IV.5 · Open & Closed Sets — Problem 5 #

← Back to IV. Open & Closed Sets, Problem 5

Problem. Let \(U, V \subseteq S\) in a metric space. (a) Decide, with proof or counterexample, whether (i) \(\overline{U} \cup \overline{V} = \overline{U \cup V}\) and (ii) \(\overline{U} \cap \overline{V} = \overline{U \cap V}\). (b) Show that \(\big(\overline{T}\big)^c = I(T^c)\), where \(I(A)\) is the set of points \(p \in A\) having a ball \(B_\epsilon(p) \subset A\).

Hint #

Two facts do most of the work: closure is monotone (\(A \subseteq B\) implies \(\overline{A} \subseteq \overline{B}\)), and \(\overline{T}\) is the smallest closed set containing \(T\) (Problem IV.3). For the one that fails, look for two sets in \(\mathbb{R}\) that are disjoint but whose closures touch. For (b), unwind what it means not to be in \(T\) and not to be a limit point of \(T\).

Solution #

(a)(i) True. Since \(U \subseteq U \cup V\), every limit point of \(U\) is a limit point of \(U \cup V\), so \(\overline{U} \subseteq \overline{U \cup V}\); likewise \(\overline{V} \subseteq \overline{U \cup V}\). Hence \(\overline{U} \cup \overline{V} \subseteq \overline{U \cup V}\). For the reverse, \(\overline{U} \cup \overline{V}\) is a union of two closed sets, hence closed, and it contains \(U \cup V\). Since \(\overline{U \cup V}\) is the smallest closed set containing \(U \cup V\), we get \(\overline{U \cup V} \subseteq \overline{U} \cup \overline{V}\).

(a)(ii) False. In \((\mathbb{R}, |\cdot|)\) take \(U = (0,1)\) and \(V = (1,2)\). Then \(U \cap V = \varnothing\), so \(\overline{U \cap V} = \varnothing\), while \(\overline{U} \cap \overline{V} = [0,1] \cap [1,2] = \{1\}\). (The containment \(\overline{U \cap V} \subseteq \overline{U} \cap \overline{V}\) does always hold, by monotonicity.)

(b) Let \(p \in S\). Then \(p \notin \overline{T}\) means \(p \notin T\) and \(p\) is not a limit point of \(T\). The latter says there is an \(\epsilon > 0\) such that \(B_\epsilon(p)\) contains no point of \(T\) other than possibly \(p\) itself; since \(p \notin T\), this means \(B_\epsilon(p) \cap T = \varnothing\), i.e. \(B_\epsilon(p) \subset T^c\). So \(p \notin \overline{T}\) iff \(p \in T^c\) and some ball about \(p\) lies in \(T^c\), which is exactly \(p \in I(T^c)\).

Long solution #

Strategy. We use the definition \(\overline{T} = T \cup T'\), where \(T'\) is the set of limit points, together with two consequences proved earlier: \(\overline{T}\) is closed, and it is the smallest closed set containing \(T\) (Problem IV.3). The union identity in (a)(i) is proved by two containments, one from monotonicity and one from minimality. The intersection identity fails because closure can create new points of contact between sets that were disjoint. Part (b) is a direct translation of definitions.

Lemma (monotonicity). If \(A \subseteq B\) then \(\overline{A} \subseteq \overline{B}\).

Proof. If \(p \in A\) then \(p \in B \subseteq \overline{B}\). If \(p \in A'\), then every ball \(B_\epsilon(p)\) contains a point of \(A\) other than \(p\); that point lies in \(B\), so \(p \in B' \subseteq \overline{B}\). \(\square\)

Part (a)(i). First containment. Since \(U \subseteq U \cup V\) and \(V \subseteq U \cup V\), the lemma gives \(\overline{U} \subseteq \overline{U \cup V}\) and \(\overline{V} \subseteq \overline{U \cup V}\), hence \(\overline{U} \cup \overline{V} \subseteq \overline{U \cup V}\).

Second containment. The set \(\overline{U} \cup \overline{V}\) is a union of two closed sets. A finite union of closed sets is closed: its complement \(\big(\overline{U}\big)^c \cap \big(\overline{V}\big)^c\) is a finite intersection of open sets, which is open. Moreover \(U \cup V \subseteq \overline{U} \cup \overline{V}\). By Problem IV.3, \(\overline{U \cup V}\) is the smallest closed set containing \(U \cup V\), so \(\overline{U \cup V} \subseteq \overline{U} \cup \overline{V}\).

The two containments give equality. \(\blacksquare\)

If you prefer to avoid IV.3, here is a direct argument for the second containment. Let \(p \in \overline{U \cup V}\) and suppose \(p \notin \overline{U}\) and \(p \notin \overline{V}\). Then \(p \notin U\), \(p \notin V\), and there are radii \(\epsilon_1, \epsilon_2 > 0\) with \(B_{\epsilon_1}(p) \cap U = \varnothing\) and \(B_{\epsilon_2}(p) \cap V = \varnothing\). With \(\epsilon = \min\{\epsilon_1, \epsilon_2\}\), the ball \(B_\epsilon(p)\) misses \(U \cup V\) entirely, so \(p\) is neither in \(U \cup V\) nor a limit point of it, contradicting \(p \in \overline{U \cup V}\). This is the same “take the smaller radius” move that shows finite intersections of open sets are open.

Part (a)(ii). The containment \(\overline{U \cap V} \subseteq \overline{U} \cap \overline{V}\) always holds: \(U \cap V\) is contained in both \(U\) and \(V\), so by the lemma its closure is contained in both \(\overline{U}\) and \(\overline{V}\). The reverse containment fails. In \((\mathbb{R}, |\cdot|)\) let \(U = (0,1)\) and \(V = (1,2)\). These are disjoint, so \(U \cap V = \varnothing\) and \(\overline{U \cap V} = \varnothing\). But \(1\) is a limit point of both \(U\) and \(V\): every ball \((1-\epsilon, 1+\epsilon)\) contains points of \((0,1)\) and points of \((1,2)\). So \(1 \in \overline{U} \cap \overline{V}\), and the two sides differ.

A more dramatic example: \(U = \mathbb{Q}\) and \(V = \mathbb{R} \setminus \mathbb{Q}\) are disjoint, yet both are dense, so \(\overline{U} \cap \overline{V} = \mathbb{R}\) while \(\overline{U \cap V} = \varnothing\).

Part (b). We show each side is contained in the other.

\(\big(\overline{T}\big)^c \subseteq I(T^c)\). Let \(p \notin \overline{T}\). Then \(p \notin T\), so \(p \in T^c\), and \(p \notin T'\), so there is an \(\epsilon > 0\) such that \(B_\epsilon(p)\) contains no point of \(T\) different from \(p\). As \(p \notin T\), the ball contains no point of \(T\) at all: \(B_\epsilon(p) \subset T^c\). Thus \(p \in I(T^c)\).

\(I(T^c) \subseteq \big(\overline{T}\big)^c\). Let \(p \in I(T^c)\), with \(B_\epsilon(p) \subset T^c\). Then \(p \in T^c\), so \(p \notin T\); and \(B_\epsilon(p)\) contains no point of \(T\), so \(p\) is not a limit point of \(T\). Hence \(p \notin T \cup T' = \overline{T}\). \(\blacksquare\)

Remark. Part (b) is the identity \(\big(\overline{T}\big)^c = \operatorname{int}(T^c)\), the formal statement that closure and interior are dual under complementation. Taking complements once more gives \(\overline{T} = \big(\operatorname{int}(T^c)\big)^c\), and swapping the roles of \(T\) and \(T^c\) gives \(\operatorname{int}(T) = \big(\overline{T^c}\big)^c\). This duality turns any fact about closures into a fact about interiors: for instance (a)(i) becomes \(\operatorname{int}(U \cap V) = \operatorname{int}(U) \cap \operatorname{int}(V)\), and the failure in (a)(ii) becomes the failure of \(\operatorname{int}(U \cup V) = \operatorname{int}(U) \cup \operatorname{int}(V)\).