# Limit Point of a Set

Let $$X$$ be a topological space with topology $$\tau $$, and $$A$$ be a subset of $$X$$. A point $$x \in X$$ is said to be the limit point or accumulation point or cluster point of $$A$$ if each open set containing$$x$$ contains at least one point of $$A$$ different from $$x$$.

In other words, a point $$x$$ of a topological space $$X$$ is said to be the limit point of a subset $$A$$ of $$X$$ if for every open set $$U$$ containing $$x$$ we have

\[ \left\{ {A \cap U} \right\}\backslash \left\{ x \right\} = \phi \]

It is clear from the above definition that the limit point of a set $$A$$ may or may not be the point of $$A$$.

Let $$X = \left\{ {a,b,c} \right\}$$ with topology $$\tau = \left\{ {\phi ,\left\{ {a,b} \right\},\left\{ c \right\},X} \right\}$$ and$$A = \left\{ a \right\}$$, then $$b$$ is the only limit point of $$A$$, because the open sets containing $$b$$, namely $$\left\{ {a,b} \right\}$$ and $$X$$, also contain a point $$a$$ of $$A$$.

On the other hand, $$a$$ and $$b$$ are not limit points of $$C = \left\{ c \right\}$$, because the open set $$\left\{ {a,b} \right\}$$ containing these points does not contain any points of $$C$$. Point $$c$$ is also not a limit point of $$C$$, because the open set $$\left\{ c \right\}$$ containing $$c$$ does not contain any other point of $$C$$ different from $$c$$. Thus, the set $$C = \left\{ c \right\}$$ has no limit points.

As another example, let $$X = \left\{ {a,b,c,d,e} \right\}$$ with topology $$\tau = \left\{ {\phi ,\left\{ a \right\},\left\{ {c,d} \right\},\left\{ {a,c,d} \right\},\left\{ {b,c,d,e} \right\},X} \right\}$$. Let $$A = \left\{ {a,b,c} \right\}$$ then $$a$$ is not a limit point of $$A$$, because the open set $$\left\{ a \right\}$$ containing $$a$$ does not contain any other point of $$A$$ different from $$a$$. $$b$$ is a limit point of $$A$$, because the open sets $$\left\{ {b,c,d,e} \right\}$$ and $$X$$ containing $$b$$ also contain a point of $$A$$ different from $$b$$. Similarly, $$d$$ and $$e$$ are also limit points of $$A$$. This illustration suggests that a set can have more than one limit point.

**Derived Set**

Let $$\left( {X,\tau } \right)$$ be a topological space, and let $$A$$ be a subset of $$X$$. The set of all limit points of $$A$$ is said to be the derived set and is denoted by $$D\left( A \right)$$ or $${A^d}$$. In the above example, $$D\left( A \right) = \left\{ {b,d,e} \right\}$$.

**Remark:**

It may be noted that under usual topology, consider the subsets $$\left[ {a,b} \right]$$, $$\left( {a,b} \right)$$, $$\left[ {a,b} \right)$$, $$\left( {a,b} \right]$$ of real, then all the points of these intervals are limits points.

samina semab

August 8@ 6:30 pm{Intersection of A and U}/{x} = empty set

This definition is wrong, because according to the def of limit point x is the limit point of A if for each open set U containing x must contains a point of A which is different from x.

So, {intersection of A and U}/{x} is not equal to empty set.

Zahid Ali

February 8@ 10:02 amYes. There is a mistake in definition but overall illustration of limit point is tremendous…

Jam Azhar

May 19@ 2:55 pmHow we can find limit point of an infinite set

Wanjala Nelson

October 2@ 12:05 pm{intersection of A and U}\{x} does not give you an empty set. Instead, we should consider the definition to be the difference of the open set, say U, and the limit point {x} of the set then you intersect with the subset, A, of the set X.

i.e (U\{x} intersection A)