Relationship between Limit Point Types

Theorem
Let $$X$$ be a topological space.

Let $$A \subseteq X$$.

Let:
 * $$C$$ be the set of condensation points of $$A$$;


 * $$W$$ be the set of $\omega$-accumulation points of $$A$$;


 * $$L$$ be the set of limit points of $$A$$;


 * $$D$$ be the set of adherent points of $$A$$.

Then:
 * $$C \subseteq W \subseteq L \subseteq D$$

That is:


 * Every condensation point is an $\omega$-accumulation point;


 * Every $\omega$-accumulation point is a limit point;


 * Every limit point is an adherent point;

In general, the inclusions do not hold in the other direction.

Proof
Let $$x \in C$$.

By definition of condensation point, every open set of $$X$$ containing $$x$$ also contains an uncountable number of points of $$A$$.

As an uncountable number is also an infinite number, we could also say that every open set of $$X$$ containing $$x$$ also contains an infinite number of points of $$A$$.

That is, $$x$$ is also by definition an $\omega$-accumulation point.

So $$x \in W$$ and by definition of subset:
 * $$C \subseteq W$$

Note that if $$x \in W$$ then it could be that there exists an open set $$U$$ of $$X$$ containing $$x$$ with a countably infinite number of points of $$A$$.

In that case $$x \notin C$$.

That is, not every $\omega$-accumulation point is necessarily a condensation point.

Let $$x \in W$$.

By definition of $\omega$-accumulation point, every open set $$U$$ of $$X$$ containing $$x$$ also contains an infinite number of points of $$A$$.

So every open set $$U$$ of $$X$$ such that $$x \in U$$ contains some point of $$A$$ other than $$x$$ (an infinite number, indeed).

That is, $$x$$ is also by definition a limit point.

So $$x \in L$$ and by definition of subset:
 * $$W \subseteq L$$

Note that if $$x \in L$$ then it could be that there exists an open set $$U$$ of $$X$$ containing $$x$$ which contains only a finite number of points of $$A$$.

In that case $$x \notin W$$.

That is, not every limit point is necessarily an $\omega$-accumulation point.

Let $$x \in L$$.

By definition of limit point, every open set $$U$$ of $$X$$ containing $$x$$ also contains some point of $$A$$ other than $$x$$.

So every open set $$U$$ of $$X$$ such that $$x \in U$$ contains some point of $$A$$.

That is, $$x$$ is also by definition an adherent point.

So $$x \in L$$ and by definition of subset:
 * $$L \subseteq D$$

Note that if $$x \in D$$ then it could be that there exists an open set $$U$$ of $$X$$ containing $$x$$ in which the only point of $$A$$ is $$x$$ itself.

In that case $$x \notin L$$.

That is, not every adherent points is necessarily a limit point.

Hence the result.