Mathematics Lecture Notes

Theorem1.3Characterizations of the Closure

Let $(X, \tau)$ be a topological space and $A \subseteq X$ . The following are equivalent for a point $x \in X$ :

$x \in \overline{A}$ (i.e., $x$ belongs to the closure of $A$ ).
Every open set $U$ containing $x$ satisfies $U \cap A \neq \emptyset$ .
Every neighborhood $N$ of $x$ satisfies $N \cap A \neq \emptyset$ .
$x \in A$ or $x$ is a limit point of $A$ (i.e., $x \in A \cup A'$ ).

Moreover, the following characterizations hold for $\overline{A}$ as a set:

$\overline{A}$ is the intersection of all closed sets containing $A$ .
$\overline{A} = X \setminus \operatorname{int}(X \setminus A)$ .

Proof

(1 $\Leftrightarrow$ 5) This is the definition: $\overline{A} = \bigcap \{C : C \text{ closed}, A \subseteq C\}$ .

(1 $\Rightarrow$ 2) Suppose $x \in \overline{A}$ and let $U$ be an open set containing $x$ . If $U \cap A = \emptyset$ , then $A \subseteq X \setminus U$ , and $X \setminus U$ is closed. By (5), $\overline{A} \subseteq X \setminus U$ , so $x \notin U$ , a contradiction.

(2 $\Rightarrow$ 1) Suppose every open set containing $x$ meets $A$ . Let $C$ be any closed set containing $A$ . If $x \notin C$ , then $U = X \setminus C$ is an open set containing $x$ with $U \cap A \subseteq U \cap C = \emptyset$ , contradicting (2). So $x \in C$ for every closed $C \supseteq A$ , giving $x \in \overline{A}$ .

(2 $\Leftrightarrow$ 3) Immediate, since every open set containing $x$ is a neighborhood of $x$ , and every neighborhood of $x$ contains an open set containing $x$ .

(2 $\Leftrightarrow$ 4) If $x \in A$ , then $U \cap A \neq \emptyset$ trivially. If $x \notin A$ but every open set $U$ containing $x$ satisfies $U \cap A \neq \emptyset$ , then $(U \setminus \{x\}) \cap A = U \cap A \neq \emptyset$ (since $x \notin A$ ), so $x$ is a limit point. Conversely, if $x \in A$ or $x \in A'$ , then every open neighborhood of $x$ meets $A$ .

(1 $\Leftrightarrow$ 6) We show $X \setminus \overline{A} = \operatorname{int}(X \setminus A)$ . We have: $x \notin \overline{A} \iff \exists U \in \tau : x \in U \text{ and } U \cap A = \emptyset \iff \exists U \in \tau : x \in U \subseteq X \setminus A \iff x \in \operatorname{int}(X \setminus A).$

Taking complements: $\overline{A} = X \setminus \operatorname{int}(X \setminus A)$ .

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Theorem1.4Properties of the Closure Operator

Let $(X, \tau)$ be a topological space and $A, B \subseteq X$ . Then:

$\overline{\emptyset} = \emptyset$ .
$A \subseteq \overline{A}$ .
$\overline{\overline{A}} = \overline{A}$ .
$\overline{A \cup B} = \overline{A} \cup \overline{B}$ .
If $A \subseteq B$ , then $\overline{A} \subseteq \overline{B}$ .
$\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$ (equality need not hold in general).
$A$ is closed if and only if $A = \overline{A}$ .
$A$ is dense in $X$ if and only if $\overline{A} = X$ .

Proof

We prove property (4) and (6), as the others follow directly from the definitions.

(4) Since $A \subseteq A \cup B$ and $B \subseteq A \cup B$ , property (5) gives $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$ . For the reverse, $\overline{A} \cup \overline{B}$ is closed (finite union of closed sets) and contains $A \cup B$ , so by minimality $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$ .

(6) Since $A \cap B \subseteq A$ and $A \cap B \subseteq B$ , we get $\overline{A \cap B} \subseteq \overline{A}$ and $\overline{A \cap B} \subseteq \overline{B}$ , hence $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$ .

For a counterexample to equality: in $\mathbb{R}$ , let $A = (0, 1)$ and $B = (1, 2)$ . Then $A \cap B = \emptyset$ , so $\overline{A \cap B} = \emptyset$ , but $\overline{A} \cap \overline{B} = [0, 1] \cap [1, 2] = \{1\}$ .

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Theorem1.5Closure in Subspace Topology

Let $Y$ be a subspace of $X$ and $A \subseteq Y$ . Then the closure of $A$ in $Y$ is $\operatorname{cl}_Y(A) = \overline{A} \cap Y$ , where $\overline{A}$ denotes the closure of $A$ in $X$ .

Proof

The closed sets in $Y$ are exactly sets of the form $C \cap Y$ where $C$ is closed in $X$ . Thus: $\operatorname{cl}_Y(A) = \bigcap\{C \cap Y : C \text{ closed in } X, A \subseteq C \cap Y\} = \left(\bigcap\{C : C \text{ closed in } X, A \subseteq C\}\right) \cap Y = \overline{A} \cap Y.$

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ExampleClosure Depends on the Ambient Space

Let $A = \{1/n : n \in \mathbb{Z}^+\}$ and consider it as a subset of different spaces:

In $\mathbb{R}$ : $\overline{A} = A \cup \{0\}$ .
In $(0, \infty)$ : $\operatorname{cl}_{(0,\infty)}(A) = \overline{A} \cap (0, \infty) = A$ (so $A$ is closed in $(0, \infty)$ ).
In $(0, 1]$ : $\operatorname{cl}_{(0,1]}(A) = A$ (closed in this subspace as well).

This illustrates that closure is relative to the ambient space.

RemarkDuality of Interior and Closure

The interior and closure operators are dual via complementation: $\operatorname{int}(A) = X \setminus \overline{X \setminus A}, \qquad \overline{A} = X \setminus \operatorname{int}(X \setminus A).$ This duality allows every statement about interiors to be translated into a statement about closures, and vice versa.

Math Notes

Closure Characterizations

Main Theorem

Proof

Derived Properties

Closure in Subspaces