The definitions I am working with say that:
Adherent Point : Let $(X, d)$ be a metric space, let $E\subseteq X$ and let $x_0\in X$, then we can say that $x_0$ is an adherent point of $E$ if $\forall r>0 : B(x_0, r)\cap E \neq \emptyset$
Limit Point : Let $(X, d)$ be a metric space, let $E\subseteq X$ and let $x_0\in X$, then we can say that $x_0$ is an limit point of $E$ if $\forall r>0 : B(x_0, r)\setminus\{x_0\} \cap E \neq \emptyset$
Closure : $\overline{A}$ is defined as the set of all adherent points of a set $A$
Knowing these things, a couple things are immediately obvious:
- All limits points are adherent points [but not the other way around]
- $\overline{A}$ will contain all limit points of $S$
And then there are derived results which say:
- A set $A$ is closed if and only if $A=\overline{A}$
- A set $A$ is closed if and only if it contains all of its limit points
I can't grasp how (4) is true. Couldn't it be the case that a set contains all its limit points, but not all its adherent point and thus be not closed under (3)?
Any and all help will be greatly appreciated
Best Answer
Every adherent point $x_0$ of $E$ that is not a limit point of $E$ is automatically an element of $E$.
Some $r>0$ exists such that $(B(x_0,r)-\{x_0\})\cap E=\varnothing$ because $x_0$ is not a limit point.
But next to that we have $B(x_0,r)\cap E\neq\varnothing$ because $x_0$ is an adherent point.
This together implies that $B(x_0,r)\cap E=\{x_0\}$ showing that $x_0\in E$.
So if $E$ contains all its limit points then it will also contain all its adherent points because: $$\{\text{adherent points}\}=\{\text{limit points}\}\cup\{\text{points that are adherent and not limit}\}$$