**limit point of \( A \)**when every neighborhood of \( x \) intersects \( A \) in some point other than \( x \) itself

Limit Point

Suppose that \( A \) is a subset of a topological space \( X \), then a point \( x \in X \) is called a **limit point of \( A \)** when every neighborhood of \( x \) intersects \( A \) in some point other than \( x \) itself

Limit Point iff element of Closure minus a Point

\( x \) is a limit point iff \( x \in \overline { A \setminus \left\{ x \right\} } \)

Limit Points are a Subset of Closure

Suppose that \( A \) is a subset of a topological space \( X \), if \( A ^ \prime \) is the set of limit points of \( A \), then \( A ^ \prime \subseteq \overline{ A } \)

Suppose that \( x \in A ^ \prime \), then every neighborhood of \( x \) intersects \( A \) therefore \( x \in \overline{ A } \) as needed.

Closure Equals Limit Points Union Itself

Let \( A \) be a subset of a topological space \( X \). If \( A ^\prime \) is the set of all limit points of \( A \), then \( \overline{ A } = A^\prime \cup A \)

Since \( A \subseteq \overline{ A } \) and \( A ^ \prime \subseteq \overline{ A } \), then \( A ^ \prime \cup A \subseteq \overline{ A } \)

Suppose that \( x \in \overline{ A } \), if \( x \in A \), then we would be done, so we assume that \( x \notin A \) and our goal remains to show that \( x \in A ^ \prime \), so let \( U \) be a neighborhood of \( x \), we want to show that it intersects \( A \) at a point different than \( x \), since \( x \in \overline{ A } \), then we know that \( U \) intersects \( A \), since \( x \notin A \) then \( U \) cannot intersect \( A \) at that point so we've shown that \( x \in A ^ \prime \) as needed.

A Subset of a Topological Space is Closed if and only if it contains all it's Limit Points

Suppose \( A \) is a subset of a topological space \( X \) then it is closed if and only if it contains it's limit points

A set is closed iff \( A = \overline{ A } \) since \( \overline{ A } = A \cup A ^ \prime \), then this is true iff \( A ^ \prime \subseteq A \).