Open Sets and Closed Sets

A great deal of analysis on $$\mathbb R$,$ $$\mathbb C$,$ and $\mathbb R^k$ depends not on their full algebraic structure, but only on the notion of distance between points. This suggests isolating the notion of distance from the particular nature of the points themselves. Accordingly, let $$M$$ be a nonempty set. We temporarily forget what its elements are and retain only a function $d:M\times M\to\mathbb R,$ where $$d(x,y)$$ is intended to represent the distance between $$x$$ and $$y$.$ Not every such function can reasonably be interpreted as a distance. A distance should be nonnegative; distinct points should have positive distance from one another; the distance from $$x$$ to $$y$$ should be the same as the distance from $$y$$ to $$x$;$ and traveling directly from $$x$$ to $$z$$ should be no longer than traveling from $$x$$ to $$z$$ through an intermediate point $$y$.$ These requirements are abstracted into the following axioms.

Definition 2.1.1. Suppose that $$M$$ is a nonempty set. A function $d:M^2\to \mathbb R$ is called a metric on $$M$$ if, for all $$x,y,z\in M$,$

$0\le d(x,y),$
$$d(x,y)=0$$ if and only if $$x=y,$$
$$d(x,y)=d(y,x),$$
$d(x,z)\le d(x,y)+d(y,z).$

Then, the set $$M$$ equipped with the metric $$d$$ is called a metric space.

Example 2.1.2. The fundamental example is Euclidean space. For $$k\in\mathbb N^{+}$,$ define $d:(\mathbb R^k)^2\to\mathbb R$ by

\[d(x,y)=\|x-y\|.\]

The usual properties of the Euclidean norm imply that $$d$$ satisfies the metric axioms, so $(\mathbb R^k,d)$ is a metric space. Moreover, if $$A\subseteq\mathbb R^k$,$ then the restriction $d\vert{}_{A\times A}:A\times A\to\mathbb R$ is again a metric. Hence every subset of Euclidean space becomes a metric space when equipped with the restricted Euclidean metric. More generally, if $$(M,d)$$ is any metric space and $$N\subseteq M$,$ then $d|_{N\times N}$ is a metric on $$N$,$ called the restricted metric. With the restricted metric, $$N$$ thus can be regarded as a metric space on its own right.

What we are primarily interested in is the behavior of all points in the distance neighborhood of a given point. Therefore, we consider a set that refers to all points within a distance $$r$$ from a point in a space.

Definition 2.1.3. Let $$M$$ be a metric space, let $$a\in M$,$ and let $$r>0$.$ The open ball of radius $$r$$ centered at $$a$$ is the set $$B_r(a)$$ defined by

\[B_r(a) = \lbrace x\in M : d(x,a)<r\rbrace.\]

The closed ball of radius $$r$$ centered at $$a$$ is the set $\overline B_r(a)$ defined by

\[\overline B_r(a) = \lbrace x\in M : d(x,a)\le r\rbrace.\]

The punctured open ball of radius $$r$$ centered at $$a$$ is the set $B^{\ast}_r(a)$ defined by

\[B^{\ast}_r(a) = \lbrace x\in M : 0<d(x,a)<r\rbrace.\]

The balls can describe the local relation between a point $$x$$ and a set $$A$$ in two related ways. The first asks whether points sufficiently close to $$x$$ are in $$A$,$ in other words, whether a sufficiently small ball lies in $$A$.$

Definition 2.1.4. Let $$M$$ be a metric space and let $$A\subseteq M$.$ $x\in M$ is an interior point of $$A$$ in $$M$$ if

\[B_r(x)\subseteq A\]

for some $$r>0$.$ Conversely, $x\in M$ is an exterior point of $$A$$ in $$M$$ if

\[B_r(x)\subseteq A^c\]

for some $$r>0$.$ The remaining possibility is that no ball centered at $$x$$ lies entirely in $$A$$ or entirely in $$A^c$.$ That is, if

\[B_r(x)\cap A \ne \varnothing \quad \text{and} \quad B_r(x) \cap A^c\ne \varnothing\]

for all $$r>0$,$ then $$x$$ is called a boundary point of $$A$$ in $$M$.$ $\operatorname{int} A$ denotes the interior of $$A$$ in $$M$,$ the set of all interior points of $$A$$ in $$M$;$ $\operatorname{ext} A$ denotes the exterior of $$A$$ in $$M$,$ the set of all exterior points of $$A$$ in $$M$;$ $\partial A$ denotes the boundary of $$A$$ in $$M$,$ the set of all boundary points of $$A$$ in $$M$.$

Theorem 2.1.5. Let $$M$$ be a metric space and let $A\subseteq M.$ Every point of $$M$$ is either an interior point of $$A,$$ an exterior point of $$A,$$ or a boundary point of $$A,$$ and hence

\[M = \operatorname{int}A \ \dot{\cup} \ \partial A\ \dot{\cup} \ \operatorname{ext}A.\]

Proof. If $x\in \operatorname {int} A,$ then $B_r(x)\subseteq A$ for some $$r>0.$$ Since $x\in \operatorname{int} A$ implies $x\in A,$ $$B_r(x)$$ does not meet $A^{c}.$ Hence $\operatorname{int} A \cap \partial A = \varnothing.$ If $$0<s<r$$ then $B_s(x)\subseteq B_r(x)\subseteq A.$ If $$s>r$$ then $B_r(x)\subseteq B_s(x)$ so that $B_s(x)\cap A\ne \varnothing.$ Hence $\operatorname{int} A \cap \operatorname{ext} A = \varnothing.$ Lastly, since $x\in \partial A$ implies $x\notin \operatorname{ext} A,$ $\partial A \cap \operatorname{ext} A = \varnothing.$ Therefore $\operatorname{int} A, \partial A,$ and $\operatorname{ext} A$ are mutually disjoint. We claim that $\partial A \cup \operatorname{ext} A = \left(\operatorname{int} A\right)^{c}.$ Indeed, if $x\in \partial A$ or $x\in \operatorname{ext} A$ then $$x\notin \operatorname{int} A$,$ because $\operatorname{int} A, \partial A,$ and $\operatorname{ext} A$ are mutually disjoint. Therefore $\partial A \cup \operatorname{ext} A \subseteq \left( \operatorname{int} A\right) ^{c}.$ Conversely, suppose $x\notin \operatorname{int} A,$ which means that every ball around $$x$$ meets $A^{c}.$ There may exist some $$r>0$$ such that $B_r(x)\subseteq A^c,$ which means $x\in \operatorname{ext} A.$ Otherwise, no such $$r$$ exists, which means that every ball around $$x$$ meets $$A.$$ Thus $x\in \partial A.$ Therefore $\left(\operatorname{int} A\right)^{c}\subseteq \partial A \cup \operatorname{ext} A.$ It is concluded that $\partial A \cup \operatorname{ext} A = \left(\operatorname{int} A\right)^{c}.$ $\square$

Closure point and limit point answer the second question: can $$A$$ be completely avoided by restricting our attention to a sufficiently small ball of $$x$?$

Definition 2.1.6. Let $$M$$ be a metric space and let $$A\subseteq M$,$ and let $$x\in M$.$ $$x$$ is a closure point of $$A$$ in $$M$$ if

\[B_r(x)\cap A\neq\varnothing\]

for all $$r>0$.$ In words, no matter how tightly we localize around $$x$,$ $$x$$ can never get completely away from $$A$.$ The closure of $$A$$ in $$M$$ is the set of all closure points of $$A$,$ denoted by $$\overline A$.$ This includes every point $x\in A$ automatically. A limit point deliberately removes this trivial possibility by puncturing the ball: $$x$$ is a limit point of $$A$$ in $$M$$ if

\[B_r^{\ast}(x)\cap A\neq\varnothing\]

for all $$r>0$.$ The derived set of $$A$$ in $$M$$ is the set of all limit points of $$A$,$ denoted by $$A'$.$

Remark 2.1.7. Since every ball $$B_r(x)$$ contains $$x$$ and hence nonempty, no $x\in M$ can be an interior point, boundary point, limit point, or closure point of $$\varnothing$.$ On the other hand, every $x\in M$ is an exterior point of $$\varnothing$.$ In the case of $$M$,$ every $x\in M$ is an interior point and closure point of $$M$,$ while no $x\in M$ can be a boundary point or an exterior point of $$M$.$

Theorem 2.1.8. Let $$M$$ be a metric space, and let $$A\subseteq M$.$ If $$a$$ is a limit point of $$A$,$ then every open ball of $$a$$ has infinitely many points of $$A$.$

Proof. Suppose, for contradiction, that there exists an punctured open ball of $$a$$ which contains only a finite number of points of $$A$.$ Since $$a$$ is a limit point of $$A$,$ we may let $p_1,\ldots,p_n$ be those points, which are distinct from $$a$.$ We now define

\[r=\min_{1\leq i\leq n} d(a,p_i),\]

and it is clear that $$r>0$.$ Then, $B_r^{\ast}(a)$ contains no point of $$A$.$ Thus $$a$$ is not a limit point of $$A$,$ which contradicts our premise. $\square$

Corollary 2.1.9. Let $$M$$ be a metric space, and let $$A\subseteq M$.$ If $$A$$ is finite, then $$A$$ has no limit points in $$M$.$

Theorem 2.1.10. Let $$A$$ be a nonempty subset of $\mathbb R$ which is bounded above, and let $\alpha = \sup A.$ Then $$\alpha \in \overline A$.$

Proof. Since $\alpha$ is the supremum for $$A$,$ there exists $x\in A$ such that $\alpha - r<x\le \alpha$ for every $$r>0$.$ Hence $B_r(\alpha)$ has a point of $$A$$ for every $$r>0$.$ Therefore $\alpha$ is a closure point of $$A$.$ $\square$

Proposition 2.1.11. Let $$M$$ be a metric space and let $A\subseteq M.$ Then the following hold.

$$\overline A = M \setminus \operatorname{ext} A$.$
$$\overline A = A\cup A'$.$
$$\operatorname{int} A \subseteq A \subseteq \overline A$.$

Proof. (1) If $x\in \overline A,$ then every ball meets $$A.$$ Hence no balls lie entirely in $A^{c},$ so that $x\notin \operatorname{ext} A.$ If $x\notin \operatorname{ext} A,$ then every ball meets $$A$$ so that $x\in \overline A.$ Hence $\overline A = M \setminus \operatorname{ext} A.$ (2) Since $B_r(x) = B_r^{\ast}(x) \cup \lbrace x\rbrace,$ every ball around $$x$$ meets $$A$$ if and only if every punctured ball meets $$A$$ or $\lbrace x\rbrace$ meets $$A.$$ Hence $\overline A = A\cup A'.$ (3) If $x\in \operatorname{int} A,$ then $B_r(x)\subseteq A$ for some $$r>0.$$ Since $x\in B_r(x),$ $x\in A,$ so that $\operatorname{int} A \subseteq A.$ (2) immediately yields $A \subseteq \overline A.$ Therefore $\operatorname{int} A \subseteq A \subseteq \overline A.$ $\square$

So far, our notions have described the local relations of an individual point to a set. We now turn these pointwise relations into global properties of a subset of a metric space as a whole. Recall that, for a subset $$A$$ of a metric space $$M$,$

\[\operatorname{int} A \subseteq A \subseteq \overline A.\]

The set $$A$$ is open in $$M$$ when the first inclusion is an equality, and closed in $$M$$ when the second inclusion is an equality.

Definition 2.1.12. Let $$M$$ be a metric space and let $$A\subseteq M$.$ $$A$$ is open in $$M$$ if every point of $$A$$ is an interior point of $$A$.$ Equivalently,

\[\mathmakebox[9em][r]{A \text{ is open in } M} \mathmakebox[3em][c]{\iff} \mathmakebox[9em][l]{\operatorname{int} A = A.}\]

$$A$$ is closed in $$M$$ if it contains all of its limit points. Equivalently,

\[\mathmakebox[9em][r]{A \text{ is closed in } M} \mathmakebox[3em][c]{\iff} \mathmakebox[9em][l]{\overline A = A.}\]

Since openness was defined in terms of open balls, the first basic fact to verify is that open balls are themselves open.

Theorem 2.1.13. Every open ball is open.

Proof. Let $$M$$ be a metric space. Define $$B_r(a)$$ for some $a\in M$ and $$r>0$.$ If $x\in B_r(a)$ then $$d(x,a)<r.$$ We choose $$s>0$$ so that $$s<r-d(x,a).$$ Since, for any $y\in M,$

\[d(y,x)<s \quad \Longrightarrow \quad d(y,a)\le d(y,x)+d(x,a)<s+d(x,a)<r,\]

we have $B_s(x)\subseteq B_r(a).$ Hence the ball $$B_r(a)$$ is open. $\square$

Theorem 2.1.14. Let $$M$$ be a metric space, and let $$A\subseteq M$.$ $$A$$ is open if and only if $$A^c$$ is closed.

Proof. Suppose that $$A$$ is open and $x\in A.$ Then we can choose $$r>0$$ so that $B_r(x)\subseteq A.$ It follows that $B^{\ast}_r(x)\cap A^c = \varnothing,$ so $$x$$ is not a limit point of $$A^c.$$ Hence every limit point of $$A^c$$ is a member of $$A^c,$$ and thus $$A^c$$ is closed. Conversely, suppose that $$A^c$$ is closed and let $x\in A.$ Then $$x$$ is not a limit point of $$A^c.$$ Hence we can choose $$r>0$$ so that $B^{\ast}_r(x)\cap A^c=\varnothing.$ It follows that $B_r(x)\subseteq A.$ Therefore $$A$$ is open. $\square$

Complementation turns the duality between open and closed sets into corresponding rules for unions and intersections.

Theorem 2.1.15. Let $$M$$ be a metric space.

If $\lbrace U_i\rbrace_{i\in I}$ is a collection of open sets, then $\bigcup_{i\in I } U_i$ is open.
If $\lbrace F_i\rbrace_{i\in I}$ is a collection of closed sets, then $\bigcap_{i\in I } F_i$ is closed.
If $\lbrace U_i\rbrace_{i\in I}$ is a finite collection of open sets, then $\bigcap_{i\in I } U_i$ is open.
If $\lbrace F_i\rbrace_{i\in I}$ is a finite collection of closed sets, then $\bigcup_{i\in I } F_i$ is closed.

Proof. (1) Let $x\in \bigcup_{i\in I } U_i.$ Then we may let $x\in U_i$ for some $i\in I.$ Since $$x$$ is an interior point of $$U_i$,$ $$x$$ is also an interior point of $$\bigcup_{i\in I } U_i$.$ Therefore $\bigcup_{i\in I } U_i$ is open. (2) By Theorem 2.1.14, $$F_i^c$$ is open for every $i\in I.$ Then $\bigcup_{i\in I } F^c_i$ is open by (1). Since

\[\bigcup_{i\in I } F^c_i = \lowparen{\bigcap_{i\in I } F_i}^{\!\!c},\]

$\bigcap_{i\in I } F_i$ is closed by Theorem 2.1.14. (3) With the standard convention $$\bigcap_{\varnothing}U_i = M$,$ if $I=\varnothing$ then $\bigcap_{\varnothing}U_i$ is open. Suppose $I\ne \varnothing$ and $x\in \bigcap_{i\in I } U_i.$ Then we have $x\in U_i$ for every $$i\in I$.$ Since $$U_i$$ is open, there exists $$r_i>0$$ such that $B_{r_i}(x)\subseteq U_i.$ Since $$I$$ is finite, we may let

\[r = \min_{i\in I}r_i.\]

Then $$B_r(x)\subseteq \bigcap_{i\in I } U_i$.$ Hence $\bigcap_{i\in I } U_i$ is open. (4) By Theorem 2.1.14, $$F_i^c$$ is open for every $i\in I.$ Then $\bigcap_{i\in I } F^c_i$ is open by (3). Since

\[\bigcap_{i\in I } F^c_i = \lowparen{\bigcup_{i\in I } F_i}^{\!\!c},\]

$\bigcup_{i\in I } F_i$ is closed by Theorem 2.1.14. $\square$

Theorem 2.1.16. Let $$M$$ be a metric space and let $$A\subseteq M$.$ Then,

the interior of $$A$$ is the union of all subsets of $$A$$ open in $$M$;$
the closure of $$A$$ is the intersection of all supersets of $$A$$ closed in $$M$.$

Proof. (1) Let $$U$$ be the union of all subsets of $$A$$ open in $$M$.$ If $$x\in \operatorname{int} A$,$ then $B_r(x)\subseteq A$ for some $$r>0$.$ Since $$B_r(x)$$ is open in $$M$,$ we have $$x\in B_r(x)\subseteq U$.$ Conversely, if $$x\in U$,$ then there is a subset $$U_i$$ of $$A$$ open in $$M$.$ Then $$x$$ has an open ball $$B_r(x)$$ contained in $$U_i$.$ It follows that $$B_r(x)\subseteq A$,$ so $$x\in \operatorname{int} A$.$ Consequently, $$\operatorname{int} A = U$.$ (2) Let $$F$$ be the intersection of all supersets of $$A$$ closed in $$M$.$ If $$x\in \overline A$,$ then every open ball of $$x$$ meets $$A$.$ Since $$A\subseteq F$,$ $$x$$ is a closure point of $$F$.$ Since $$F$$ is closed, $$x\in F$.$ Conversely, if $$x$$ is not a closure point of $$A$,$ then $B_r(x)\cap A = \varnothing$ for some $$r>0$.$ It follows that $$A\subseteq M\setminus B_r(x)$.$ Thus the set $M\setminus B_r(x)$ is a superset of $$A$$ closed in $$M$.$ Since $F \subseteq M\setminus B_r(x)$ and $$x\notin M\setminus B_r(x)$,$ we have $$x\notin F$.$ Consequently, $$\overline A = F$.$ $\square$

Corollary 2.1.17. Let $$M$$ be a metric space and let $$A\subseteq M$.$ The interior of $$A$$ is open in $$M$,$ and the closure of $$A$$ is closed in $$M$.$

So far, all notions have been considered within one ambient space $$M$.$ If $N\subseteq M$ is equipped with the restricted metric, however, a set $A\subseteq N$ may be studied either in $$M$$ or in $$N$.$ We first prove that an open set in a subspace is simply the visible portion, inside that subspace, of an open set in the larger metric space.

Theorem 2.1.18. Let $$M$$ be a metric space, and regard $N\subseteq M$ as a metric space with the restricted metric. Then a subset $$A$$ of $$N$$ is open in $$N$$ if and only if $A=U\cap N$ for some $$U$$ open in $$M$.$

Proof. Suppose $A\subseteq N$ is open in $$N$.$ Then, every $x\in A$ has an open ball $B_{r_x}^N(x)$ contained in $$A$.$ Define $$U = \bigcup_{x\in A} B^{M}_{r_x}(x)$.$ Since

\[B^N_{r_x}(x) = B^M_{r_x}(x) \cap N,\tag{$\ast$}\]

we have

\[\begin{aligned} U\cap N &=\lowparen{\bigcup_{x\in A}B_{r_x}^M(x)}\cap N\\ &=\bigcup_{x\in A}\left(B_{r_x}^M(x)\cap N\right)\\ &=\bigcup_{x\in A}B_{r_x}^N(x). \end{aligned}\]

It is clear that $$A\subseteq \bigcup_{x\in A}B_{r_x}^N(x)$.$ Since each $B_{r_x}^N(x)$ is contained in $$A$,$ we also have $$\bigcup_{x\in A}B_{r_x}^N(x) \subseteq A$.$ Therefore $$A = \bigcup_{x\in A}B_{r_x}^N(x) = U \cap N$.$ Conversely, suppose that $A=U\cap N$ for some $U\subseteq M$ open in $$M$.$ If $x\in A$ then $$x\in U$.$ Thus every $x\in A$ has an open ball $$B^M_r(x)$$ contained in $$U$.$ Since the intersection of $$U$$ with $$N$$ equals $$A$,$ the intersection of $$B^M_r(x)$$ with $$N$$ is contained in $$A$.$ Thus, by $$(\ast)$,$ we have $$B^N_r(x)\subseteq A$.$ Therefore $$A$$ is open in $$N$.$ $\square$

Theorem 2.1.19. Let $$M$$ be a metric space, and regard $N\subseteq M$ as a metric space with the restricted metric. Then, for a set $$A\subseteq N$,$

\[\overline A^{\raise{-0.5em}{N}} = \overline A^{\raise{-0.5em}{M}} \cap N.\]

Proof. If $$x\in \overline A^{\raise{-0.5em}{N}}$,$ then every open ball $$B^N_r(x)$$ of $$x$$ meets $$A$.$ Thus $$B^M_r(x)$$ meets $$A$.$ Hence $x\in \overline A^{\raise{-0.5em}{M}}$ and thus $$x\in \overline A^{\raise{-0.5em}{M}} \cap N$.$ If $$x\in \overline A^{\raise{-0.5em}{M}} \cap N$,$ then every open ball $$B^M_r(x)$$ of $$x$$ meets $$A$.$ Since $$A\subseteq N$,$ the intersection point must be in $$N$.$ Thus, even if we take the intersection of $$B^M_r(x)$$ with $$N$,$ it still meets $$A$;$ that is, $$B^N_r(x)$$ meets $$A$.$ Therefore $$x\in \overline A^{\raise{-0.5em}{N}}$.$ $\square$

Remark 2.1.20. An important asymmetry is that, in general,

\[\operatorname{int}_N A\neq \operatorname{int}_M A \cap N.\]

For instance, if $$M=\mathbb R$,$ $$N=[0,1]$,$ and $$A=[0,\tfrac12)$,$ then $$0\in\operatorname{int}_N A$,$ even though $$0\notin\operatorname{int}_{M}A$.$ Thus shrinking the ambient space may create new interior points, while closure in the smaller space is obtained simply by intersecting the ambient closure with that space.

References

Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.
Munkres, J. R. (2000). Topology (2nd ed.). Pearson.

External

2.1. Open Sets and Closed Sets

References