# General Topology: Closed Sets

## Closure, Interior, and Boundary

Let $(X, \mathcal{T}))$ be a topological space, and let $A \subseteq X$.

• The closure of $A$, denoted $\overline{A}$, is the intersection of all closed sets that contain $A$. Formally, $\overline{A} = \bigcap \{x : A \subseteq x \text{ and } x^c \in \mathcal{T} \}$.
• The interior of $A$, denoted by $A^{\circ}$, is the union of all open subsets of $A$. Formally, $A^{\circ} = \bigcup \{x : x \subseteq A \text{ and } x \in \mathcal{T} \}$.
• The boundary of $A$, denoted by $\mathcal{\partial} A$, is the set of all points in the closure of $A$ that are not in the interior of $A$. Formally, $\partial A = \overline{A} - A^{\circ}$.

## Problems

1. Show that $x \in \overline{A}$ if and only if every neighborhood of $x$ intersects $A$.

Proof by contraposition: We show that $x \notin \overline{A}$ if and only if there exists a neighborhood of $x$ that does not intersect $A$.

$\implies$: Assume that $x \notin \overline{A}$. Then $x \in (\overline{A})^c$. Because $\overline{A}$ is closed, $(\overline{A})^c$ is open and thus a neighborhood of $x$. Because $A \subseteq \overline{A}$, $A \bigcap (\overline{A})^c = \varnothing$.

$\impliedby$: Assume there exists a neighborhood of $x$ that does not intersect $A$, call it $B$. Then $B^c$ is closed and contains $A$. By definition of closure, $\overline{A} \bigcap B^c = \overline{A}$. Therefore $\overline{A} \subseteq B^c$, so $x \notin B$.

2. Let $A$ and $B$ be subsets of a topological space $X$. Show that if $A \subset B$, then $\overline{A} \subseteq \overline{B}$.

Proof by contraposition: We show that if $\overline{A} \not\subset \overline{B}$, then $A \not\subset B$.

If $\overline{A} \not\subset \overline{B}$, then there exists some $x \in \overline{A}$ such that $x \notin \overline{B}$. Because $x \in \overline{A}$, then by the theorem in problem 1 there exists some neighborhood of $x$, call it $N_0$, such that $N_0 \bigcap B = \varnothing$. By the same theorem, we see that $N_0 \bigcap A \neq \varnothing$. Thus there is some element $y \in A$ that is not in $B$, therefore $A \not\subset B$.