Are interior points ever limit points as well? 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?
 A: 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!)
A: 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.
