Interior, Closure, and Boundary

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Before we take up continuous functions, sequences, and related notions in earnest, we complete the introduction of the language used in topology.

Closed Sets

Definition 1 For a topological space $X$, a set $A$ is called a closed set if its complement $A^c=X\setminus A$ is an open set.

In any topology $\mathcal{T}$ on $X$, the sets $\emptyset$ and $X$ are simultaneously open and closed, and under the discrete topology every subset is both open and closed. Thus, closed sets and open sets are not opposite concepts; rather, they are nearly the same thing expressed in a different way. For instance, a topology $\mathcal{T}$ can in fact be defined using closed sets as follows.

Proposition 2 Let $\mathcal{C}$ be a collection of subsets of a set $X$ satisfying the following conditions.

$\emptyset$, $X\in\mathcal{C}$
$\mathcal{C}$ is closed under arbitrary intersections.
$\mathcal{C}$ is closed under finite unions.

Then there exists a unique topology $\mathcal{T}$ whose open sets are the complements of the elements of $\mathcal{C}$.

Proof

This is immediate from the following De Morgan laws (§Union and Intersection, ⁋Proposition 8 (De Morgan’s law))

\[\left(\bigcap A_i\right)^c=\bigcup A_i^c,\quad\left(\bigcup A_i\right)^c=\bigcap A_i^c\]

We can refine the third condition of the preceding proposition further.

Definition 3 Let a topological space $X$ be given, and let $(A_i)_{i\in I}$ be a family of subsets of $X$. Then $(A_i)$ is called locally finite if for every $x\in X$ there exists a neighborhood $V$ such that the set of indices $i$ with $V\cap A_i\neq\emptyset$ is finite.

That any finite family is locally finite is obvious, so the above definition may be regarded as a generalization of finite families. The following holds.

Proposition 4 Let a topological space $X$ be given. If $(A_i)_{i\in I}$ is a locally finite collection of closed sets, then $A=\bigcup A_i$ is a closed set.

Proof

To show this, it suffices to prove that $A^c$ is an open set. Let $x\in A^c$. Then $x\in A_i^c$ holds for all $i$. On the other hand, since $(A_i)$ is locally finite, there exists a neighborhood $V$ of $x$ such that the set of indices $i$ satisfying $V\cap A_i\neq\emptyset$ is finite. Let $J$ be the subset of $I$ consisting of such indices. Then for every $j\in J$, the set $A_j^c$ is open, and therefore the set

\[V\cap\bigcap_{j\in J} A_j^c\]

is a neighborhood of $x$ contained in $A^c$. Hence $A^c$ is open, and so $A$ is closed.

Interior and Closure

Let a topological space $(X,\mathcal{T})$ be given. For any subset $A$ of $X$, there always exist a closed set containing $A$ and an open set contained in $A$ (namely $X$ and $\emptyset$). On the other hand, since arbitrary intersections of closed sets are closed and arbitrary unions of open sets are open, both the smallest closed set containing $A$ and the largest open set contained in $A$ exist. We define them as follows.

Definition 5 For any subset $A$ of a topological space $X$, the smallest closed set containing $A$ is called the closure of $A$, and the largest open set contained in $A$ is called the interior of $A$; we denote them by $\cl(A)$ and $\interior(A)$, respectively.

Defined in this way, it is obvious that the two operators $\cl$ and $\interior$ preserve inclusion.

Let us prove the identity

\[\interior(A^c)=(\cl(A))^c\]

By definition, $\interior(A^c)$ is the largest open set contained in $A^c$, which is the same as saying the largest open set not containing $A$. On the other hand, $\cl(A)$ is the smallest closed set containing $A$, so $(\cl(A))^c$ is the largest open set not containing $A$; hence the two must coincide. We call this set the exterior of $A$.

Through the above argument, we see that having any one of interior, closure, or exterior allows us to construct the other two.

Consider the interior of a set $A$. The statement $x\in\interior(A)$ means that there exists an open set $U$ containing $x$ and contained in $A$, which is equivalent to saying that $A$ is a neighborhood of $x$. Therefore, for any two sets $A$ and $B$, the statement $x\in\interior(A\cap B)$ is equivalent to $x\in\interior(A)\cap\interior(B)$. (The second condition of §Open Sets, ⁋Proposition 6) Rewriting this as a statement about closures in the manner described above yields the identity

\[\cl(A\cup B)=\cl(A)\cup\cl(B)\]

Proposition 6 For a topological space $X$ and a subset $A$, the following two conditions are equivalent:

$x\in\cl A$;
every neighborhood $U$ of $x$ meets $A$.

Proof

It is convenient to prove the contrapositive. Suppose $x\not\in\cl A$. Then $x\in(\cl A)^c=\ext A$ is an open set containing $x$ and disjoint from $\cl A$, hence also disjoint from $A$. That is, the statement there exists some neighborhood of $x$ disjoint from $A$ is true.

Conversely, suppose there exists some neighborhood of $x$ disjoint from $A$. Then there is an open neighborhood $U$ of $x$ contained in this neighborhood that does not meet $A$, so $U\cap A=\emptyset$. Now since $U^c\cap A=A$, the set $U^c$ is a closed set containing $A$, and by minimality of the closure, $U^c$ also contains $\cl A$. Hence if $x\not\in U^c$ then $x\not\in\cl A$, and the reverse direction also holds.

Corollary 7 Let a topological space $X$ be given. For an open set $A$ and any set $B$, the following inclusion holds:

\[A\cap\cl(B)\subseteq\cl(A\cap B)\]

Proof

Let $x\in A\cap\cl(B)$. Since $A$ is an open neighborhood of $x$, for any neighborhood $V$ of $x$, the intersection $V\cap A$ is also a neighborhood of $x$. Therefore, from $x\in\cl(B)$ and Proposition 6, we know that $(V\cap A)\cap B\neq\emptyset$. But this can be interpreted as saying that the intersection of $A\cap B$ with $V$ is nonempty, and since $V$ is an arbitrary neighborhood of $x$, Proposition 6 again gives $x\in\cl(A\cap B)$.

Definition 8 For a topological space $X$ and any subset $A$ of $X$, a point $x\in X$ is called a limit point of $A$ if every neighborhood of $x$ meets $A$ at some point other than $x$ itself.

Then by definition, $\cl(A)$ is the union of $A$ and its limit points. If $x\in\cl(A)\setminus A$, then by Proposition 6, $x$ must be a limit point of $A$. On the other hand, if $x\in A$, this need not hold. If for $x\in A$ there exists a neighborhood $V$ such that $V\cap A=\{x\}$, then $x$ is called an isolated point of $A$. A closed set with no isolated points is called a perfect set.

Boundary

Definition 9 For any subset $A$ of a topological space $X$, the boundary of $A$ is the set $\partial A$ defined by the formula

\[\partial A=\cl A\setminus\interior A\]

Hence $\partial A$ is a closed set.

Dense Sets

Definition 10 For any subset $A$ of a topological space $X$, $A$ is called a dense subset if $\cl(A)=X$.

By Proposition 6, saying that $A$ is dense in $X$ means that every nonempty open set in $X$ meets $A$. Intuitively, finding a dense subset of $X$ means that with only a slight perturbation we can recover all of $X$. In more everyday language, a dense subset of $X$ can be thought of as containing “almost all” of $X$.

Meanwhile, the notion of size in topology is also given by the size of a base, as the following proposition shows.

Proposition 11 For a base $\mathcal{B}$ of a topological space $X$, there exists a dense subset $D$ of $X$ such that $\card(D)\leq\card(\mathcal{B})$.

Proof

For each $U\in\mathcal{B}$, choose an element $x_U\in U$, and let $D$ be the set of these. That $D$ is dense follows because for any open set $V$, we can express it as a union of elements of $\mathcal{B}$, and this union must contain some $x_U$, so $V\cap D\neq\emptyset$.

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Closed Sets

Interior and Closure

Boundary

Dense Sets

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