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# 3.8: Open and Closed Sets. Neighborhoods

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I. Let $$A$$ be an open globe in $$(S, \rho)$$ or an open interval $$(\overline{a}, \overline{b})$$ in $$E^{n} .$$ Then every $$p \in A$$ can be enclosed in a small globe $$G_{p}(\delta) \subseteq A($$ Figures 7 and 8$$)$$. (This would fail for "boundary" points; but there are none inside an open $$G_{q}$$ or $$(\overline{a}, \overline{b}) . )$$.  This suggests the following ideas, for any $$(S, \rho)$$.

Definition

A point $$p$$ is said to be interior to a set $$A \subseteq(S, \rho)$$ iff $$A$$ contains some $$G_{p} ;$$ i.e., $$p,$$ together with some globe $$G_{p},$$ belongs to $$A .$$ We then also say that $$A$$ is a neighborhood of $$p .$$ The set of all interior points of $$A$$ ("the interior of $$A^{\prime \prime}$$ is denoted $$A^{0} .$$ Note: $$\emptyset^{0}=\emptyset$$ and $$S^{0}=S !$$

Definition

$$\mathrm{A}$$ set $$A \subseteq(S, \rho)$$ is said to be open iff $$A$$ coincides with its interior $$\left(A^{0}=A\right) .$$ Such are $$\emptyset$$ and $$S .$$

Example $$\PageIndex{1}$$

(1) As noted above, an open globe $$G_{q}(r)$$ has interior points only, and thus is an open set in the sense of Definition $$2 .$$ (See Problem 1 for a proof.)

(2) The same applies to an open interval $$(\overline{a}, \overline{b})$$ in $$E^{n} .$$ (See Problem $$2 . )$$

(3) The interior of any interval in $$E^{n}$$ never includes its endpoints $$\overline{a}$$ and $$\overline{b}$$. In fact, it coincides with the open interval $$(\overline{a}, \overline{b}) .$$ (See Problem $$4 . )$$

(4) The set $$R$$ of all rationals in $$E^{1}$$ has no interior points at all $$\left(R^{0}=\emptyset\right)$$ because it cannot contain any $$G_{p}=(p-\varepsilon, p+\varepsilon) .$$ Indeed, any such $$G_{p}$$ contains irrationals (see Chapter 2, §§11-12, Problem 5$$),$$ so it is not entirely contained in $$R .$$

Theorem $$\PageIndex{1}$$

(Hausdorff property). Any two points $$p$$ and $$q$$ $$(p \neq q)$$ in $$(S, \rho)$$ are centers of two disjoint globes.

More precisely,

$(\exists \varepsilon>0) \quad G_{p}(\varepsilon) \cap G_{q}(\varepsilon)=\emptyset.$

Proof

As $$p \neq q,$$ we have $$\rho(p, q)>0$$ by metric axiom $$\left(\mathrm{i}^{\prime}\right) .$$ Thus we may put

$\varepsilon=\frac{1}{2} \rho(p, q)>0.$

It remains to show that with this $$\varepsilon, G_{p}(\varepsilon) \cap G_{q}(\varepsilon)=\emptyset$$.

Seeking a contradiction, suppose this fails. Then there is $$x \in G_{p}(\varepsilon) \cap G_{q}(\varepsilon)$$ so that $$\rho(p, x)<\varepsilon$$ and $$\rho(x, q)<\varepsilon .$$ By the triangle law,

$\rho(p, q) \leq \rho(p, x)+\rho(x, q)<\varepsilon+\varepsilon=2 \varepsilon ; \text{ i.e., } \rho(p, q)<2 \varepsilon,$

which is impossible since $$\rho(p, q)=2 \varepsilon . \square$$  Note. A look at Figure 9 explains the idea of this proof, namely, to obtain two disjoint globes of equal radius, it suffices to choose $$\varepsilon \leq \frac{1}{2} \rho(p, q) .$$ The reader is advised to use such diagrams in $$E^{2}$$ as a guide.

II. We can now define closed sets in terms of open sets.

Definition

A set $$A \subseteq(S, \rho)$$ is said to be closed iff its complement $$-A=S-A$$ is open, i.e., has interior points only.

That is, each $$p \in-A$$ (outside $$A )$$ is in some globe $$G_{p} \subseteq-A$$ so that

$A \cap G_{p}=\emptyset.$

Example $$\PageIndex{1}$$

(Continued).

(5) The sets $$\emptyset$$ and $$S$$ are closed, for their complements, $$S$$ and $$\emptyset,$$ are open, as noted above. Thus a set may be both closed and open ("clopen").

(6) All closed globes in $$(S, \rho)$$ and all closed intervals in $$E^{n}$$ are closed sets by Definition $$3 .$$ Indeed (see Figures 9 and $$10 ),$$ if $$A=\overline{G}_{q}(r)$$ or $$A=[\overline{a}, \overline{b}]$$, then any point $$p$$ outside $$A$$ can be enclosed in a globe $$G_{p}(\delta)$$ disjoint from $$A ;$$ so, by Definition $$3, A$$ is closed (see Problem 12$$)$$.

(7) A one-point set $$\{q\}$$ (also called "singleton") in $$(S, \rho)$$ is always closed, for any $$p$$ outside $$\{q\}(p \neq q)$$ is in a globe disjoint from $$\{q\}$$ by Theorem 1 In a discrete space (§§11,) Example (3)), $$\{q\}$$ is also open since it is an open $$g$$ lobe, $$\{q\}=G_{q}\left(\frac{1}{2}\right)($$ why? $$) ;$$ so it is "clopen. " Hence, in such a space, all sets are "clopen". For $$p \in A$$ implies $$\{p\}=G_{p}\left(\frac{1}{2}\right) \subseteq A ;$$ similarly for $$-A .$$ Thus $$A$$ and $$-A$$ have interior points only, so both are open.

(8) The interval $$(a, b]$$ in $$E^{1}$$ is neither open nor closed. (Why?)

III. (The rest of this section may be deferred until Chapter (4, §10.)

Theorem $$\PageIndex{2}$$

The union of any finite or infinite family of open sets $$A_{i}(i \in I)$$, denoted

$\bigcup_{i \in I} A_{i},$

is open itself. So also is

$\bigcap_{i=1}^{n} A_{i}$

for finitely many open sets. (This fails for infinitely many sets $$A_{i} ;$$ see Problem 11 below.)

Proof

We must show that any point $$p$$ of $$A=\bigcup_{i} A_{i}$$ is interior to $$A$$.

Now if $$p \in \bigcup_{i} A_{i}, p$$ is in some $$A_{i},$$ and it is an interior point of $$A_{i}$$ (for $$A_{i}$$ is open, by assumption). Thus there is a globe

$G_{p} \subseteq A_{i} \subseteq A,$

as required.

For finite intersections, it suffices to consider two open sets $$A$$ and $$B$$ (for $$n$$ sets, all then follows by induction). We must show that each $$p \in A \cap B$$ is interior to $$A \cap B .$$

Now as $$p \in A$$ and $$A$$ is open, we have some $$G_{p}\left(\delta^{\prime}\right) \subseteq A .$$ Similarly, there is $$G_{p}\left(\delta^{\prime \prime}\right) \subseteq B .$$ Then the smaller of the two globes, call it $$G_{p},$$ is in both $$A$$ and $$B,$$ so

$G_{p} \subseteq A \cap B$

and $$p$$ is interior to $$A \cap B,$$ indeed. $$\square$$

Theorem $$\PageIndex{3}$$

If the sets $$A_{i}(i \in I)$$ are closed, so is

$\bigcap_{i \in I} A_{i}$

(even for infinitely many sets). So also is

$\bigcup_{i=1}^{n} A_{i}$

for finitely many closed sets $$A_{i} .$$ (Again, this fails for infinitely many sets $$A_{i} . )$$

Proof

Let $$A=\bigcap_{i \in I} A_{i} .$$ To prove that $$A$$ is closed, we show that $$-A$$ is open.

Now by set theory (see Chapter 1, §§1-3, Theorem 2) ,

$-A=-\bigcap_{i} A_{i}=\bigcup_{i}\left(-A_{i}\right),$

where the $$\left(-A_{i}\right)$$ are open (for the $$A_{i}$$ are closed $$) .$$ Thus by Theorem $$2,-A$$ is open, as required.

The second assertion (as to $$\bigcup_{i=1}^{n} A_{i} )$$ follows quite similarly. $$\square$$

corollary $$\PageIndex{1}$$

$$A$$ nonempty set $$A \subseteq(S, \rho)$$ is open iff $$A$$ is a union of open globes.

For if $$A$$ is such a union, it is open by Theorem $$2 .$$ Conversely, if $$A$$ is open, then each $$p \in A$$ is in some $$G_{p} \subseteq A .$$ All such $$G_{p}(p \in A)$$ cover all of $$A,$$ so $$A \subseteq \bigcup_{p \in A} G_{p} .$$ Also, $$\bigcup_{p \in A} G_{p} \subseteq A$$ since all $$G_{p}$$ are in $$A .$$ Thus

$A=\bigcup_{p \in A} G_{p}.$

corollary $$\PageIndex{2}$$

Every finite set $$F$$ in a metric space $$(S, \rho)$$ is closed.

Proof

If $$F=\emptyset, F$$ is closed by Example $$(5) .$$ If $$F \neq \emptyset,$$ let

$F=\left\{p_{1}, \ldots, p_{n}\right\}=\bigcup_{k=1}^{n}\left\{p_{k}\right\}.$

Now by Example $$(7),$$ each $$\left\{p_{k}\right\}$$ is closed; hence so is $$F$$ by by theorem $$3 . \square$$

Note. The family of all open sets in a given space $$(S, \rho)$$ is denoted by $$\mathcal{G}$$; that of all closed sets, by $$\mathcal{F} .$$ Thus $$" A \in \mathcal{G}^{\prime \prime}$$ means that $$A$$ is open; "A $$\in \mathcal{F}^{\prime \prime}$$ means that $$A$$ is closed. By Theorems 2 and $$3,$$ we have

$(\forall A, B \in \mathcal{G}) \quad A \cup B \in \mathcal{G} \text{ and } A \cap B \in \mathcal{G};$

similarly for $$\mathcal{F} .$$ This is a kind of "closure law." We say that $$\mathcal{F}$$ and $$\mathcal{G}$$ are "closed under finite unions and intersections."

In conclusion, consider any subspace $$(A, \rho)$$ of $$(S, \rho) .$$ As we know from §11 it is a metric space itself, so it has its own open and closed sets (which must consist of points of $$A$$ only $$) .$$ We shall now show that they are obtained from those of $$(S, \rho)$$ by intersecting the latter sets with $$A .$$

Theorem $$\PageIndex{4}$$

Let $$(A, \rho)$$ be a subspace of $$(S, \rho) .$$ Then the open (closed) sets in $$(A, \rho)$$ are exactly all sets of the form $$A \cap U,$$ with $$U$$ open $$($$closed$$)$$ in $$S$$.

Proof

Let $$G$$ be open in $$(A, \rho) .$$ By Corollary $$1, G$$ is the union of some open globes $$G_{i}^{*}(i \in I)$$ in $$(A, \rho) .$$ (For brevity, we omit the centers and radii; we also omit the trivial case $$G=\emptyset .$$

$G=\bigcup_{i} G_{i}^{*}=\bigcup_{i}\left(A \cap G_{i}\right)=A \cap \bigcup_{i} G_{i},$

by set theory (see Chapter 1, §§1-3,\) Problem 9).

Again by Corollary $$1, U=\bigcup_{i} G_{i}$$ is an open set in $$(S, \rho) .$$ Thus $$G$$ has the form

$A \cap \bigcup_{i} G_{i}=A \cap U,$

with $$U$$ open in $$S,$$ as asserted.

Conversely, assume the latter, and let $$p \in G .$$ Then $$p \in A$$ and $$p \in U .$$ As $$U$$ is open in $$(S, \rho),$$ there is a globe $$G_{p}$$ in $$(S, \rho)$$ such that $$p \in G_{p} \subseteq U .$$ As $$p \in A,$$ we have

$p \in A \cap G_{p} \subseteq A \cap U.$

However, $$A \cap G_{p}$$ is a globe in $$(A, \rho),$$ call it $$G_{p}^{*} .$$ Thus

$p \in G_{p}^{*} \subseteq A \cap U=G;$

i.e., $$p$$ is an interior point of $$G$$ in $$(A, \rho) .$$ We see that each $$p \in G$$ is interior to $$G,$$ as a set in $$(A, \rho),$$ so $$G$$ is open in $$(A, \rho) .$$

This proves the theorem for open sets. Now let $$F$$ be closed in $$(A, \rho) .$$ Then by Definition $$3, A-F$$ is open in $$(A, \rho) .$$ (Of course, when working in $$(A, \rho)$$, we replace $$S$$ by $$A$$ in taking complements.) Let $$G=A-F,$$ so $$F=A-G,$$ and $$G$$ is open in $$(A, \rho) .$$ By what was shown above, $$G=A \cap U$$ with $$U$$ open in $$S$$.

Thus

$F=A-G=A-(A \cap U)=A-U=A \cap(-U)$

by set theory. Here $$-U=S-U$$ is closed in $$(S, \rho)$$ since $$U$$ is open there. Thus $$F=A \cap(-U),$$ as required.

The proof of the converse (for closed sets) is left as an exercise. $$\square$$