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

In this exercise, I need to determine the boundary (∂A) of the set $ A = (0, 1) \cup \{2\} $ in the standard topology on $\mathbb{R}$.

The boundary of a set is the difference between its closure and its interior.

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

The closure of $A$ is the smallest closed set containing $A$.

In the standard topology on $\mathbb{R}$, the closure of $(0, 1)$ is $[0, 1]$. Including the singleton $\{2\}$, the closure of $A$ becomes $[0, 1] \cup \{2\}$.

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

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

In the standard topology on $\mathbb{R}$, the set $(0, 1)$ is already open. However, $\{2\}$ contains no open set as it is a singleton.

Thus, the interior of $ A $ is:

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

Now that we know the closure and the interior of $ A $, we can calculate the boundary $ \partial A $:

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

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

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

In conclusion, the boundary of $ A $ consists of the three elements $ \partial A = \{0, 1, 2\} $.

Alternative Method

We can also determine the boundary of $ A $ by finding the intersection of the closure of $ A $ with the closure of its complement $ \mathbb{R}-A $.

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

We already know the closure of the set $ A $ in this topology is:

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

The complement of $A$ is:

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

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

The closure of the complement of $A$ is the smallest closed set containing the complement.

The closure of $(-\infty, 0]$ is $(-\infty, 0]$.
The closure of $[1, 2)$ is $[1, 2]$.
The closure of $(2, \infty)$ is $(2, \infty)$.

Therefore, the closure of the complement of $ A $ is:

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

Now, we find the intersection of the closures:

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

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

The intersection of $[0, 1]$ with $(-\infty, 0]$ is $\{0\}$. The intersection of $[0, 1]$ with $[1, 2]$ is $\{1\}$. The intersection of $\{2\}$ with $[1, 2]$ is $\{2\}$.

Therefore, the boundary of $ A $ in the standard topology on $\mathbb{R}$ is:

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