The Closure of a Set is the Union of the Set and Its Limit Points

The closure of a set \( A \) in a topological space \( X \), denoted by \(\text{Cl}(A)\), is the union of the set \( A \) and the set \( A' \) of its limit points. $$ \text{Cl}(A) = A \cup A' $$

This theorem explains the concept of the closure of a subset \( A \) in a topological space \((X, \tau)\).

The closure of a set \( A \) includes all points that are "close" to \( A \) in the topological sense.

It is important to note that the limit points are not necessarily elements of the set \( A \).

From this theorem, we can infer that a set \( A \) is closed if and only if it contains all its limit points. $$ A \text{ is closed } \ \Leftrightarrow \ A = A \cup A' = \text{Cl}(A) $$ In other words, a set is closed if and only if it is equal to its closure.

A Practical Example

Consider the set \( A = (0, 1) \) in the Euclidean space \(\mathbb{R}\) with the standard topology.

$$ A = (0,1) $$

This set consists of all real numbers between 0 and 1, excluding 0 and 1 themselves.

Let's identify the limit points:

• Every point \( x \) in the interval (0,1) is a limit point because for any point \( x \in (0,1) \), any neighborhood around it (x-ε, x+ε) contains other points of the set \( A \).
• The endpoint 0 is a limit point of \( A \) because any neighborhood of this point (0,0+ε) contains other points of the set \( A \).
• The endpoint 1 is another limit point of the set \( A \) because any neighborhood around it (1-ε,1) contains other points belonging to the set \( A \).

Therefore, the set \( A' \) of limit points of \( A \) is:

$$ A' = [0,1] $$

The union of the points in the set \( A = (0,1) \) and its limit points \( A' = [0,1] \) is the closure of the set.

$$ \text{Cl}(A) = A \cup A' = [0,1] $$

Since the closure of \( A \) does not coincide with the set \( A \), we can conclude that the set \( A \) is not closed in this topology.

$$ A \ne \text{Cl}(A) $$

Example 2

Consider the set \( B = [0, 1] \) in the real number space \( \mathbb{R} \) with the standard topology.

$$ B = [0,1] $$

The set \( B = [0, 1] \) is the closed interval containing all real numbers \( x \) such that \( 0 \leq x \leq 1 \).

Let's identify the limit points of \( B \)

A point \( x \) is a limit point of \( B \) if every neighborhood of \( x \) contains at least one point of \( B \) other than \( x \) itself.

• If \( x \in (0, 1) \), then for any neighborhood \( U \) of \( x \), there is a \( y \in B \) such that \( y \neq x \). Therefore, every point in \( (0, 1) \) is a limit point.
• If \( x = 0 \) or \( x = 1 \), every neighborhood of \( x \) will contain points of \( B \) (because the neighborhoods of \( 0 \) and \( 1 \) in \(\mathbb{R}\) always include parts of the closed interval \( [0, 1] \)). Therefore, \( 0 \) and \( 1 \) are limit points of \( B \).

Hence, the set of limit points of \( B \) is:

$$ B' = [0, 1] \setminus \{0, 1\} = (0, 1) \cup \{0, 1\} = [0, 1] $$

Now, let's calculate the closure of \( B \).

The closure of \( B \), denoted \( \text{Cl}(B) \), is the union of \( B \) with the set of its limit points.

$$ \text{Cl}(B) = B \cup B' = [0, 1] \cup [0, 1] = [0, 1] $$

In this case, the set \( B \) is closed because it coincides with its closure.

$$ B = \text{Cl}(B) = [0, 1] $$

This example confirms that a set is closed if and only if it is equal to its closure.

The Proof

Here is a proof that the closure of a set \( A \) in a topological space \( X \) is the union of \( A \) and the set \( A' \) of its limit points, i.e., \( \text{Cl}(A) = A \cup A' \).

First, let's define the terms closure and limit point:

• Closure of \( A \): \( \text{Cl}(A) \) is the intersection of all closed sets that contain \( A \).
• Limit point: A point \( x \in X \) is a limit point of \( A \) if every neighborhood of \( x \) contains at least one point of \( A \) different from \( x \).

We need to prove that \( \text{Cl}(A) = A \cup A' \), where \( A' \) is the set of limit points of \( A \).

We will divide this proof into three steps:

1] The Union \( A \cup A' \) is a Subset of the Closure of \( A \)

By definition, \( \text{Cl}(A) \) is the intersection of all closed sets that contain \( A \).

$$ A \subseteq \text{Cl}(A) $$

Consider a point \( x \in A' \), the set of limit points of \( A \).

By the definition of a limit point, every neighborhood of \( x \) contains a point of \( A \) different from \( x \).

Assume, for contradiction, that \( x \notin \text{Cl}(A) \), meaning it does not belong to the closure of \( A \).

If \( x \notin \text{Cl}(A) \), then there exists a neighborhood \( U \) of \( x \) such that \( U \cap \text{Cl}(A) = \emptyset \).

Since \( A \subseteq \text{Cl}(A) \), we have \( U \cap A = \emptyset \).

This contradicts the fact that \( x \in A' \) is a limit point because, by definition, every neighborhood of \( x \) must intersect \( A \) in at least one point other than \( x \).

Therefore, the assumption \( x \notin \text{Cl}(A) \) is false, and consequently, the opposite statement \( x \in \text{Cl}(A) \) is true.

Thus, the set \( A' \) of limit points is a subset of the closure of \( A \).

$$ A' \subseteq \text{Cl}(A) $$

We have now shown that both \( A \) and \( A' \) are subsets of the closure of \( A \).

$$ A \subseteq \text{Cl}(A) $$

$$ A' \subseteq \text{Cl}(A) $$

Therefore, their union \( A \cup A' \) is also a subset of the closure of \( A \).

$$ A \cup A' \subseteq \ text{Cl}(A) $$

2] The Closure of \( A \) is a Subset of the Union \( A \cup A' \)

Consider any point \( x \in \text{Cl}(A) \).

Assume for the sake of argument that \( x \notin A \).

Since \( x \in \text{Cl}(A) \) but \( x \notin A \), every neighborhood \( U \) of \( x \) must intersect \( A \).

If there existed a neighborhood \( U \) of \( x \) such that \( U \cap A = \emptyset \), \( x \) would not be in \( \text{Cl}(A) \).

Therefore, for every neighborhood \( U \) of \( x \), \( U \cap A \neq \emptyset \), meaning \( x \) is a limit point of \( A \).

Hence, \( x \in A' \).

Thus:

\[ x \in A \cup A' \]

We have shown that:

\[ \text{Cl}(A) \subseteq A \cup A' \]

3] Conclusion

Knowing from the results obtained that:

$$ A \cup A' \subseteq \text{Cl}(A) $$

$$ \text{Cl}(A) \subseteq A \cup A' $$

We can conclude that the closure of a set \( A \) in a topological space \( X \) is exactly the union of \( A \) with the set of its limit points.

\[ \text{Cl}(A) = A \cup A' \]

And so on.