[Munkres Topology] 17. Closed Sets and Limit Points

 This article is one of the posts in the Textbook Commentary Project.

Now that we have a few examples at hand, we can introduce some of the basic concepts associated with topological spaces. In this section, we treat the notions of closed set, closure of a set and limit point as we saw in Analysis course. These lead naturally to consideration of a certain axiom for topological spaces called the Hausdorff axiom.

Closed Sets

A subset $A$ of a topological space $X$ is said to be closed is the set $X-A$ is open.

The collection of closed subsets of a space $X$ has properties similar to those satisfied by the collection of open subsets of $X$:

Theorem 17.1. Let $X$ be a topological Then the following conditions hold:

(1) $\varnothing$ and $X$ are closed.

(2) Arbitrary intersections of closed sets are closed.

(3) Finite unions of closed sets are closed.

Proof. (1) $\varnothing$ and $X$ are closed because they are complements of the open sets $X$ and $\varnothing$, respectively.

(2) Given a collection of closed sets $\{A_\alpha\}_{\alpha \in J}$, we apply DeMorgan's law, $$X-\bigcap_{\alpha \in J} A_\alpha=\bigcup_{\alpha \in J} (X-A_\alpha).$$ 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. 

(3) Similarly, if $A_i$ is closed for $i=1,\cdots,n$, consider the equation $$X-\bigcup_{i=1}^n A_i = \bigcap_{i=1}^n (X-A_i).$$ 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.

Instead of using open sets. one could just as well specify a topology on a space by giving a collection of sets (to be called 'closed sets') satisfying the three properties of this theorem. One could then define open sets as the complements of closed sets and proceed just as before. This procedure has no particular advantage over the one we have adopted, and most mathematicians prefer to use open sets to define topologies.

Now when dealing with subspaces, one needs to be careful in using the term 'closed set.' 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 $Y$). We have the following theorem:

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. [$C$ is closed $\Rightarrow$ $A$ is closed] Assume that $A=C\cap Y$, where $C$ is closed in $X$. (See Figure 17.1.) 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 $Y$. [$A$ is closed $\Rightarrow$ $X-U$ is closed] Conversely, assume that $A$ is closed in $Y$. (See Figure 17.2.) 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.

A set $A$ that is closed in the subspace $Y$ may or may not be closed in the larger space $X$. As was the case with open sets, there is a criterion for $A$ to be closed in $X$.

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 $X$.

Closure and Interior of a Set

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 $\mbox{Int }A$ and the closure of $A$ is denoted by $\mbox{Cl }A$ or by $\bar{A}$. Obviously $\mbox{Int }A$ is an open set and $\bar{A}$ is a closed set; furthermore, $$\mbox{Int }A \subset A \subset \bar{A}.$$ If $A$ is open, $A=\mbox{Int }A$; while if $A$ is closed, $A=\bar{A}$.

We shall note make mush use of the interior of a set, but the closure of a set will be quite important.

When dealing with a topological space $X$ and a subspace $Y$, one needs to exercise care in taking closure of sets. If $A$ is a subset of $Y$, the closure of $A$ in $Y$ and the closure of $A$ in $X$ will in general be different. In such a situation, we reserve the notation $\bar{A}$ to stand for the closure of $A$ in $X$. The closure of $A$ in $Y$ can be expressed in terms of $\bar{A}$, as the following theorem shows:

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. [$B\subset (\bar{A}\cap Y)$] 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 $A$, we must have $B\subset (\bar{A}\cap Y)$.

[$(\bar{A}\cap Y) \subset B$] 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 $(\bar{A}\cap Y)\subset (C\cap Y)=B$.

The definition of the closure of a set does not give us a convenient way for actually finding the closure of specific sets, since the collection of all closed sets in $X$, like the collection of all open sets, is usually much too big to work with. Another way of describing the closure of a set, useful because it involves only a basis for the topology of $X$, is given in the following theorem.

First, let us introduce some convenient terminology. 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 $X$.

(a) Then $x\in \bar{A}$ if and only if every open set $U$ containing $x$ intersects $A$.

(b) 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. 

Consider the statement in (a). Let us transform each implication to its contrapositive: $$x\notin \bar{A} \Leftrightarrow \mbox{there exists an open set }U\mbox{ containing }x \mbox{ that does not intersect }A.$$

[$\Rightarrow$] 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. [$\Leftarrow$] 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 $\bar{A}$.

Statement (b) follows readily from (a). [$\Rightarrow$] If every open set containing $x$ intersects $A$, so does every basis element $B$ containing $x$, because $B$ is an open set. [$\Leftarrow$] 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 $x$. Then by (a), $x\in \bar{A}$.

Mathematicians often use some special terminology there. They shorten the statement '$U$ is an open set containing $x$ to the phrase $$\mbox{'}U\mbox{ is a neighborhood of } x\mbox{.'}$$ Using this terminology, one can write the first half of the preceding theorem as follows:

If $A$ is a subset of the topological space $X$, then $x\in \bar{A}$ if and only if every neighborhood of $x$ intersects $A$.

Some mathematicians use the term 'neighborhood' differently. They say that $A$ is a neighborhood of $x$ if $A$ merely contains an open set containing $x$. We shall not follow this practice.

Limit Points

There is yet another way of describing the closure of a set, a way that involves the important concept of limit point, which we consider now.

If $A$ is as 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 $A'$ be the set of all limit points of $A$. Then $$\bar{A}=A\cup A'.$$

Proof. [$\bar{A} \supset A\cup A'$] If $x$ is in $A'$, every neighborhood of $x$ intersects $A$ (in a point different from $x$). Therefore, by Theorem 17.5, $x$ belongs to $\bar{A}$. Hence $A'\subset \bar{A}$. Since by definition $A\subset \bar{A}$, it follows that $A\cap A'\subset \bar{A}$.

[$\bar{A} \subset A\cup A'$] We let $x$ be a point of $\bar{A}$ and show that $x\in A \cap A'$. If $x$ happens to lie in $A$, it is trivial that $x \in A \cup A'$; 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 $x\in A'$ so that $x\in A \cup A'$, as desired.

Corollary 17.7. A subset of a topological space is closed if and only if 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 $A'\subset A$.

Hausdorff Spaces

One's experience with open and closed sets and limit points in the real line and the plane can be misleading when one considers more general topological spaces. For example, in the space $\mathbb{R}$ and $\mathbb{R}^2$, each one-point set $\{x_0\}$ closed. This fact is easily proved; every point different from $x_0$ has a neighborhood not interesting $\{x_0\}$, so that $\{x_0\}$ is its own closure. But this fact is not true for arbitrary topological spaces. Consider the topology on the three-point set $\{a,b,x\}$ indicated in Figure 17.3. In this space, the one-point set $\{b\}$ is not closed, for its complement is not open.

Similarly, one's experience with the properties of convergent sequences in $\mathbb{R}$ and $\mathbb{R}^2$ can be misleading when one deals with more general topological spaces. In an arbitrary topological space, one says that a sequence $x_1,x_2,\cdots$ 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 $n\ge N$. In $\mathbb{R}$ and $\mathbb{R}^2$, a sequence cannot converge to more than one point, but in an arbitrary space, it can. In the space indicated in Figure 17.3, for example, the sequence defined by setting $x_n=b$ for all $n$ converges not only to the point $b$, but also to the point $a$ and to the point $c$!

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.

 

Definition. 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.

Theorem 17.8. Every finite point set in a Hausdorff space $X$ is closed.

Proof. It suffices to show that every one-point set $\{x_0\}$ is closed. If $x$ is a (any) point of $X$ different from $x_0$, then $x$ and $x_0$ have disjoint neighborhoods $U$ and $V$, respectively. Since $Y$ does not intersect $\{x_0\}$ the point $x$ cannot belong to the closure of the set $\{x_0\}$, At a result, the closure of the set $\{x_0\}$ is $\{x_0\}$ itself, so that it is closed.

The condition that finite point sets be closed is in fact weaker than the Hausdorff condition. For example, the real line $\mathbb{R}$ in the finite complement topology is not a Hausdorff space, but it is a space in which finite point sets are closed. The condition that finite point sets be closed has been given a name of its own; it is called the $T_1$ axiom. (Look at Chapter 4)

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 is every neighborhood of $x$ contains infinitely many points of $A$.

Proof. [$\Leftarrow$] 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 $A$.

[$\Rightarrow$] 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 $\{x_1,\cdots ,x_m\}$ be the points of $U\cap (A-\{x\})$. The set $X-\{x_1,\cdots ,x_m\}$ is an open set of $X$m since the finite point set $\{x_1,\cdots, x_m\}$ is closed; then $$U\cap (X-\{x_1,\cdots ,x_m\})$$ 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 $A$.

One reason for our lack of interest in the $T_1$ axiom is the fact that many of the interesting theorems of topology require not just that axiom, but the full strength of the Hausdorff axiom. Furthermore, most of the spaces that important to mathematicians are Hausdorff spaces. The following two theorems give some substance to these remarks.

Theorem 17.10. If $X$ is a Hausdorff space, then a sequence of point 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 $y\ne x$, 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 converges to $y$.

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 $x_n$.

Theorem 17.11. Every simply ordered set if 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.