Theorem 8.8.
Let \(X\) be a metric space. A subset \(O\) of \(X\) is open if and only if every point of \(O\) is an interior point of \(O\text{.}\)
Section The Interior of a Set
Open sets can be characterized in terms of their interior points. By definition, every open set is a neighborhood of each of its points, so every point of an open set \(O\) is an interior point of \(O\text{.}\) Conversely, if every point of a set \(O\) is an interior point, then \(O\) is a neighborhood of each of its points and is open. This argument is summarized in the next theorem.
The collection of interior points in a set form a subset of that set, called the interior of the set.
Definition 8.9.
The interior of a subset \(A\) of a metric space \(X\) is the set
\begin{equation*}
\Int(A) = \{a \in A \mid a \text{ is an interior point of } A\}\text{.}
\end{equation*}
Activity 8.7.
Determine \(\Int(A)\) for each of the sets \(A\text{.}\)
(a)
\(A = (0,1]\) in \((\R, d_E)\)
(b)
\(A = [0,1]\) in \((\R, d_E)\)
(c)
\(A = \{-2\} \cup [0,5] \cup \{7,8,9\}\) in \((\R,d_E)\)
One might expect that the interior of a set is an open set. This is true, but we can say even more. As Theorem 8.10 will show, if \(A\) is a subset of a metric space \(X\text{,}\) not only is \(\Int(A)\) an open set, but every open set that is contained in \(A\) is a subset of \(\Int(A)\text{.}\) So \(\Int(A)\) is the largest, in the sense of containment, open subset of \(X\) that contains \(A\text{.}\)
Theorem 8.10.
Let \((X,d)\) be a metric space, and let \(A\) be a subset of \(X\text{.}\) Then interior of \(A\) is the largest open subset of \(X\) contained in \(A\text{.}\)
Proof.
Let \((X,d)\) be a metric space, and let \(A\) be a subset of \(X\text{.}\) We need to prove that \(\Int(A)\) is an open set in \(X\text{,}\) and that \(\Int(A)\) is the largest open subset of \(X\) contained in \(A\text{.}\) First we demonstrate that \(\Int(A)\) is an open set. Let \(a \in \Int(A)\text{.}\) Then \(a\) is an interior point of \(A\text{,}\) so \(A\) is a neighborhood of \(a\text{.}\) This implies that there exists an \(\epsilon \gt 0\) so that \(B(a, \epsilon) \subseteq A\text{.}\) But \(B(a, \epsilon)\) is a neighborhood of each of its points, so every point in \(B(a, \epsilon)\) is an interior point of \(A\text{.}\) It follows that \(B(a, \epsilon) \subseteq \Int(A)\text{.}\) Thus, \(\Int(A)\) is a neighborhood of each of its points and, consequently, \(\Int(A)\) is an open set.
The proof that \(\Int(A)\) is the largest open subset of \(X\) contained in \(A\) is left for the next activity.
Activity 8.8.
Let \((X,d)\) be a metric space, and let \(A\) be a subset of \(X\text{.}\)
(a)
What will we have to show to prove that \(\Int(A)\) is the largest open subset of \(X\) contained in \(A\text{?}\)
(b)
Suppose that \(O\) is an open subset of \(X\) that is contained in \(A\text{,}\) and let \(x \in O\text{.}\) What does the fact that \(O\) is open tell us?
(c)
Complete the proof that \(O \subseteq \Int(A)\text{.}\)
One consequence of Theorem 8.10 is the following.
Corollary 8.11.
A subset \(O\) of a metric space \(X\) is open if and only if \(O = \Int(O)\text{.}\)
The proof is left for Exercise 2.
