I've read the definition and have seen videos where they all say the same thing. We have a $\epsilon$ neighborhood around $l$ and if it contains another point, let's say set $A$, other then $l$, then we call $l$ an accumulation point of $A$. But what's the point of this? why do we need accumulation points? and is my description correct?
What are accumulation points in easy terms? and how do we use them
general-topologyreal-analysis
Related Solutions
An accumulation point of a sequence is a more general concept that a limit. For example, the sequence $+1,-1,+1,-1,...$ has no limit, but two accumulation points, $\pm 1$.
(If you prefer to have distinct points, take the sequence $+1+\frac{1}{1},-1+\frac{1}{2},+1+\frac{1}{3},-1+\frac{1}{4},...$)
Think of accumulation points as limits of subsequences. A point is an accumulation point of a sequence iff you can find a subsequence converging to that point.
It should be clear that if $a_n \to A$, then all subsequences must also converge to $A$.
Suppose $J$ is an infinite set, and $U$ an open set containing $A$. Since $a_n \to A$, we have some $N$ such that $a_n \in U$ for all $n \ge N$. Then we see that the set $J'=J \cap \{N,N+1,...\}$ is also infinite (otherwise a quick contradiction), and for all $n \in J'$, $a_n \in U$. Hence $A$ is an accumulation point of the subsequence $a_n$, $n \in J$.
There is no necessary relationship between the two sets.
Let $A$ be a set in a topological space $X$. A point $x$ is in the interior of $A$ if there is an open set $U$ such that $x\in U\subseteq A$; in particular this implies that $x\in A$. A point $x$ is an accumulation point of $A$ if for each open set $U$ containing $x$, $U\cap(A\setminus\{x\}\ne\varnothing$; in words, if every open nbhd of $x$ contains at least one point of $A$ different from $x$. This does not imply that $x\in A$.
Take the space $\Bbb R$ with the usual topology as a familiar example. The set $\Bbb Z$ has empty interior: for any $n\in\Bbb Z$, no matter how small an $\epsilon>0$ you take, $(n-\epsilon,n+\epsilon)\nsubseteq\Bbb Z$, so $n$ is not in the interior of $\Bbb Z$. $\Bbb Z$ also has no accumulation points: if $x\in\Bbb R\setminus Z$, there is an integer $n$ such that $n<x<n+1$, and $(n,n+1)$ is then an open nbhd of $n$ that contains no point of $\Bbb Z$; and if $n\in\Bbb Z$, $(n-1,n+1)$ is an open nbhd of $n$ that contains no point of $\Bbb Z\setminus\{n\}$.
The set $\left\{\frac1n:n\in\Bbb Z^+\right\}$, on the other hand, has empty interior and exactly one accumulation point, $0$. In this case the interior of the set is its set of accumulation points. But now consider the following sets:
$\Bbb Q$ also has empty interior, and every real number is an accumulation point of $\Bbb Q$.
- $[0,1]$, $[0,1)$, and $(0,1)$ all have interior $(0,1)$, and all have $[0,1]$ as their set of accumulation points.
(In each case you should try to prove the assertions.)
If we look at $\Bbb Z$ as a space in its own right, with the discrete topology, then every subset of $\Bbb Z$ is open, and no point of $\Bbb Z$ is an accumulation point of any subset of $\Bbb Z$. Thus, if $A\subseteq\Bbb Z$, then the interior of $A$ is $A$ itself, but $A$ has no accumulation points.
Best Answer
Here's how I've attempted to conceptualize the notion of an accumulation point and for simplicity let's take the metric case setting.
The idea behind accumulation points is that they are arbitrarily "close" to elements of the $X$ while not necessarily being an element of the set itself. The definition is literally saying that no matter how small a neighbourhood we consider around an accumulation point, we always have at least one point of the set contained in the neighbourhood.
To make the usefulness of accumulation points more clear, it might help to think of the very related definition of an accumulation point of a sequence.
That is to say an accumulation point of a sequence is a point where infinitely many sequence terms cluster to. The way I think of their usefulness is that if you didn't have an accumulation point, how could you even talk about a sequence $(x_n)$ converging to a point? If we can't get arbitrarily close to that point, i.e if we don't have this accumulation of sequence terms arbitrarily close to the point, how can we start to think about convergence?
Another perspective that you might have already encountered; if you think about general limits of functions, say $f:X \rightarrow \mathbb{R}$ and you want to talk about $\lim_{x \rightarrow c} f(x)$ for this to make sense $c$ must be an accumulation point of the set $X$, because otherwise how can we even think about getting arbitrarily close to the point?