Boundary of the Set A=(0,1)U{2} in the Lower Limit Topology

To determine the boundary ($\partial A$) of the set $A = (0,1) \cup \{2\}$ in the lower limit topology on $\mathbb{R}$, we need to find the closure $\text{Cl}(A)$ and the interior $\text{Int}(A)$ of $A$, and then calculate the boundary as the difference between the closure and the interior:

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

In the lower limit topology, the basis for open sets consists of intervals of the form $[a, b)$ where $a < b$.

The set $A = (0,1) \cup \{2\}$ includes all points between 0 and 1 (excluding 0 and 1) and the point 2.

To find the closure of $A$, we need to include all points in $A$ as well as the limit points of $A$.

In the lower limit topology, the closure of $(0,1)$ is $[0,1]$ while the point $\{2\}$ is already closed.

$$\text{Cl}(A) = [0,1] \cup \{2\}$$

The interior of $A$ is the union of all open sets contained within $A$.

In the lower limit topology, $(0,1)$ is not an open set.

The open sets that can be contained in $(0,1)$ are of the form $[a,b)$ with $0 < a < b \leq 1$.

The set $\{2\}$ does not have an interval of the form $[2, b)$ contained within $A$, so it has no open interior.

Thus, the interior of $A$ is the union of all intervals $[a, b)$ that are contained within $(0,1)$:

$$\text{Int}(A) = \bigcup_{0 < a < b < 1} [a, b)$$

$$\text{Int}(A) = (0,1)$$

Now, we calculate the difference between the closure and the interior of $A$:

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

$$\partial A = ([0,1] \cup \{2\}) - (0,1)$$

This means we need to remove the points in the interior from the closed set:

$$\partial A = \{0 , 1 \} \cup \{2\}$$

Therefore, the boundary of $A = (0,1) \cup \{2\}$ in the lower limit topology on $\mathbb{R}$ is $\partial A = \{0, 1, 2\}$

Alternative Method

We can also calculate the boundary ($\partial A$) of the set $A = (0,1) \cup \{2\}$ in the lower limit topology on $\mathbb{R}$ by using the definition of the boundary as the intersection of the closure of $A$ and the closure of the complement of $A$:

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

We already know that the closure of $A$ is:

$$\text{Cl}(A) = [0,1] \cup \{2\}$$

The complement of $A$ is:

$$X - A = \mathbb{R} - ((0,1) \cup \{2\})$$

$$X - A = (-\infty, 0] \cup [1,2) \cup (2, \infty)$$

To find the closure of this set in the lower limit topology, we consider the limits:

• $(-\infty, 0]$ is already closed.
• $[1, 2)$ is open in this topology because it includes all points up to but not including 2. Its closure is $[1,2]$.
• $(2, \infty)$ is closed in this topology

Thus:

$$\text{Cl}(X - A) = (-\infty, 0] \cup [1, 2] \cup (2, \infty)$$

Now, we find the intersection of the closures:

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

$$\partial A = \left( [0,1] \cup \{2\} \right) \cap \left( (-\infty, 0] \cup [1, 2] \cup (2, \infty) \right)$$

The intersection of these sets is:

$$\partial A = \{0 , 1, 2 \}$$

Therefore, in the lower limit topology on $\mathbb{R}$, the boundary of $A = (0,1) \cup \{2\}$ is $\partial A = \{0, 1, 2\}$

And so on.