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

In this exercise, we have a closed interval $ A = [-1, 1] $ from -1 to 1 in the lower limit topology, and we need to determine its boundary.

The real line $\mathbb{R}_l$ in the lower limit topology differs from the standard topology.

In this topology, open sets are of the form $[a, b)$, meaning they include the lower bound (point a) but not the upper bound (point b).

The boundary of $A$ is the set of points that belong to the closure of $A$ but not to the interior of $A$.

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

The closure of $A$ is the set of all points in $A$ plus the accumulation points of $A$.

In the lower limit topology, the closure of $[-1, 1]$ remains $[-1, 1]$, i.e., $\text{Cl}(A) = [-1, 1]$.

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

The interior of $A$ is the set of all points in $A$ that have a neighborhood entirely contained within $A$.

In the lower limit topology, the interior of $[-1, 1]$ is $[-1, 1)$, i.e., $\text{Int}(A) = [-1, 1)$ because any open interval within $ A $ starting from -1 or any point greater than -1 and ending before 1 is contained in $[-1, 1]$. However, we cannot include 1 because there is no open interval ending exactly at 1 that meets the lower limit topology criteria.

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

In this case, $\partial A = \{ 1 \}$, because 1 is the only point not in the interior of $A$ but in the closure of $A$.

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

And so on.