Boundary of $A = (-1,1]$ in the Lower Limit Topology

In this exercise, we need to determine the boundary of $A = (-1, 1]$ within the context of the lower limit topology on $ \mathbb{R} $.

In the lower limit topology, open sets are of the form $[a, b)$ where $a < b$.

The boundary of $A$ is defined as the difference between the closure of $A$ and the interior of $A$. Formally,

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

The interior of $A = (-1, 1]$ in the lower limit topology is the largest open set contained within $A$.

To find the interior, observe that any interval $[a, b)$ for $-1 < a < b \leq 1$ is contained within $(-1, 1]$.

However, we cannot include the right endpoint $1$ in a set of the form $[a, b)$.

Thus, the interior of $A$ is $(-1, 1)$.

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

The closure of $A = (-1, 1]$ is the smallest closed set that contains $A$.

A closed set in the lower limit topology is the complement of an open set of the form $[a, b)$.

The set $(-1, 1]$ is already closed in this topology because it includes all its limit points, and no open set of the form $[a, b)$ can have a complement that excludes points from $(-1, 1]$.

Therefore, the closure of $A$ is $(-1, 1]$.

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

Given the interior $ \text{Int}(A) = (-1,1) $ and the closure $ \text{Cl}(A) = (-1,1] $, we can determine the boundary of $A$.

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

$$ \partial A = (-1,1] - (-1,1) $$

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

Therefore, the boundary of the set $A = (-1, 1]$ in the lower limit topology on $\mathbb{R}$ is $\{1\}$.

And so on.

Boundary of \(A = (-1,1]\) in the Lower Limit Topology