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From my understanding of limit points and interior points there is somewhat of an overlap and that a lot of the time interior points are also limit points.

For the reals, a neighborhood, $r>0$, around a point must contain only a single point of the set in question to determine if it's a limit point or not. However, an interior must be completely contained in the set in question, meaning it has a neighborhood that contains at least a point in the set, which also makes it a limit point.

For example: $[0,1)$ in the reals.

From what I understand, the set $(0,1)$ is the set of all interior points, the set $[0,1]$ is the set of all limit points, and the set $(0,1)$ is the set of all points which are both interior and limit points.

Is this correct or are interior points always not limit points for some reason?

Jake Tae
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allthosethings
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2 Answers2

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You are right that interior points can be limit points.

Your example was a perfect one: The set $[0,1)$ has interior $(0,1)$, and limit points $[0,1]$. So actually all of the interior points here are also limit points.

Here is an example of an interior point that's not a limit point:

Let $X$ be any set and consider the discrete topology $\mathcal{T} = \mathcal{P}(X)$ on $X$. Then let $A$ be any non-empty subset of $X$. For any $a \in A$, $\{a \}$ is an open set containing $a$, and $a \in \{a \} \subseteq A$, so $a$ is an interior point for $A$. However, $a$ is not a limit point because we can find an open neighborhood of $a$ that doesn't contain any points from $A$ other than $a$ -- namely, $\{a \}$ is the open neighborhood that satisfies this. So $a$ is not a limit point of $A$. (By this same argument, we can show that $A$ has no limit points!)

layman
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  • You are saying {a} is an open set containing a, but it seems this is not an open set since there is no interval around a that is also in A. – jeffery_the_wind Sep 15 '19 at 11:30
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    @jeffery_the_wind You are speaking from the perspective of the Euclidean topology that we are all so used to. But in my answer, I mentioned that ${a}$ is open in the discrete topology, which is the topology that consists of all possible subsets of $X$. So every subset of $X$ is open, including ${a}$. So the question of which sets are open and which are closed really must be asked/framed in the context of a specific topology, since the answer changes if you change the topology. ${a}$ is not open in the Euclidean topology, but it is in the discrete topology. – layman Sep 16 '19 at 03:23
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A point $x$ of $A$ can be of one of two mutually exclusive types: a limit point of $A$ or an isolated point of $A$.

If the latter, it means that there exists some open $O$ in $X$ such that $\{x\} = O \cap A$. The negation of this is exactly that every open set $O$ that contains $x$ always intersects points of $A$ unequal to $x$ as well, and this means exactly that $x$ is a limit point of $A$.

E.g. $A = (0,1) \cup \{2,3\}$ (usual topology of the reals) has two isolated points $2$ and $3$ (which are not interior points of $A$), and the rest are limit points of $A$ as well as interior points. There are also limit points $0,1$ that are not in $A$ (showing $A$ is not closed).

So if $A$ has no isolated point, all of the points of $A$ (in particular all its interior points) are limit points of $A$. So often there will quite an overlap between interior and limit points.

Henno Brandsma
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