What is the intuition behind these three points in a topological space. Can you provide a diagram also, preferably in the plane.
Def. Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $x \in X$. Then, x is an adherent point of $A$ iff. every open neighborhood of $x$ contains at least one point of $A$.
Def. Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $x \in X$. Then, x is a limit point of $A$ iff. every open neighborhood of $x$ contains at least one point of $A$ distinct of $x$.
Def. Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $x \in X$. Then, x is an isolated point of $A$ iff. there is an open neighborhood of $x$ containing no points of $A$ other than $x$.
So as I understand it, an adherent point can be in $A$ but does not have to be. Thus, if I were to draw a set in the plane, then all the adherent points would be the interior united with the boundary? Is my intuition correct here? Diagrammatically speaking.
Then, a limit point is an adherent point but not conversely. Taking the plane again and an arbitrary set in the plane, then intuitively are the limit points those points inside the set but not on the boundary?
And then the isolated points in the plane would be all points on the boundary? Or perhaps boundary and exterior?
I can't seem to figure the two-dimensional intuition behind these points even though I can work with the definitions.
Best Answer
Adherent points are either limit points or isolated points. Consider the set $X = (2, 3) \cup \{6\}$.
If $y$ is an isolated point of $X$, then $y\in X$ and we must be able to find an open neighborhood that contains $y$ but no other element of $X$. So in the example above, $y = \{6\}$ is an isolated point because we can find an interval, say $I = (5,7)$, which contains $I$ but no other element of $X$.
If $y$ is a limit point of $X$, then each one of the open neighborhoods that contain it must also contain some point of $X$ that is not equal to $y$. So any points in the interval $I = [2,3]$ are limit points of $X$. The limit point of a set may or may not be a member of it. The point $\{2.5\}\in X$ but $\{2\} \not\in X$, but both are limit points.
It's close. Let's look at the points inside (but not on the boundary of) a circle at the origin with radius 1 and some weird point outside the circle such as $(3,3)$. The interior points and the ones right on the boundary are the limit points of the circle, since even the smallest open neighborhood containing them must overlap or be in the circle. In this case, the limit points nearly capture the closure of the circle, but we also have to consider the isolated point $(3,3)$. When we look at the union of the limit and isolated points, we get all the adherent points, which make up the closure of the circle. Adherent points are also called closure points.