Boundary of the Interval [-1, 1] in Discrete Topology

In this exercise, we consider the set $ A = [-1, 1] $ within the real numbers $ \mathbb{R} $.

We need to determine the boundary $ \partial A $ of the set $ A $.

The boundary of $ A $ is defined as the set of points that lie in the closure of $ A $ but not in its interior.

$$ \partial A = \text{Cl}(A) - \text{Int}(A) $$

The discrete topology on a set is a structure where all subsets are considered open.

In this topology, every subset of $ A $ is open. This implies that any set, including $ A $, is open by definition.

The interior of $ A $ consists of the points within $ A $ that are "surrounded" by $ A $.

Thus, in discrete topology, the interior of $ A $ is $ A $ itself.

$$ \text{Int}(A) = [-1, 1] $$

Explanation. The interior of a set $ A $ is defined as the set of points in $ A $ that are contained in an open neighborhood entirely within $ A $. In discrete topology, every point of $ A $ is an interior point because we can always find a neighborhood (in this case, the point itself) that is contained in $ A $. Hence, Int(A)=A.

The closure of $ A $ is the set $ A $ itself along with any of its limit points.

In discrete topology, the closure of $ A $ remains $ A $.

$$ \text{Cl}(A) = [-1, 1] $$

Explanation. The closure of a set $ A $ includes all points in $ A $ plus all its limit points. In discrete topology, there are no limit points because each point is isolated. Thus, the closure of $ A $ is simply $ A $ itself, Cl(A)=A.

Since $ \text{Int}(A) = \text{Cl}(A) $, there are no points that lie in the closure of $ A $ but not in its interior. Therefore, the boundary $ \partial A $ is empty ($ \varnothing $).

$$ \partial A = \text{Cl}(A) - \text{Int}(A) $$

$$ \partial A = \emptyset $$

Alternative Solution

The closure of $ A $ is equal to the interior of $ A $.

$$ \text{Int}(A) = \text{Cl}(A) $$

Therefore, the set $[-1, 1]$ is both open and closed in discrete topology.

Explanation. In discrete topology, every subset is considered open. This means that any set, including $ A $, is open by definition. Additionally, every open set is also closed. This is because the complement of an open set is closed and vice versa. Since every subset of $[-1, 1]$ is open, its complement is also open, making every set closed as well.

According to a property of boundaries, a set that is both open and closed has an empty boundary.

$$ \partial A = \emptyset $$

And so on.