This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. 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 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 \(X\), \(\emptyset\) and \(X\) are simultaneously open and closed, and with the discrete topology every subset becomes both 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 Let \(\mathcal{C}\) be a collection on 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 \(\mathcal{T}\) having the complements of the elements of \(\mathcal{C}\) as its open sets.
Proof
This is obvious from the following De Morgan laws ([Set Theory] §Unions and Intersections, ⁋Proposition 8)
\[\left(\bigcap A_i\right)^c=\bigcup A_i^c,\quad\left(\bigcup A_i\right)^c=\bigcap A_i^c\]We can refine the third condition of the preceding proposition.
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.
It is obvious that any finite family is locally finite, 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 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 \(j\in J\), \(A_j^c\) are all open sets, and therefore the following set
\[V\cap\bigcap_{j\in J} A_j^c\]is a neighborhood of \(x\) and 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
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 \(X\), 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 \(A\). \(x\in\interior(A)\) 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 \(B\), \(x\in\interior(A\cap B)\) is equivalent to \(x\in\interior(A)\cap\interior(B)\). (The second condition of §Open Sets, ⁋Proposition 6) Rewriting this as a statement about closure in the manner described above gives 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. Let \(x\not\in\cl A\). 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
Conversely, suppose there exists some neighborhood of \(x\) disjoint from \(A\). Then there exists an open neighborhood \(U\) of \(x\) contained in this neighborhood that does not meet \(A\), so \(U\cap A=\emptyset\). Now since \(U^c\cap A=A\), \(U^c\) is a closed set containing \(A\), and by minimality of the closure, \(U^c\) also contains \(\cl A\). That is, if \(x\not\in U^c\) then \(x\not\in\cl A\), so 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 \(x\), \(V\cap A\) 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 be the case. 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 \(X\) must meet \(A\). 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 collection 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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