Math for Non-Geeks/Accumulation points of sets

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Motivation

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 $M\subseteq ℝ$ "accumulate" there. Let us make this intuition mathematically precise.

What does it mean that elements of

$M$

"accumulate" around

$p\in ℝ$

? When we put a small interval around

$p$

, there should still be an element of the set inside it. I this was not the case, we could choose some

$ϵ>0$

, such that each

$m\in M$

is outside the interval

$(p-ϵ,p+ϵ)$

, i.e.

$|p-m|\ge ϵ$

.

point

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

$p$

.

So we could use the following (preliminary) definition: We have a kind of accumulation point $p\in ℝ$ of $M$ , whenever for each $ϵ>0$ there is an $m\in M$, such that $|p-m|<ϵ$. 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: $M=\{1\}$ . The above definition tells us that $p=1$ is a kind of accumulation point of $M$. This can easily be checked: Let $ϵ>0$. As $p\in M$ , we can directly take $m=p\in M$ as the element of $M$ which is included inside the y$ϵ$-interval. There is $|p-p|=0<ϵ$ and $p=1$ 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 $p$ (i.e. "more people").

We therefore introduce a new definition: we call $p\in ℝ$ an accumulation point of $M\subset ℝ$, whenever for each $ϵ>0$ there is another point $m\in M$ not equal to $p$ , such that $|p-m|<ϵ$ . With this definition, the kind of accumulation point $p=1$ is no longer a real accumulation point of $M=\{1\}$ . For instance, take $ϵ=1/2$ . The set $M$ has no element $m\ne p=1$, so in particular, there cannot be any element $m\ne p$ in $M$ with $|p-m|<1/2$ . This argument would also work with any other $ϵ>0$. 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 $M$ 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. $M=\{1\}\cup [2,3]$. This set contains an interval with infinitely many points. Following the above arguments, the "extra single point" $p=1$ is not an accumulation point, since there are no points around it.

And how about the other points? Let us consider some

$p\in [2,3]$

with

$\max (|2-p|,|3-p|)>ϵ>0$

. If we choose

$1/2>ϵ>0$

arbitrarily small, the point

$p-ϵ/2$

or

$p+ϵ/2$

will always be an element of

$[2,3]$

. These points also lie inside the interval, if we choose

$ϵ>0$

arbitrarily big. So every point

$p\in [2,3]$

has another point of the interval in each

$ϵ$

-neighbourhood around it and must be an accumulation point.

An accumulation point of the set M

Those are exactly all accumulation points, since all points outside of

$M$

cannot be approached by points in

$M$

(You may shortly think about this fact by choosing a suitably small

$ϵ>0$

). Similarly, any closed interval

$[a,b]$

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 $f:M\to ℝ$ at $x\in M$ . This can be done if $x$ is an accumulation point within the domain of definition, so it can be approached by other points, which are not $x$.

Accumulation point of a set

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

Math for Non-Geeks: Template:Definition

Properties of accumulation points

First, we note that accumulation points of sets and sequences are closely related: Math for Non-Geeks: Template:Satz

The next theorem justifies the name "accumulation point" $p$ by proving that points indeed "accumulate" around $p$. More precisely, there are infinitely many points accumulating around $p$ Math for Non-Geeks: Template:Satz

What if $M$ has only finitely many elements? Then each $ϵ$-neighbourhood ${\mathbb{N}}_{ϵ}$ around $p$ with elements in $M$ has only finitely many elements. So the condition for the second theorem is violated and we cannot have an accumulation point: Math for Non-Geeks: Template:Satz 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

Sometimes, we have points $p\in M$, which are no accumulation point, but still included in $M$. So there is a sequence in $M$ converging to $p$, e.g. the constant sequence $(m_n{)}_{n\in \mathbb{N}},m_n=p$. 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.

Naturally, all accumulation points are adherent points: Whenever there is a sequence $(m_n{)}_{n\in \mathbb{N}},m_n\in M,m_n\ne p$ converging to $p$, then we automatically have a sequence in $M$ converging to $p$.

Properties of adherent points

Let us verify some properties of adherent points. We have already seen that each point of a set $p\in M\subset ℝ$ is an adherent point, since the constant sequence $m_n=p$ converges to $p$. Let us put this into a theorem: Math for Non-Geeks: Template:Satz

We have also argued above that each accumulation point is an adherent point: "accumulation point" means limit of a sequence of elements in $M$, which are not equal to $p$ and "adherent point" mean just limit of a sequence in $M$. Let us also put this into a theorem: Math for Non-Geeks: Template:Satz

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 $p$ out of $M$ and ask again, whether there is a sequence in $M\setminus \{p\}$ converging to $p$. Math for Non-Geeks: Template:Satz

Examples

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