# Math for Non-Geeks/Accumulation points of sets
## 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 $M\subseteq \mathbb {R}$ "accumulate" there. Let us make this intuition mathematically precise.
What does it mean that elements of $M$ "accumulate" around $p\in \mathbb {R}$? 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 $\epsilon >0$ , such that each $m\in M$ is outside the interval $(p-\epsilon ,p+\epsilon )$, i.e. $|p-m|\geq \epsilon$.
. 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 $p\in \mathbb {R}$ of $M$ , whenever for each $\epsilon >0$ there is an $m\in M$, such that $|p-m|<\epsilon$. 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 $\epsilon >0$. As $p\in M$ , we can directly take $m=p\in M$ as the element of $M$ which is included inside the y$\epsilon$-interval. There is $|p-p|=0<\epsilon$ 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 $p\in \mathbb {R}$ an accumulation point of $M\subset \mathbb {R}$, whenever for each $\epsilon >0$ there is another point $m\in M$ not equal to $p$ , such that $|p-m|<\epsilon$ . With this definition, the kind of accumulation point $p=1$ is no longer a real accumulation point of $M=\{1\}$ . For instance, take $\epsilon =1/2$ . The set $M$ has no element $m\neq p=1$, so in particular, there cannot be any element $m\neq p$ in $M$ with $|p-m|<1/2$ . This argument would also work with any other $\epsilon >0$. 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 $\max(|2-p|,|3-p|)>\epsilon >0$ . If we choose $1/2>\epsilon >0$ arbitrarily small, the point $p-\epsilon /2$ or $p+\epsilon /2$ will always be an element of $[2,3]$ . These points also lie inside the interval, if we choose $\epsilon >0$ arbitrarily big. So every point $p\in [2,3]$ has another point of the interval in each $\epsilon$-neighbourhood around it and must be an accumulation point.
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 $\epsilon >0$ ). 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 $f\colon M\to \mathbb {R}$ 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.
Definition (accumulation point of a set)
A number $p\in \mathbb {R}$ is called accumulation point of a set $M$, if for each $\epsilon >0$ there is some $m\in M$ with $m\neq p$ and $|m-p|<\epsilon$.

## Properties of accumulation points

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 $p\in \mathbb {R}$ is exactly an accumulation point of a set $M$, if there is a sequence $(m_{n})_{n\in \mathbb {N} }$ in $M$ which are distinct from $p$ (i.e. $m_{n}\neq p$ for all $n\in \mathbb {N}$) and $\lim _{n\to \infty }m_{n}=p$ .
Proof (Accumulation points of a set are limits of set elements sequences)
"exactly" means an equivalence "$\Leftrightarrow$", so we need to show both the directions "$\Rightarrow$" and "$\Leftarrow$".
"$\Rightarrow$": Let $p\in \mathbb {R}$ be an accumulation point of the set $M$. For a fixed $n\in \mathbb {N}$ we set $\epsilon _{n}:=1/n$. Since $p$ is an accumulation point, there must be an element $m\in M$ with $m\neq p$ and $|p-m|<\epsilon _{n}=1/n$. This element depends on $\epsilon _{n}$ and hence on $n$ , so we can call it $m_{n}$. Taking such an element for each $n\in \mathbb {N}$, we obtain a sequence $(m_{n})_{n\in \mathbb {N} }$. This sequence converges to $p$ (see also the definition of convergence). Let $\epsilon >0$ arbitrary. If we choose an $N\in \mathbb {N}$ with $N>1/\epsilon$ , then for $n\geq N$ there is also $1/n\leq 1/N<\epsilon$. Hence, the distance of $m_{n}$ to the suspected limit $p$ is $|m_{n}-p|<\epsilon _{n}=1/n\leq 1/N<\epsilon$. As this holds for all $n\geq N$, we have proven convergence of $(m_{n})_{n\in \mathbb {N} }$ to $p$.
Now, we prove the other direction.
"$\Leftarrow$": Let $\epsilon >0$ be fixed and $(m_{n})_{n\in \mathbb {N} }$ a sequence of elements in $M$ converging to $p\in \mathbb {R}$ and not including $p$( i.e. $m_{n}\neq p$ for all $n\in \mathbb {N}$) Since $\lim _{n\to \infty }m_{n}=p$ , we may choose some $n_{0}\in \mathbb {N}$ with $|m_{n}-p|<\epsilon$. This is exactly the $m_{n}\neq p$ we need for the $\epsilon >0$. And since we can find such an $m_{n}$ for each $\epsilon >0$, the limit $p$ is an accumulation point.
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 $p$
Theorem (Sequence elements accumulate around accumulation points)
Let $p\in \mathbb {R}$ be an accumulation point of the set $M\subset \mathbb {R}$. For each $\epsilon >0$ there is a set $N\subset M$, such that for all $m\in N$ there is $|m-p|<\epsilon$, and $N$ 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 $\epsilon >0$, such that each set $N\subset M$ containing infinitely elements contains some $m\in N$ with $|p-m|\geq \epsilon$. Or equivalently, for some $\epsilon >0$, the set $N_{\epsilon }:=\{m\in M:|m-p|<\epsilon ,m\neq p\}$ (all elements in $M$, which are closer to $p$ than $\epsilon$) is finite. Why are both equivalent? If $N_{\epsilon }$ had infinitely many elements, then there would have to be some $m\in N_{\epsilon }$ with $|p-m|\geq \epsilon$ (further away than $\epsilon$), which is not allowed by the definition of $N_{\epsilon }$ . Now we choose ${\tilde {\epsilon }}:=\min\{|p-m|:m\in N_{\epsilon }\}$ (the closest distance of any element in $M$ to $p$). Since the set $N_{\epsilon }$ is finite and $p\notin N_{\epsilon },$ this must be ${\tilde {\epsilon }}>0$ (so the minimal distance is strictly bigger than zero). Hence, for all $m\in M$ with $m\neq p$ there is $|p-m|\geq {\tilde {\epsilon }}$, which contradicts $p$ being an accumulation point of $M$. ↯
What if $M$ has only finitely many elements? Then each $\epsilon$-neighbourhood $\mathbb {N} _{\epsilon }$ 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:
Theorem (Finite sets cannot have accumulation points)
A set $M$ 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

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.
Definition (adherent point)
Let $M\subset \mathbb {R}$ be a set. A number $p\in \mathbb {R}$ is called 'adherent point of $M$, if there is a sequence in $M$ converging to $p$.
Naturally, all accumulation points are adherent points: Whenever there is a sequence $(m_{n})_{n\in \mathbb {N} },m_{n}\in M,m_{n}\neq p$ 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 $p\in M\subset \mathbb {R}$ is an adherent point, since the constant sequence $m_{n}=p$ converges to $p$. Let us put this into a theorem:
Theorem (Every point of a set is an adherent point)
Every point $p$ of a set $M\subset \mathbb {R}$ is an adherent point of $M$.
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 $M$, 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:
Theorem (Every accumulation point is an adherent point)
Every accumulation point $p$ of a set $M\subset \mathbb {R}$ is an adherent point of $M$.
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 $p$ out of $M$ and ask again, whether there is a sequence in $M\setminus \{p\}$ converging to $p$.
Theorem (Condition that an adherent point is also an accumulation point)
Every point $p$ of a set $M\subset \mathbb {R}$ is an accumulation point of $M$, wif and only if it is an adherent point of $M\setminus \{p\}$.
Proof (Condition that an adherent point is also an accumulation point)
If $p$ is an adherent point of $M\setminus \{p\}$ , then there is a sequence in $M\setminus \{p\}$, converging to $p$ . Hence, $p$ is an accumulation point of $M$ . Conversely, let $p$ be an accumulationn point of $M$ . Then, there is a sequence $(m_{n})_{n\in \mathbb {N} }$ in $M$ with $m_{n}\neq p$ for all $n\in \mathbb {N}$ and $\lim \nolimits _{n\to \infty }m_{n}=p$ . Therefore, $p$ is an adherent point of $M\setminus \{p\}$.

## Examples

Example (accumulation points and adherent points of intervals)
Let $I=(a,b)\subset \mathbb {R}$ ($a<b$) be an open interval. The set of its accumulation points is the closed interval $[a,b]$, 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 $(1,2]$ are given by the closed interval $[1,2]$. The set $\{4\}$ consisting of only a single point has only $4$ as adherent point and no accumulation point. If we take the union $(1,2]\cup \{4\}$ then the adherent points are given by the union of both adherent point sets $[1,2]\cup \{4\}$. The same holds for accumulation points: they are given by the union of $[1,2]$ with the empty set $\emptyset$, which is again $[1,2]$.
Theorem (accumulation points of the rational numbers)
The set of accumulation points of the rational numbers $\mathbb {Q}$ is given by the real numbers $\mathbb {R}$.
Proof (accumulation points of the rational numbers)
Let $p\in \mathbb {R}$ and $\epsilon >0$ be given. The rational numbers $\mathbb {Q}$ are dense in $\mathbb {R}$ , which means that there has to be an $r\in \mathbb {Q}$ with $r\neq p$ and $|r-p|<\epsilon$. Therefore, $p$ is an accumulation point of $\mathbb {Q}$ . So the set of accumulation points (and hence also adherent points) of $\mathbb {Q}$ is given by the real numbers $\mathbb {R}$.
Question: What are the adherent and accumulation points of the following sets in $\mathbb {R}$?
* $(4,\infty )$
* $\mathbb {N}$
* $\{1/n:n\in \mathbb {N} \}$
* $\mathbb {R} \setminus \mathbb {Q}$
* adherent points: $[4,\infty )$, accumulation points: $[4,\infty )$
* adherent points: $\mathbb {N}$, accumulation points: $\emptyset$
* adherent points: $\{{\tfrac {1}{n}}:n\in \mathbb {N} \}\cup \{0\}$, accumulation points: $\{0\}$
* adherent points: $\mathbb {R}$, accumulation points: $\mathbb {R}$