Question about Accumulation points

[TheChemist_]

So we just recently did accumulation points in my maths class for chemists. I understood everything that was taught but ever since I was trying to find a reasonable explanation if the sequence a_n = (-1)ⁿ has 2 accumulation points (-1,1) or if it doesn't have any at all. I mean it's clear that its divergent (?), but that doesn't solve the problem. Because the Set of values for the sequence would be just {1,-1}, no values in between. So the general definition of accumulation points doesn't quite fit, or does it? Can somebody explain it maybe in words and in a mathematical way? Thx

 

[fresh_42]

What is your definition of an accumulation point?

 

[TheChemist_]

Say D⊂ℝ. Then a ∈ ℝ is called accumulation point of D, when there is a sequence (a_n) in D "without" {a} and a_n→a (→ means goes towards)

thats how the prof defined it

 

[PeroK]

Say D⊂ℝ. Then a ∈ ℝ is called accumulation point of D, when there is a sequence (a_n) in D "without" {a} and a_n→a (→ means goes towards)

thats how the prof defined it

The clause "without a" would seem to make things clear. Or not?

 

[fresh_42]

There is a slightly difference between an accumulation point of a set $D$, which your prof defined, and an accumulation point of a sequence $(a_n)_{n \in \mathbb{N}}$, which you were asking for. I'm not sure whether they have different names in English. I think the accumulation points of sets are called limit points, and those of sequences accumulation or cluster points.

The limit points $a$ of a set are defined as above, or one can say, that every neighborhood of $a$ contains a point of $D-\{a\}$.

A cluster point or accumulation point of a sequence is a point $a$, for which there is a subsequence $(a_{n_k})_{k \in \mathbb{N}} \subseteq (a_n)_{n \in \mathbb{N}}$ such that $\lim_{k \rightarrow \infty}a_{n_k} = a$. This includes constant subsequences.

So both terms are related but not the same.

 

[Svein]

The standard definition for a cluster point is:

• a is a cluster point for the sequence $\{x_{n}\}$ if $(\forall \epsilon >0)(\forall N)(\exists n>N)(\vert a - x_{n}\vert <\epsilon )$

 

[Stephen Tashi]

Say D⊂ℝ. Then a ∈ ℝ is called accumulation point of D, when there is a sequence (a_n) in D "without" {a} and a_n→a (→ means goes towards)

thats how the prof defined it

To repeat what fresh_42 said, be alert to the fact your course materials may use similar terminology for two different concepts. It may discuss "accumulation of a set" and also "accumulation point of a sequence".

The definition you gave above is for "accumulation point of a set" even though it mentions a sequence. An alternate definition of "accumulation point of a set" is "any open set (or 'open ball') that contains $a$ contains at least one element of $D$ different from $a$".

For example, on the real number line, 1 is an accumulation point of the set (0,1).

Notice that if we test the professors definition for accumulation point of a set, we can find the sequence $a[j] = (j+1)/(j+2)$ that contains only numbers in (0,1) and converges to 1 but does not contain 1 as a term of the sequence.

The definition for "accumulation point of a sequence" is a different concept. It's true that the terms in a sequence can be viewed as a set, but a sequence has the additional aspect of having an order of terms and the fact that repeated terms have some significance.

On the real number line, the set of the two numbers $D = \{-1,1\}$ has no accumulation points. (For example you can find the open interval (-3/2, 0) that contains no point of $D$ different that (-1), so (-1) is not an accumulation point of $D$.

However, (-1) is an accumulation point of the sequence {a} = -1,1,-1,1,-1,... because there exists a subsequence of {a} that converges to (-1) - namely the subsequence -1,-1,-1,... Notice that in the definition for accumulation point of a sequence there is nothing preventing the accumulation point itself from being equal to a term of the sequence.

 

[Svein]

The definition you gave above is for "accumulation point of a set" even though it mentions a sequence. An alternate definition of "accumulation point of a set" is "any open set (or 'open ball') that contains aa contains at least one element of D different from a".

It seems to me that your definition of "accumulation point of a set" is very close to the definition of "a point of closure" of a set (in a metric space).

 

[Stephen Tashi]

It seems to me that your definition of "accumulation point of a set" is very close to the definition of "a point of closure" of a set (in a metric space).

The professors definition and my version of it may be equivalent to that definition. I'm not familiar with the terminology "point of closure", but (in the terminology I know) to take the "closure of a set", one forms the union of the set and its "limit points", which are also (I think) called its "accumulation points".

 

[Svein]

OK. The definitions of open and closed sets in a metric space are.

• A set O is called open if $(\forall x \in O)(\exists \delta >0)(\forall y: d(y, x)<\delta \Rightarrow y\in O)$
• A point z is called a point of closure of a set E if $(\forall \delta >0)(\exists x \in E)(d(z, x)<\delta)$
• The collection of all points of closure of a set E is denoted $\bar{E}$. Thus $E \subset \bar{E}$

 

[Stephen Tashi]

• A point z is called a point of closure of a set E if $(\forall \delta >0)(\exists x \in E)(d(z, x)<\delta)$

Let $E$ be $(0,1) \cup \{2\}$. As I read that definition, ${2}$ is a point of closure of the set $E$ (we can take $x = 2$ because we are not required to make $z$ and $x$ distinct points. ) The closure of the set $E$ is $[0,1] \cup \{2\}$.

By the professor's or my definition of "accumulation point", ${2}$ is not an accumulation point of $E$, but the "closure of $E$" is still $[0,1] \cup \{2\}$.

 

[Svein]

Let $E$ be $(0,1) \cup \{2\}$. As I read that definition, ${2}$ is a point of closure of the set $E$ (we can take $x = 2$ because we are not required to make $z$ and $x$ distinct points. ) The closure of the set $E$ is $[0,1] \cup \{2\}$.

By the professor's or my definition of "accumulation point", ${2}$ is not an accumulation point of $E$, but the "closure of $E$" is still $[0,1] \cup \{2\}$.

Yes. My old book on real analysis (Royden) defines accumulation point:

• A point x is called an accumulation point of a set E if it is a point of closure of $E\sim \{x\}$ (setwise difference)