Intuition behind adherent, limit, and isolated points in a topological space

Question

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.

Answer 1

To answer your questions: (i) $\bar{A}=\mathring{A}\sqcup\partial A$ is true. this is in fact how we define the boundary $\partial A$ in general; (ii) all limit points are adherent points; (iii) a point is a limit point if and only if it is not an isolated point (whether or not it's contained in $A$) (this is a great exercise of taking the negation of a proposition by sending $\forall\rightarrow\exists$ and vice versa); (iv) those points inside the set but not in the boundary are called interior points; (v) those points inside the boundary but not in the set are limit points, but the converse is not true (there are limit points contained in the set); (vi) points on the exterior of the closure of $A$ are isolated points, but the converse is not true. there are isolated points contained in $A$.

Answer 2

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.

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?

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.

Answer 3

I'm so annoyed that no one explicitly tells that in a set of continuous interval or ball, either open or close, has its limit points the same with its adherent points. Limit points and adherent points differ only when the set has isolated points.

Consider the set $S = (2, 3) \cup \{6\}$.

$\{2\}$ and $\{3\}$ are limit points and also adherent points, and of course all points in $(2, 3)$ are limit points and also adherent points, because their neighborhood contains infinite number of points from the set $S$ including and not including themselves.

$\{6\}$ is an isolated point because there is a neighborhood contains no point at all from the set $S$.

$\{6\}$ is not a limit point because there is a neighborhood contains no points from the set $S$ other than $\{6\}$ itself.

But $\{6\}$ is an adherent point because there is a neighborhood contains exactly one point from the set $S$ which is $\{6\}$ itself.