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

In this exercise, we consider the interval $ A=[-1,1] $ as a subset of the real line $ \mathbb{R} $, equipped with the discrete topology.

Our goal is to determine the boundary of the set $ A $.

Recall that the boundary of a set is defined as the difference between its closure and its interior:

$$ \partial A=\operatorname{Cl}(A)\setminus\operatorname{Int}(A) $$

To compute the boundary, we first determine the interior and the closure of $ A $.

The Interior of the Set
The Closure of the Set
Computing the Boundary
Alternative Solution

The Interior of the Set

In the discrete topology, every subset of the space is open.

Since $ A $ is itself a subset of $ \mathbb{R} $, it is automatically an open set.

Consequently, every point of $ A $ is an interior point. In fact, for every $ x\in A $, the singleton $ \{x\} $ is an open neighborhood entirely contained in $ A $.

Therefore,

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

Explanation. The interior of a set is the largest open set contained in it. Since every subset is open in the discrete topology, the set $ A $ is already open. As a result, its interior is simply the set itself.

The Closure of the Set

Next, we compute the closure of $ A $.

The closure of a set is the smallest closed set containing it.

In the discrete topology, every subset is closed as well as open, because the complement of any subset is again open.

Since $ A $ is already closed, taking its closure does not add any new points.

Thus,

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

Explanation. Every point in a discrete topological space is isolated. As a result, a set has no limit points outside itself, so its closure coincides with the set.

Computing the Boundary

Now substitute the interior and the closure into the definition of the boundary:

$$ \partial A =\operatorname{Cl}(A)\setminus\operatorname{Int}(A) =[-1,1]\setminus[-1,1]. $$

Hence,

$$ \partial A=\emptyset. $$

The interval $ [-1,1] $ has no boundary in the discrete topology because its closure and its interior are identical.

Alternative Solution

There is an even shorter argument.

Every subset of a discrete topological space is both open and closed. Such sets are called clopen.

Since the interval $ [-1,1] $ is clopen, its boundary must be empty.

Explanation. A standard result in topology states that a set has an empty boundary if and only if it is both open and closed. Because every subset of a discrete space is clopen, every subset has an empty boundary.

Therefore,

$$ \partial A=\emptyset. $$