I don't understand the difference between the interior of a set, and the set of all its accumulation points.

My understanding of an accumulation point is any point in a set which has an epsilon neighborhood around it, which is contained in the set- not necessarily implying that the accumulation point itself is in the set.

From what I can gather, the interior is identical. Can someone explain the difference?

  • $\begingroup$ If $N$ is a nbhd of $x$, then by definition $x\in N$, so if $N\subseteq A$, then $x\in A$: if a nbhd of $x$ is contained in the set $A$, then necessarily $x$ belongs to the set $A$. $\endgroup$ Oct 31 '13 at 12:14
  • $\begingroup$ not necessarily? for example, in the set [0,2]\ {1}, 1 is an accumulation point but is not contained in the set. $\endgroup$
    – user102447
    Oct 31 '13 at 12:19
  • $\begingroup$ Your example has nothing to do with my assertion. I did not say that an accumulation point of $A$ is necessarily in $A$; that is of course false. I said that if $x$ is in the interior of $A$, then $x$ is necessarily in $A$. $\endgroup$ Oct 31 '13 at 12:26

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.


I assume that you're working with subsets of the real line with the usual topology.

The definition you gave for accumulation point is actually the definition for interior point, and it does imply that the point lies in the set. Indeed, an $\epsilon$-neighborhood of a point $x$ is an interval of the form $(x-\epsilon,x+\epsilon)$ with $\epsilon>0,$ and $x$ readily lies in any such interval, so $x$ lies in any set that contains such an interval.

We say that a point $x$ is an accumulation point of a set $A$ if every $\epsilon$-neighborhood of $x$ intersects $A$ at a point distinct from $x$. This does not mean that every epsilon neighborhood of $x$ is contained in $A$--it's possible that no $\epsilon$-neighborhood of $x$ is contained in $A,$ which means that accumulation points need not be interior points.

For an example to see that the interior and set of accumulation points are not the same, consider $[0,2]\setminus\{1\}$. The interior of the set is $(0,1)\cup(1,2),$ but the set of accumulation points is $[0,2]$. The point $1$ is an example of a non-interior accumulation point.

For another example, consider the set $\{0\}\cup\{\frac1n:n\in\Bbb Z^+\}.$ No point of this set has a neighborhood contained in the set, so the interior is empty, but it can be shown that $0$ is the unique accumulation point. So, even if an accumulation point is in the set, it need not be an interior point.


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