary) definition: We have a *kind of accumulation point* of , whenever for each there is an , such that . Does this *kind of accumulation point* match our intuition of an *accumulation point*?

Let us take the set only consisting of the number 1 as an example: . The above definition tells us that is a *kind of accumulation point* of . This can easily be checked: Let . As , we can directly take as the element of which is included inside the y-interval. There is and is a *kind of accumulation point* of the set. But intuitively, the sequence elements do not "accumulate" around this single point, like a single person would also not be considered to be an "accumulation of people". We need a new definition with more points different from (i.e. "more people").

We therefore introduce a new definition: we call an *accumulation point* of , whenever for each there is another point *not equal to* , such that . With this definition, the *kind of accumulation point* is no longer a real *accumulation point* of . For instance, take . The set has no element , so in particular, there cannot be any element in with . This argument would also work with any other . We will later see that each *accumulation point* has to have *infinitely many* points around it. So if there is just one point or a finite amount of points, we can never have an accumulation point.

If the above has no accumulation points, what does a set with an accumulation point look like, then? Let us take a set with infinitely many points, e.g. . This set contains an interval with infinitely many points. Following the above arguments, the "extra single point" is not an accumulation point, since there are no points around it.

And how about the other points? Let us consider some with . If we choose arbitrarily small, the point or will always be an element of . These points also lie inside the interval, if we choose arbitrarily big. So every point has another point of the interval in each -neighbourhood around it and must be an accumulation point.

Those are exactly all accumulation points, since all points outside of cannot be approached by points in (You may shortly think about this fact by choosing a suitably small ). Similarly, any closed interval has all of its points being *accumulation points*.

Later, the notion of an accumulation point will even become useful, if we mathematically define derivatives of a function at . This can be done if is an accumulation point within the domain of definition, so it can be approached by other points, which are not .

## Accumulation point of a set[Bearbeiten]

Now, we formulate the thoughts above in a mathematical way.

**Definition** (accumulation point of a set)

A number is called accumulation point of a set , if for each there is some with and .

## Properties of accumulation points[Bearbeiten]

First, we note that accumulation points of sets and sequences are closely related:

**Theorem** (Accumulation points of a set are limits of set elements sequences)

A point is exactly an accumulation point of a set , if there is a sequence in which are distinct from (i.e. for all ) and .

**Proof** (Accumulation points of a set are limits of set elements sequences)

"exactly" means an equivalence "", so we need to show both the directions "" and "".

"": Let be an accumulation point of the set . For a fixed we set . Since is an accumulation point, there must be an element with and . This element depends on and hence on , so we can call it . Taking such an element for each , we obtain a sequence . This sequence converges to (see also the definition of convergence). Let arbitrary. If we choose an with , then for there is also . Hence, the distance of to the suspected limit is . As this holds for all , we have proven convergence of to .

Now, we prove the other direction.

"": Let be fixed and a sequence of elements in converging to and not including ( i.e. for all ) Since , we may choose some with . This is exactly the we need for the . And since we can find such an for each , the limit is an accumulation point.

The next theorem justifies the name "accumulation point" by proving that points indeed "accumulate" around . More precisely, there are infinitely many points accumulating around

**Theorem** (Sequence elements accumulate around accumulation points)

Let be an accumulation point of the set . For each there is a set , such that for all there is , and contains infinitely many (!) elements.

**Proof** (Sequence elements accumulate around accumulation points)

The proof goes by contradiction. We assume that the statement of the proof is not true, meaning that there is an , such that each set containing infinitely elements contains some with . Or equivalently, for some , the set (all elements in , which are closer to than ) is finite. Why are both equivalent? If had infinitely many elements, then there would have to be some with (further away than ), which is not allowed by the definition of . Now we choose (the closest distance of any element in to ). Since the set is finite and this must be (so the minimal distance is strictly bigger than zero). Hence, for all with there is , which contradicts being an accumulation point of . ↯

What if has only finitely many elements? Then each -neighbourhood around with elements in has only finitely many elements. So the condition for the second theorem is violated and we cannot have an accumulation point:

**Theorem** (Finite sets cannot have accumulation points)

A set with finitely many elements has no accumulation points.

The proof is virtually given above: We assume that there was an accumulation point and show that the second theorem leads to a contradiction. For an exercise, you may think about the arguments needed and write them down in a mathematically understandable way.

## Adherent points[Bearbeiten]

Sometimes, we have points , which are no accumulation point, but still included in . So there is a sequence in converging to , e.g. the constant sequence . This is, for instance, the case for all elements of a finite set. We will introduce a different notation for these points and call them *adherent points*.

**Definition** (adherent point)

Let be a set. A number is called '*adherent point* of , if there is a sequence in converging to .

Naturally, all accumulation points are adherent points: Whenever there is a sequence converging to , then we automatically have a sequence in converging to .

## Properties of adherent points[Bearbeiten]

Let us verify some properties of adherent points. We have already seen that each point of a set is an adherent point, since the constant sequence converges to . Let us put this into a theorem:

**Theorem** (Every point of a set is an adherent point)

Every point of a set is an adherent point of .

**Proof** (Every point of a set is an adherent point)

See above.

We have also argued above that each accumulation point is an adherent point: "accumulation point" means limit of a sequence of elements in , which are not equal to and "adherent point" mean just limit of a sequence in . Let us also put this into a theorem:

**Theorem** (Every accumulation point is an adherent point)

Every accumulation point of a sequence is an accumulation point of .

**Proof** (Every accumulation point is an adherent point)

See above.

But is an adherent point also an accumulation point? Certainly not: for a set of finitely many elements, each point is an adherent point (as it is contained in the set), but no point is an accumulation point. Now if we have an adherent point, how can we make sure that it is also an accumulation point? The answer is not to difficult: we just take out of and ask again, whether there is a sequence in converging to .

**Theorem** (Condition that an adherent point is also an accumulation point)

Every point of a set is an accumulation point of , wif and only if it is an adherent point of .

**Proof** (Condition that an adherent point is also an accumulation point)

If is an adherent point of , then there is a sequence in , converging to . Hence, is an accumulation point of . Conversely, let be an accumulationn point of . Then, there is a sequence in with for all and . Therefore, is an adherent point of .

## Examples[Bearbeiten]

**Example** (accumulation points and adherent points of intervals)

Let () be an open interval. The set of its accumulation points is the closed interval , which is also the set of its adherent points.

**Example** (accumulation points and adherent points of a union of sets)

The set of adherent points and accumulation points of the set are given by the closed interval . The set consisting of only a single point has only as adherent point and no accumulation point. If we take the union then the adherent points are given by the union of both adherent point sets . The same holds for accumulation points: they are given by the union of with the empty set , which is again .

**Theorem** (accumulation points of the rational numbers)

The set of accumulation points of the rational numbers is given by the real numbers .

**Proof** (accumulation points of the rational numbers)

Let and be given. The rational numbers are dense in , which means that there has to be an with and . Therefore, is an accumulation point of . So the set of accumulation points (and hence also adherent points) of is given by the real numbers .

Question: What are the adherent and accumulation points of the following sets in ?

- adherent points: , accumulation points:
- adherent points: , accumulation points:
- adherent points: , accumulation points:
- adherent points: , accumulation points: