Boundary of \(A = (-1,1]\) in the Lower Limit Topology on \(\mathbb{R}\)
In this example, we find the boundary of the set \(A = (-1,1]\) in the lower limit topology on \(\mathbb{R}\).
Recall that in the lower limit topology, the basis consists of all half-open intervals of the form \([a,b)\), where \(a< b\). This topological space is commonly known as the Sorgenfrey line.
To determine the boundary, we first compute the interior and the closure of the set. The boundary is then given by
$$ \partial A=\operatorname{Cl}(A)\setminus\operatorname{Int}(A). $$
Step 1. Find the interior
The interior of a set is the largest open set contained in it. Equivalently, a point is an interior point if it has a basic open neighborhood lying entirely inside the set.
Every point \(x\) with \(-1<x<1\) satisfies this condition. In fact, we can always choose a basic open interval \([x,b)\) with \(x< b\leq1\), which is completely contained in \(A\).
The point \(1\), however, is not an interior point. Every basic open neighborhood of \(1\) has the form \([1,b)\), where \(b>1\), so it always contains points that are not in \(A\).
Therefore,
$$ \operatorname{Int}(A)=(-1,1). $$
Step 2. Find the closure
A point belongs to the closure of a set if every open neighborhood of that point intersects the set.
The endpoint \(-1\) belongs to the closure because every basic open neighborhood of \(-1\) is of the form \([-1,b)\), with \(b>-1\), and always intersects \(A\).
Every point of \(A\) is, of course, contained in its own closure.
On the other hand, if \(x<-1\), we can choose a basic open neighborhood that lies entirely to the left of \(-1\), so it does not intersect \(A\). Likewise, if \(x>1\), every sufficiently small basic open neighborhood of \(x\) is disjoint from \(A\).
Hence,
$$ \operatorname{Cl}(A)=[-1,1]. $$
Step 3. Compute the boundary
Now subtract the interior from the closure:
$$ \partial A =\operatorname{Cl}(A)\setminus\operatorname{Int}(A) =[-1,1]\setminus(-1,1) =\{-1,1\}. $$
Therefore, the boundary of the set \(A=(-1,1]\) in the lower limit topology on \(\mathbb{R}\) is
$$ \boxed{\partial A=\{-1,1\}.} $$
