§ 17. Closed Sets and Limit Points¶
Definition 17.1 : Closed Set
A subset \(A\) of a topological space \(X\) is said to be closed if the set \(X-A\) is open.
Theorem 17.1
Let \(X\) be a topological space. Then the following conditions hold:
- \(\varnothing\) and \(X\) are closed.
- Arbitrary intersections of closed sets are closed.
- Finite unions of closed sets are closed.
Proof
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\(\varnothing\) and \(X\) are closed because they are the complements of the open sets \(X\) and \(\varnothing\), respectively.
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Given a collection of closed sets \(\left\{A_{\alpha}\right\}_{\alpha \in J}\), we apply DeMorgan's law,
\[ X-\bigcap_{\alpha \in J} A_{\alpha}=\bigcup_{\alpha \in J}\left(X-A_{\alpha}\right) \]Since the sets \(X-A_{\alpha}\) are open by definition, the right side of this equation represents an arbitrary union of open sets, and is thus open. Therefore, \(\bigcap A_{\alpha}\) is closed.
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Similarly, if \(A_{i}\) is closed for \(i=1, \ldots, n\), consider the equation
\[ X-\bigcup_{i=1}^{n} A_{i}=\bigcap_{i=1}^{n}\left(X-A_{i}\right) \]The set on the right side of this equation is a finite intersection of open sets and is therefore open. Hence \(\bigcup A_{i}\) is closed.
Definition 17.2 : Closed In
If \(Y\) is a subspace of \(X\), we say that a set \(A\) is closed in \(Y\) if \(A\) is a subset of \(Y\) and if \(A\) is closed in the subspace topology of \(Y\) (that is, if \(Y-A\) is open in \(Y\) ).
Theorem 17.2
Let \(Y\) be a subspace of \(X\). Then a set \(A\) is closed in \(Y\) if and only if it equals the intersection of a closed set of \(X\) with \(Y\).
Proof
Assume that \(A=C \cap Y\), where \(C\) is closed in \(X\). Then \(X-C\) is open in \(X\), so that \((X-C) \cap Y\) is open in \(Y\), by definition of the subspace topology. But \((X-C) \cap Y=Y-A\). Hence \(Y-A\) is open in \(Y\), so that \(A\) is closed in \(Y\).
Conversely, assume that \(A\) is closed in \(Y\). Then \(Y-A\) is open in \(Y\), so that by definition it equals the intersection of an open set \(U\) of \(X\) with \(Y\). The set \(X-U\) is closed in \(X\), and \(A=Y \cap(X-U)\), so that \(A\) equals the intersection of a closed set of \(X\) with \(Y\), as desired.
Theorem 17.3
Let \(Y\) be a subspace of \(X\). If \(A\) is closed in \(Y\) and \(Y\) is closed in \(X\), then \(A\) is closed in \(X\).
Proof
Since \(A\) is closed in \(Y\), \(Y-A\) is open in \(Y\), and \(Y-A = Y \cap B\) for some set \(B\) open in \(X\). Also, \(Y\) is closed in \(X\), so \(X-Y\) is open in \(X\). Since \(B\) and \(X-Y\) are both open in \(X\), \((X-Y) \cup B = (X-Y) \cup (Y-A) = X-A\) is open in \(X\), and \(A\) is closed in \(X\).
Definition 17.3 : Interior / Closure
Given a subset \(A\) of a topological space \(X\), the interior of \(A\) is defined as the union of all open sets contained in \(A\),
and the closure of \(A\) is defined as the intersection of all closed sets containing \(A\).
The interior of \(A\) is denoted by \(\text{Int} A\) and the closure of \(A\) is denoted by \(\text{Cl} A\) or by \(\bar{A}\).
Obviously \(\text{Int} A\) is an open set and \(\bar{A}\) is a closed set; furthermore,
If \(A\) is open, \(A=\operatorname{Int} A\); while if \(A\) is closed, \(A=\bar{A}\).
Theorem 17.4
Let \(Y\) be a subspace of \(X\); let \(A\) be a subset of \(Y\); let \(\bar{A}\) denote the closure of \(A\) in \(X\). Then the closure of \(A\) in \(Y\) equals \(\bar{A} \cap Y\).
Proof
Let \(B\) denote the closure of \(A\) in \(Y\). The set \(\bar{A}\) is closed in \(X\), so \(\bar{A} \cap Y\) is closed in \(Y\) by Theorem 17.2. Since \(\bar{A} \cap Y\) contains \(A\), and since by definition \(B\) equals the intersection of all closed subsets of \(Y\) containing \(A\), we must have \(B \subset(\bar{A} \cap Y)\).
On the other hand, we know that \(B\) is closed in \(Y\). Hence by Theorem 17.2, \(B=C \cap Y\) for some set \(C\) closed in \(X\). Then \(C\) is a closed set of \(X\) containing \(A\); because \(\bar{A}\) is the intersection of all such closed sets, we conclude that \(\bar{A} \subset C\). Then \((\bar{A} \cap Y) \subset(C \cap Y)=B\).
Definition 17.4 : Intersect
We shall say that a set \(A\) intersects a set \(B\) if the intersection \(A \cap B\) is not empty.
Theorem 17.5
Let \(A\) be a subset of the topological space \(X\).
- Then \(x \in \bar{A}\) if and only if every open set \(U\) containing \(x\) intersects \(A\).
- Supposing the topology of \(X\) is given by a basis, then \(x \in \bar{A}\) if and only if every basis element \(B\) containing \(x\) intersects \(A\).
Proof
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The theorem's logically equivalent contrapositive is the following:
\(x \notin \bar{A} \Longleftrightarrow\) there exists an open set \(U\) containing \(x\) that does not intersect \(A\).
In this form, our theorem is easy to prove. If \(x\) is not in \(\bar{A}\), the set \(U=X-\bar{A}\) is an open set containing \(x\) that does not intersect \(A\), as desired.
Conversely, if there exists an open set \(U\) containing \(x\) which does not intersect \(A\), then \(X-U\) is a closed set containing \(A\). By definition of the closure \(\bar{A}\), the set \(X-U\) must contain \(\bar{A}\); therefore, \(x\) cannot be in \(\bar{A}\). -
If every open set containing \(x\) intersects \(A\), so does every basis element \(B\) containing \(x\), because \(B\) is an open set.
Conversely, if every basis element containing \(x\) intersects \(A\), so does every open set \(U\) containing \(x\), because \(U\) contains a basis element that contains \(x\).
Definition 17.5 : Neighborhood
We shorten to statement "\(U\) is an open set containing \(x\)" to the phrase "\(U\) is a neighborhood of \(x\)".
Definition 17.6 : Limit Point
If \(A\) is a subset of the topological space \(X\) and if \(x\) is a point of \(X\), we say that \(x\) is a limit point (or "cluster point," or "point of accumulation") of \(A\) if every neighborhood of \(x\) intersects \(A\) in some point other than \(x\) itself. Said differently, \(x\) is a limit point of \(A\) if it belongs to the closure of \(A-\{x\}\). The point \(x\) may lie in \(A\) or not; for this definition it does not matter.
Theorem 17.6
Let \(A\) be a subset of the topological space \(X\); let \(A^{\prime}\) be the set of all limit points of \(A\). Then
Proof
If \(x\) is in \(A^{\prime}\), every neighborhood of \(x\) intersects \(A\) (in a point different from \(x\) ). Therefore, by Theorem 17.5, \(x\) belongs to \(\bar{A}\). Hence \(A^{\prime} \subset \bar{A}\). Since by definition \(A \subset \bar{A}\), it follows that \(A \cup A^{\prime} \subset \bar{A}\).
To demonstrate the reverse inclusion, we let \(x\) be a point of \(\bar{A}\) and show that \(x \in A \cup A^{\prime}\). If \(x\) happens to lie in \(A\), it is trivial that \(x \in A \cup A^{\prime}\); suppose that \(x\) does not lie in \(A\). Since \(x \in \bar{A}\), we know that every neighborhood \(U\) of \(x\) intersects \(A\); because \(x \notin A\), the set \(U\) must intersect \(A\) in a point different from \(x\). Then \(x \in A^{\prime}\), so that \(x \in A \cup A^{\prime}\), as desired.
Corollary 17.7
A subset of a topological space is closed if and only if it contains all its limit points.
Proof
The set \(A\) is closed if and only if \(A=\bar{A}\), and the latter holds if and only if \(A^{\prime} \subset A\).
Definition 17.7 : Converge
In an arbitrary topological space, one says that a sequence \(x_{1}, x_{2}, \ldots\) of points of the space \(X\) converges to the point \(x\) of \(X\) provided that, corresponding to each neighborhood \(U\) of \(x\), there is a positive integer \(N\) such that \(x_{n} \in U\) for all \(n \geq N\).
Definition 17.8 : Hausdorff Space
A topological space \(X\) is called a Hausdorff space if for each pair \(x_{1}, x_{2}\) of distinct points of \(X\), there exist neighborhoods \(U_{1}\), and \(U_{2}\) of \(x_{1}\) and \(x_{2}\), respectively, that are disjoint.
Note
Topologies in which one-point sets are not closed, or in which sequences can converge to more than one point, are considered by many mathematicians to be somewhat strange. They are not really very interesting, for they seldom occur in other branches of mathematics. And the theorems that one can prove about topological spaces are rather limited if such examples are allowed. Therefore, one often imposes an additional condition that will rule out examples like this one, bringing the class of spaces under consideration closer to those to which one's geometric intuition applies. The condition was suggested by the mathematician Felix Hausdorff, so mathematicians have come to call it by his name.
Theorem 17.8
Every finite point set in a Hausdorff space \(X\) is closed.
Proof
It suffices to show that every one-point set \(\left\{x_{0}\right\}\) is closed. If \(x\) is a point of \(X\) different from \(x_{0}\), then \(x\) and \(x_{0}\) have disjoint neighborhoods \(U\) and \(V\), respectively. Since \(U\) does not intersect \(\left\{x_{0}\right\}\), the point \(x\) cannot belong to the closure of the set \(\left\{x_{0}\right\}\). As a result, the closure of the set \(\left\{x_{0}\right\}\) is \(\left\{x_{0}\right\}\) itself, so that it is closed.
Definition 17.9 : \(T_1\) axiom
The condition that finite point sets be closed has been given a name of its own: it is called the \(T_1\) axiom.
Theorem 17.9
Let \(X\) be a space satisfying the \(T_{1}\) axiom; let \(A\) be a subset of \(X\). Then the point \(x\) is a limit point of \(A\) if and only if every neighborhood of \(x\) contains infinitely many points of \(A\).
Proof
If every neighborhood of \(x\) intersects \(A\) in infinitely many points, it certainly intersects \(A\) in some point other than \(x\) itself, so that \(x\) is a limit point of \(A\).
Conversely, suppose that \(x\) is a limit point of \(A\), and suppose some neighborhood \(U\) of \(x\) intersects \(A\) in only finitely many points. Then \(U\) also intersects \(A-\{x\}\) in finitely many points; let \(\left\{x_{1}, \ldots, x_{m}\right\}\) be the points of \(U \cap(A-\{x\})\). The set \(X-\left\{x_{1}, \ldots, x_{m}\right\}\) is an open set of \(X\), since the finite point set \(\left\{x_{1}, \ldots, x_{m}\right\}\) is closed; then
is a neighborhood of \(x\) that intersects the set \(A-\{x\}\) not at all. This contradicts the assumption that \(x\) is a limit point of \(A\).
Theorem 17.10
If \(X\) is a Hausdorff space, then a sequence of points of \(X\) converges to at most one point of \(X\).
Proof
Suppose that \(x_{n}\) is a sequence of points of \(X\) that converges to \(x\). If \(y \neq x\), let \(U\) and \(V\) be disjoint neighborhoods of \(x\) and \(y\), respectively. Since \(U\) contains \(x_{n}\) for all but finitely many values of \(n\), the set \(V\) cannot. Therefore, \(x_{n}\) cannot converge to \(y\).
Definition 17.10 : Limit
If the sequence \(x_{n}\) of points of the Hausdorff space \(X\) converges to the point \(x\) of \(X\), we often write \(x_{n} \rightarrow x\), and we say that \(x\) is the limit of the sequence \(x_{n}\).
Theoren 17.11
Every simply ordered set is a Hausdorff space in the order topology.
The product of two Hausdorff spaces is a Hausdorff space.
A subspace of a Hausdorff space is a Hausdorff space.