# Dense Sets in Topology

In a topological space X, a subset A is a **dense set** if its closure encompasses the whole space X. $$ Cl(A)=X $$

Essentially, a dense set intersects every part of the space, meaning every point within the space either belongs to the subset or is a limit point of the subset.

The closure of A encompasses all points within A as well as its limit points.

## Practical Examples

**Example 1**

In the standard topology on \( \mathbb{R} \), the subset consisting of all rational numbers (\( \mathbb{Q} \subset \mathbb{R} \)) is considered dense.

This occurs because between any two distinct real numbers, there are rational numbers. Consequently, every real number can be approached infinitely closely by rational numbers.

In this scenario, the closure of the rational numbers is precisely the entire space \( \mathbb{R} \).

$$ Cl ( \mathbb{Q} ) = \mathbb{R} $$

Thus, it meets the criteria for a dense set.

**Note**. By the same token, in the standard topology on \( \mathbb{R} \), the set of irrational numbers \( \mathbb{I} \subset \mathbb{R} \) is dense in \( \mathbb{R} \) for analogous reasons. Every real number can be approximated arbitrarily closely by irrational numbers. Therefore, enclosing the set of irrational numbers requires including every point from the set of real numbers. $$ Cl ( \mathbb{I} ) = \mathbb{R} $$

**Example 2**

In the finite complement topology on \( \mathbb{R} \), the set \( \mathbb{R} \setminus \{0\} \) is dense.

Under this topology, a set is open if its complement within \( \mathbb{R} \) is finite.

Given that the complement of \( \mathbb{R} \setminus \{0\} \) consists only of the finite set \{ 0 \}, it follows that \( \mathbb{R} \setminus \{0\} \) is an open set.

To determine the closure of \( \mathbb{R} \setminus \{0\} \), all accumulation points must be considered.

Since adding \( 0 \) back to \( \mathbb{R} \setminus \{0\} \) results in the entire \( \mathbb{R} \), it is clear that the only closed set that can fully contain \( \mathbb{R} \setminus \{0\} \) is \( \mathbb{R} \) itself.

Thus, the closure aligns perfectly with the entire topological space.

$$ Cl( \mathbb{R} \setminus \{0\} ) = \mathbb{R} $$

This indicates that the set \( \mathbb{R} \setminus \{0\} \) is dense in \( \mathbb{R} \).

**Note**. This characteristic of density illustrates the distinctive nature of the finite complement topology, in which all infinite sets are inherently dense. In this topology, the closed sets are the finite sets. Therefore, the only closed set capable of containing an infinite set is the entire set \( \mathbb{R} \).

**Example 3**

In the standard topology on \( \mathbb{R} \), the interval (0,1) is not dense.

The closure of the interval (0,1) is the interval [0,1], including the endpoints 0 and 1, since any neighborhood around these points intersects the interval (0,1).

Thus, the closure does not span the entire set \( \mathbb{R} \).

Consequently, the interval (0,1) does not qualify as a dense set in the standard topology on \( \mathbb{R} \).

**Note**. However, if considered as a subset within the standard topology of the subspace [0,1], then (0,1) becomes dense in [0,1] with the induced topology, as its closure in this subspace precisely matches the entire set [0,1]. This example highlights how the concept of density can vary depending on the context and the space considered: while (0,1) is not dense in \( \mathbb{R} \), it is dense within the subspace [0,1].

And so forth.