위상수학
Interior, Closure, and Boundary
Basic concepts in topology
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 begin treating continuous functions, sequences, and related notions in earnest, we introduce the remaining language of 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 \(X\), both \(\emptyset\) and \(X\) are simultaneously open and closed, and if the discrete topology is given then every subset is simultaneously open and closed. Thus closed sets and open sets are not opposite concepts; rather, they are closer to being the same thing expressed in different ways. For instance, a topology \(\mathcal{T}\) can in fact be defined using closed sets as follows.
Proposition 2 Suppose a collection \(\mathcal{C}\) on a set \(X\) is given 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 exactly the complements of the elements of \(\mathcal{C}\).
Proof
From De Morgan’s laws ([Set Theory] §Union and Intersection, ⁋Proposition 8 (De Morgan’s law))
\[\left(\bigcap A_i\right)^c=\bigcup A_i^c,\quad\left(\bigcup A_i\right)^c=\bigcap A_i^c\]the correspondence taking complements reverses inclusion and translates conditions 1, 2, and 3 into the three axioms for \(\mathcal{T}=\{C^c\mid C\in\mathcal{C}\}\) to be a topology, and since the collection of open sets is determined by this \(\mathcal{T}\), uniqueness follows as well.
The third condition of the preceding proposition can be refined 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 can 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 indices \(i\) satisfying \(V\cap A_i\neq\emptyset\) are only finitely many. Let \(J\) be the subset of \(I\) consisting of such indices. Then for every \(j\in J\), each \(A_j^c\) is an open set, and therefore the following set
\[V\cap\bigcap_{j\in J} A_j^c\]is a neighborhood of \(x\) and is a subset of \(A^c\). From this we see that \(A^c\) is an open set, and hence \(A\) is a closed set.
Interior and Closure of a Set
Let a topological space \((X,\mathcal{T})\) be given. For any subset \(A\) of \(X\), there always exist
Definition 5 For any subset \(A\) of a topological space \(X\), we call the smallest closed set containing \(A\) the closure of \(A\), and the largest open set contained in \(A\) the interior of \(A\), and denote them by \(\cl(A)\) and \(\interior(A)\), respectively.
With this definition, 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 disjoint from \(A\). On the other hand, \(\cl(A)\) is the smallest closed set containing \(A\), so \((\cl(A))^c\) is the largest open set disjoint from \(A\); hence the two must be equal. We call this set the exterior of \(A\).
The same argument shows that if we have any one of interior, closure, or exterior, we can 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 \(A,B\), the condition \(x\in\interior(A\cap B)\) is equivalent to \(x\in\interior(A)\cap\interior(B)\). (The second condition of §Open Sets, ⁋Proposition 6.) Translating this into a proposition about closure via the method explained above, we obtain the equality
\[\cl(A\cup B)=\cl(A)\cup\cl(B).\]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 \((\cl A)^c=\ext A\) contains \(x\), is an open set disjoint from \(\cl A\), and hence also disjoint from \(A\). That is, the statement
Conversely, suppose there exists a neighborhood of \(x\) that does not meet \(A\). Then there is an open neighborhood \(U\) of \(x\) contained in this neighborhood such that \(U\) does not meet \(A\), so \(U\cap A=\emptyset\). Now \(U^c\cap A=A\), so \(U^c\) is a closed set containing \(A\), and by the minimality of the closure, \(U^c\) also contains \(\cl A\). Hence, if \(x\not\in U^c\) then \(x\not\in\cl A\), and therefore the reverse direction also holds.
Corollary 7 Let a topological space \(X\) be given. For an open set \(A\) and an arbitrary set \(B\), the following identity
\[A\cap\cl(B)\subseteq\cl(A\cap B)\]holds.
Proof
Suppose \(x\in A\cap\cl(B)\). Since \(A\) is an open neighborhood of \(x\), for any neighborhood \(V\) of \(x\), the intersection \(V\cap A\) is also a neighborhood of \(x\). Thus, from the fact that \(x\in\cl(B)\) and Proposition 6, we know that \((V\cap A)\cap B\neq\emptyset\). However, this can also be interpreted as saying that the intersection of \(A\cap B\) and \(V\) is nonempty, and since \(V\) is an arbitrary neighborhood of \(x\), we again have \(x\in\cl(A\cap B)\) by Proposition 6.
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 \(\cl(A)\) is the union of \(A\) and the limit points of \(A\). If \(x\in\cl(A)\setminus A\), then by Proposition 6 the point \(x\) must be a limit point of \(A\); conversely, any limit point of \(A\) belongs to \(\cl(A)\) by Proposition 6 again, because every neighborhood of it meets \(A\). 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 we call \(x\) an isolated point of \(A\). A closed set with no isolated points is called a perfect set.
Boundary of a Set
Definition 9 For any subset \(A\) of a topological space \(X\), the boundary of \(A\) is the set \(\partial A\) defined by the equation
\[\partial A=\cl A\setminus\interior A\]Thus \(\partial A\) is a closed set.
Dense Sets
Definition 10 A subset \(A\) of a topological space \(X\) is called a dense subset if \(\cl(A)=X\).
By Proposition 6, the condition that \(A\) is dense in \(X\) means that every nonempty open subset of \(X\) must intersect \(A\). Intuitively, one may think that if we find a dense subset of \(X\), then we can recover all of \(X\) with only a slight perturbation. In more everyday language, a dense subset of \(X\) can be thought of as containing “almost all” of \(X\).
On the other hand, in topology the notion of size is also given by the cardinality of a base, as shown in the following proposition.
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 nonempty \(U\in\mathcal{B}\), choose an element \(x_U\in U\), and let \(D\) be the collection of these elements. That \(D\) is dense follows because for any nonempty open set \(V\), we can express \(V\) as a union of elements of \(\mathcal{B}\), and this union must contain some \(x_U\), so \(V\cap D\neq\emptyset\).
References
[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.
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