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

In this example, we will compute the boundary \(\partial A\) of the set \(A = (0,1) \cup \{2\}\) in the standard topology on \(\mathbb{R}\).

A convenient way to find the boundary of a set is to subtract its interior from its closure:

$$ \partial A = \operatorname{Cl}(A) \setminus \operatorname{Int}(A). $$

Let's determine each of these two sets step by step.

Step 1. Find the Closure of the Set

The closure of a set is the smallest closed set that contains it.

In the standard topology on \(\mathbb{R}\), the closure of the open interval \((0,1)\) is the closed interval \([0,1]\). The singleton \(\{2\}\) is already closed, so its closure is simply \(\{2\}\).

Therefore,

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

Step 2. Find the Interior of the Set

The interior of a set is the largest open set contained within it.

The interval \((0,1)\) is open, so every point in this interval is an interior point of \(A\). The isolated point \(2\), however, is not an interior point because no open interval centered at \(2\) is entirely contained in \(A\).

Hence,

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

Step 3. Compute the Boundary

Now subtract the interior from the closure:

$$ \partial A = \operatorname{Cl}(A) \setminus \operatorname{Int}(A) $$

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

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

Therefore, the boundary of \(A\) consists of the three points \(0\), \(1\), and \(2\).

Alternative Method

You can also compute the boundary by using the following identity:

$$ \partial A = \operatorname{Cl}(A) \cap \operatorname{Cl}(\mathbb{R} \setminus A). $$

We have already found that

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

Next, determine the complement of \(A\):

$$ \mathbb{R} \setminus A = \mathbb{R} \setminus \bigl((0,1) \cup \{2\}\bigr) $$

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

Now compute its closure. The three components become:

• \((-\infty,0]\), which is already closed.
• \([1,2]\), obtained by taking the closure of \([1,2)\).
• \([2,\infty)\), obtained by taking the closure of \((2,\infty)\).

Combining these intervals gives

$$ \operatorname{Cl}(\mathbb{R} \setminus A) = (-\infty,0] \cup [1,\infty). $$

Finally, intersect the two closures:

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

The intersection contains exactly the points \(0\), \(1\), and \(2\).

Therefore,

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

Both methods lead to the same result, confirming that the boundary of the set is

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