Boundary of the Interval [-1,1] in the Lower Limit Topology

In this example, we determine the boundary of the interval \(A=[-1,1]\) when the real line is equipped with the lower limit topology.

Unlike the usual topology on \(\mathbb{R}\), the lower limit topology uses half-open intervals of the form \([a,b)\) as its basic open sets. Each such interval contains its left endpoint but excludes its right endpoint. This seemingly small difference significantly changes many topological properties, including the notions of interior and boundary.

To find the boundary of \(A\), we use the definition

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

We therefore need to determine both the closure and the interior of the interval.

Closure

The closure of a set is the smallest closed set containing it. Equivalently, it consists of all points whose every open neighborhood intersects the set.

For the interval \([-1,1]\), the closure is simply the interval itself:

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

Interior

The interior of a set is the collection of all points that have an open neighborhood lying entirely within the set.

Every point \(x<1\) belongs to the interior because we can choose a basic open neighborhood of the form \([x,b)\) with \(b\leq1\), ensuring that

$$ [x,b)\subseteq[-1,1]. $$

The point \(1\), however, is different. Every basic open neighborhood of \(1\) has the form \([1,b)\) with \(b>1\), so it necessarily contains points outside the interval. Consequently, \(1\) is not an interior point.

Therefore,

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

Boundary

Substituting the closure and the interior into the definition of the boundary, we obtain

$$ \partial A=[-1,1]\setminus[-1,1). $$

Hence, the boundary consists of the single point

$$ \partial A=\{1\}. $$

This example illustrates one of the characteristic features of the lower limit topology: unlike the usual topology on the real line, only the right endpoint of the interval belongs to its boundary.