Definition 10.9.
The closure of a subset \(A\) of a metric space \(X\) is the set
\begin{equation*}
\overline{A} = A \cup A'\text{.}
\end{equation*}
Section The Closure of a Set
We have seen that the interior of a set is the largest open subset of that set. There is a similar result for closed sets. For example, let \(A = (0,1)\) in \((\R, d_E)\text{.}\) The set \(A\) is an open set, but if we union \(A\) with its limit points, we obtain the closed set \(C = [0,1]\text{.}\) Moreover, The set \([0,1]\) is the smallest closed set that contains \(A\text{.}\) This leads to the idea of the closure of a set.
In other words, the closure of a set is the collection of the elements of the set and the limit points of the set — those points that are on the “edge” of the set. The importance of the closure of a set \(A\) is that the closure of \(A\) is the smallest closed set that contains \(A\text{.}\)
Theorem 10.10.
Let \(X\) be a metric space and \(A\) a subset of \(X\text{.}\) The closure of \(A\) is a closed set. Moreover, the closure of \(A\) is the smallest closed subset of \(X\) that contains \(A\text{.}\)
Proof.
Let \(X\) be a metric space and \(A\) a subset of \(X\text{.}\) To prove that \(\overline{A}\) is a closed set, we will prove that \(\overline{A}\) contains its limit points. Let \(x \in \overline{A}'\text{.}\) To show that \(x \in \overline{A}\text{,}\) we proceed by contradiction and assume that \(x \notin \overline{A}\text{.}\) This implies that \(x \notin A\) and \(x \notin A'\text{.}\) Since \(x \notin A'\text{,}\) there exists a neighborhood \(N\) of \(x\) that contains no points of \(A\) other than \(x\text{.}\) But \(A \subseteq \overline{A}\) and \(x \notin \overline{A}\text{,}\) so it follows that \(N \cap A = \emptyset\text{.}\) This implies that there is an open ball \(B \subseteq N\) centered at \(x\) so that \(B \cap A = \emptyset\text{.}\) The fact that \(x \in \overline{A}'\) means that \(B \cap \overline{A}\) contains a point \(y\) in \(\overline{A}\) different from \(x\text{.}\) Since \(B \cap A = \emptyset\text{,}\) we must have \(y \in A'\text{.}\) But this, and the fact that \(B\) is a neighborhood of \(y\text{,}\) means that \(B\) must contain a point of \(A\) different than \(y\text{.}\) But \(B \cap A = \emptyset\text{,}\) so we have reached a contradiction. We conclude that \(x \in \overline{A}\) and \(\overline{A}' \subseteq \overline{A}\text{.}\) This shows that \(\overline{A}\) is a closed set.
The proof that \(\overline{A}\) is the smallest closed subset of \(X\) that contains \(A\) is left for the next activity.
Activity 10.9.
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 \(\overline{A}\) is the smallest closed subset of \(X\) that contains \(A\text{?}\)
(b)
Suppose that \(C\) is a closed subset of \(X\) that contains \(A\text{.}\) To show that \(\overline{A} \subseteq C\text{,}\) why is it enough to demonstrate that \(A' \subseteq C\text{?}\)
(c)
If \(x \in A'\text{,}\) what can we say about \(x\text{?}\)
(d)
Complete the proof that \(\overline{A} \subseteq C\text{.}\)
One consequence of Theorem 10.10 is the following.
Corollary 10.11.
A subset \(C\) of a metric space \(X\) is closed if and only if \(C = \overline{C}\text{.}\)
We can also characterize closed sets as sets that contain their boundaries.
Definition 10.12.
The boundary \(\Bdry(A)\) of a subset \(A\) of a metric space \(X\) is the set of all boundary points of \(A\text{.}\)
Theorem 10.13.
A subset \(C\) of a metric space \(X\) is closed if and only if \(C\) contains its boundary.
The proof of Theorem 10.13 is left to Exercise 10.
Recall that a boundary point of a subset \(A\) of a metric space \(X\) is a point \(x \in X\) such that every neighborhood of \(x\) contains a point in \(A\) and a point in \(X \setminus A\text{.}\) The boundary points are those that are somehow “between” a set and its complement. For example if \(A = (0,1]\) in \(\R\text{,}\) the boundary of \(A\) is the set \(\{0,1\}\text{.}\) We also have that \(\overline{A} = [0,1]\text{,}\) \(\R \setminus A = (-\infty, 0] \cup (1, \infty)\text{,}\) and \(\overline{\R \setminus A} = (-\infty, 0] \cup [1, \infty)\text{.}\) Notice that \(\Bdry(A) = \overline{A} \cap \overline{\R \setminus A}\text{.}\) That this is always true is formalized in the next theorem.
Theorem 10.14.
Let \(X\) be a metric space and \(A\) a subset of \(X\text{.}\) Then
\begin{equation*}
\Bdry(A) = \overline{A} \cap \overline{X \setminus A}\text{.}
\end{equation*}
Proof.
Let \(X\) be a metric space and \(A\) a subset of \(X\text{.}\) To prove \(\Bdry(A) = \overline{A} \cap \overline{X \setminus A}\) we need to verify the containment in each direction. Let \(x \in \Bdry(A)\) and let \(N\) be a neighborhood of \(x\text{.}\) Then \(N\) contains a point in \(A\) and a point in \(X \setminus A\text{.}\) We consider the cases of \(x \in A\) or \(x \notin A\text{.}\)
- Suppose \(x \in A\text{.}\) Then \(x \in \overline{A}\text{.}\) Also, \(x \notin X \setminus A\text{,}\) so \(N\) must contain a point in \(X \setminus A\) different from \(x\text{.}\) That makes \(x\) a limit point of \(X \setminus A\) and so \(X \in \overline{A} \cap \overline{X \setminus A}\text{.}\)
- Suppose \(x \notin A\text{.}\) Then \(x \in X \setminus A \subseteq \overline{X \setminus A}\text{.}\) Also, \(x \notin A\text{,}\) so \(N\) must contain a point in \(A\) different from \(x\text{.}\) That makes \(x\) a limit point of \(A\) and so \(X \in \overline{A} \cap \overline{X \setminus A}\text{.}\)
In either case we have \(x \in \overline{A} \cap \overline{X \setminus A}\) and so \(\Bdry(A) \subseteq \overline{A} \cap \overline{X \setminus A}\text{.}\)
For the reverse implication, refer to the next activity.
Activity 10.10.
Let \(X\) be a metric space and \(A\) a subset of \(X\text{.}\) In this activity we prove that
\begin{equation*}
\overline{A} \cap \overline{X \setminus A} \subseteq \Bdry(A)\text{.}
\end{equation*}
Let \(x \in \overline{A} \cap \overline{X \setminus A}\text{.}\)
(a)
What must be true about \(x\text{,}\) given that \(x\) is in the intersection of two sets?
(b)
Let \(N\) be a neighborhood of \(x\text{.}\) As we did in the proof of Theorem 10.14, we consider the cases \(x \in A\) and \(x \notin A\text{.}\)
(i)
Suppose \(x \in A\text{.}\) Why must \(N\) contain a point in \(A\) and a point not in \(A\text{?}\) What does this tell us about \(x\text{?}\)
(ii)
Suppose \(x \notin A\text{.}\) Why must \(N\) contain a point in \(A\) and a point not in \(A\text{?}\) What does this tell us about \(x\text{?}\)
(iii)
What can we conclude from parts (i) and (ii)?
