# Relationship between Limit Point Types

## Theorem

Let $T = \left({X, \tau}\right)$ be a topological space.

Let $A \subseteq X$.

Let:

Then:

$C \subseteq W \subseteq L \subseteq D$

That is:

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 $T$ 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 $T$ 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 $T$ 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.

$\Box$

Let $x \in W$.

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

So every open set $U$ of $T$ 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 $T$ 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.

$\Box$

Let $x \in L$.

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

So every open set $U$ of $T$ 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 $T$ containing $x$ in which the only point of $A$ is $x$ itself.

In that case $x \notin L$.

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

$\Box$

Hence the result.

$\blacksquare$