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4.10: Arcs and Curves. Connected Sets


A deeper insight into continuity and the Darboux property can be gained by generalizing the notions of a convex set and polygon-connected set to obtain so-called connected sets.

I. As a first step, we consider arcs and curves.

Definition

A set $$A \subseteq ( S , \rho )$$ is called an arc iff $$A$$ is a continuous image of a compact interval $$[ a , b ] \subset E ^ { 1 } ,$$ i.e., iff there is a continuous mapping
$f : [ a , b ] \underset { \text { onto } } { \longrightarrow } A.$
If, in addition, $$f$$ is one to one, $$A$$ is called a simple arc with endpoints $$f ( a )$$ and $$f ( b )$$.
If instead $$f ( a ) = f ( b ) ,$$ we speak of a closed curve.
A curve is a continuous image of any finite or infinite interval in $$E ^ { 1 }$$.

corollary $$\PageIndex{1}$$

Each arc is a compact (hence closed and bounded) set (by Theorem 1 of §8).

Definition

A set $$A \subseteq ( S , \rho )$$ is said to be arcwise connected iff every two points $$p , q \in A$$ are in some simple arc contained in $$A .$$ (We then also say the $$p$$ and $$q$$ can be joined by an arc in $$A . )$$

Example $$\PageIndex{1}$$

(a) Every closed line segment $$L [ \overline { a } , \overline { b } ]$$ in $$E ^ { n } \left( ^ { * } \text { or in any other normed space } \right)$$ is a simple arc (consider the map $$f$$ in Example (1) of §8).
(b) Every polygon
$A = \bigcup _ { i = 0 } ^ { m - 1 } L \left[ \overline { p } _ { i } , \overline { p } _ { i + 1 } \right]$
is an arc (see Problem 18 in §8). It is a simple arc if the half-closed segments $$L \left[ \overline { p } _ { i } , \overline { p } _ { i + 1 } \right)$$ do not intersect and the points $$\overline { p } _ { i }$$ are distinct, for then the map $$f$$ in Problem 18 of §8 is one to one.
(c) It easily follows that every polygon-connected set is also arcwise connected; one only has to show that every polygon joining two points $$\overline { p } _ { 0 } , \overline { p } _ { m }$$ can be reduced to a simple polygon (not a self-intersecting one). See Problem $$2 .$$
However, the converse is false. For example, two discs in $$E ^ { 2 }$$ connected by a parabolic arc form together an arcwise- (but not polygonwise-) connected set.
(d) Let $$f _ { 1 } , f _ { 2 } , \ldots , f _ { n }$$ be real continuous functions on an interval $$I \subseteq E ^ { 1 }$$. Treat them as components of a function $$f : I \rightarrow E ^ { n }$$,
$f=\left(f_{1}, \ldots, f_{n}\right).$
Then $$f$$ is continuous by Theorem 2 in §3. Thus the image set $$f[I]$$ is a
curve in $$E^{n} ;$$ it is an arc if $$I$$ is a closed interval.
Introducing a parameter $$t$$ varying over $$I,$$ we obtain the parametric equations of the curve, namely,
$x_{k}=f_{k}(t), \quad k=1,2, \ldots, n.$
Then as $$t$$ varies over $$I$$ , the point $$\overline{x}=\left(x_{1}, \ldots, x_{n}\right)$$ describes the curve
$$f[I] .$$ This is the usual way of treating curves in $$E^{n}\left(^{*} \text { and } C^{n}\right)$$.

It is not hard to show that Theorem 1 in §9 holds also if $$B$$ is only arcwise connected (see Problem 3 below). However, much more can be proved by introducing the general notion of a connected set. We do this next.

II. For this topic, we shall need Theorems 2-4 of Chapter 3, §12, and Problem 15 of Chapter 4, §2. The reader is advised to review them. In particular, we have the following theorem.

Theorem $$\PageIndex{1}$$

$$A$$ function $$f :(A, \rho) \rightarrow\left(T, \rho^{\prime}\right)$$ is continuous on $$A$$ iff $$f^{-1}[B]$$ is closed in $$(A, \rho)$$ for each closed set $$B \subseteq\left(T, \rho^{\prime}\right) ;$$ similarly for open sets.

Indeed, this is part of Problem 15 in §2 with $$(S, \rho)$$ replaced by $$(A, \rho)$$.

Definition

A metric space $$(S, \rho)$$ is said to be connected iff $$S$$ is not the union $$P \cup Q$$ of any two nonvoid disjoint closed sets; it is disconnected otherwise.
A set $$A \subseteq(S, \rho)$$ is called connected iff $$(A, \rho)$$ is connected as a subspace of $$(S, \rho) ;$$ i.e., iff $$A$$ is not a union of two disjoint sets $$P, Q \neq \emptyset$$ that are closed (hence also open) in $$(A, \rho),$$ as a subspace of $$(S, \rho) .$$

Note 1. By Theorem 4 of Chapter 3, §12, this means that

$P=A \cap P_{1} \text { and } Q=A \cap Q_{1}$

for some sets $$P_{1}, Q_{1}$$ that are closed in $$(S, \rho) .$$ Observe that, unlike compact sets, a set that is closed or open in $$(A, \rho)$$ need not be closed or open in $$(S, \rho) .$$

Example $$\PageIndex{1}$$

(a') $$\emptyset$$ is connected.

(b') So is any one-point set $$\{p\} .$$ (Why?)

(c') Any finite set of two or more points is disconnected. (Why?)

Other examples are provided by the theorems that follow.

Theorem $$\PageIndex{2}$$

The only connected sets in $$E^{1}$$ are exactly all convex sets, i.e., finite and infinite intervals, including $$E^{1}$$ itself.

Proof

The proof that such intervals are exactly all convex sets in $$E^{1}$$ is left as an exercise.

Seeking a contradiction, suppose $$p \notin A$$ for some $$p \in(a, b), a, b \in A .$$ Let

$P=A \cap(-\infty, p) \text { and } Q=A \cap(p,+\infty).$

Then $$A=P \cup Q, a \in P, b \in Q,$$ and $$P \cap Q=\emptyset .$$ Moreover, $$(-\infty, p)$$ and $$(p,+\infty)$$ are open sets in $$E^{1} .$$ (Why?) Hence $$P$$ and $$Q$$ are open in $$A,$$ each being the intersection of $$A$$ with a set open in $$E^{1}$$ (see Note 1 above). As $$A=P \cup Q,$$ with $$P \cap Q=\emptyset,$$ it follows that $$A$$ is disconnected. This shows that if $$A$$ is connected in $$E^{1},$$ it must be convex.

Conversely, let $$A$$ be convex in $$E^{1} .$$ The proof that $$A$$ is connected is an almost exact copy of the proof given for Theorem 1 of §9, so we only briefly
sketch it here.

If $$A$$ were disconnected, then $$A=P \cup Q$$ for some disjoint sets $$P, Q \neq \emptyset$$, both closed in $$A .$$ Fix any $$p \in P$$ and $$q \in Q .$$ Exactly as in Theorem 1 of §9, select a contracting sequence of line segments (intervals) $$\left[p_{m}, q_{m}\right] \subseteq A$$ such that $$p_{m} \in P, q_{m} \in Q,$$ and $$\left|p_{m}-q_{m}\right| \rightarrow 0,$$ and obtain a point

$r \in \bigcap_{m=1}^{\infty}\left[p_{m}, q_{m}\right] \subseteq A$

so that $$p_{m} \rightarrow r, q_{m} \rightarrow r,$$ and $$r \in A .$$ As the sets $$P$$ and $$Q$$ are closed in $$(A, \rho),$$ Theorem 4 of Chapter $$3, \ 16$$ shows that both $$P$$ and $$Q$$ must contain the common limit $$r$$ of the sequences $$\left\{p_{m}\right\} \subseteq P$$ and $$\left\{q_{m}\right\} \subseteq Q .$$ This is impossible, however, since $$P \cap Q=\emptyset,$$ by assumption. This contradiction shows that $$A$$ cannot be disconnected. Thus all is proved. $$\square$$

Note 2. By the same proof, any convex set in a normed space is connected. In particular, $$E^{n}$$ and all other normed spaces are connected themselves.

Theorem $$\PageIndex{3}$$

If a function $$f : A \rightarrow\left(T, \rho^{\prime}\right)$$ with $$A \subseteq(S, \rho)$$ is relatively continuous on a connected set $$B \subseteq A,$$ then $$f[B]$$ is a connected set in $$\left(T, \rho^{\prime}\right)$$.

Proof

By definition (§1), relative continuity on $$B$$ becomes ordinary continuity when $$f$$ is restricted to $$B .$$ Thus we may treat $$f$$ as a mapping of $$B$$ into $$f[B],$$ replacing $$S$$ and $$T$$ by their subspaces $$B$$ and $$f[B] .$$

Seeking a contradiction, suppose $$f[B]$$ is disconnected, i.e.,

$f[B]=P \cup Q$

for some disjoint sets $$P, Q \neq \emptyset$$ closed in $$\left(f[B], \rho^{\prime}\right) .$$ Then by Theorem $$1,$$ with $$T$$ replaced by $$f[B],$$ the sets $$f^{-1}[P]$$ and $$f^{-1}[Q]$$ are closed in $$(B, \rho) .$$ They also are nonvoid and disjoint (as are $$P$$ and $$Q )$$ and satisfy

$B=f^{-1}[P \cup Q]=f^{-1}[P] \cup f^{-1}[Q]$

$(see Chapter 1, §4-7, Problem 6). Thus $$B$$ is disconnected, contrary to assumption. $$\square$$ corollary $$\PageIndex{2}$$ All arcs and curves are connected sets (by Definition 2 and Theorems 2 and 3). lemma $$\PageIndex{1}$$ $$A$$ set $$A \subseteq(S, \rho)$$ is connected iff any two points $$p, q \in A$$ are in some connected subset $$B \subseteq A .$$ Hence any arcwise connected set is connected. Proof Seeking a contradiction, suppose the condition stated in Lemma 1 holds but $$A$$ is disconnected, so $$A=P \cup Q$$ for some disjoint sets $$P \neq \emptyset, Q \neq \emptyset$$ both closed in $$(A, \rho)$$. Pick any $$p \in P$$ and $$q \in Q .$$ By assumption, $$p$$ and $$q$$ are in some connected $$\operatorname{set} B \subseteq A .$$ Treat $$(B, \rho)$$ as a subspace of $$(A, \rho),$$ and let \[ P^{\prime}=B \cap P \text { and } Q^{\prime}=B \cap Q.$

Then by Theorem 4 of Chapter $$3, §12, P^{\prime}$$ and $$Q^{\prime}$$ are closed in $$B$$ . Also, they are disjoint (for $$P$$ and $$Q$$ are $$)$$ and nonvoid (for $$p \in P^{\prime}, q \in Q^{\prime} ),$$ and

$B=B \cap A=B \cap(P \cup Q)=(B \cap P) \cup(B \cap Q)=P^{\prime} \cup Q^{\prime}.$

Thus $$B$$ is disconnected, contrary to assumption. This contradiction proves the lemma (the converse proof is trivial).

In particular, if $$A$$ is arcwise connected, then any points $$p, q$$ in $$A$$ are in some arc $$B \subseteq A,$$ a connected set by Corollary $$2 .$$ Thus all is proved. $$\square$$

corollary $$\PageIndex{3}$$

Any convex or polygon-connected set $$(e . g ., a \text { globe})$$ in $$E^{n}$$ (or in any other normed space) is arcwise connected, hence connected.

Proof

Use Lemma 1 and Example $$(\mathrm{c})$$ in part I of this section. $$\square$$

Caution: The converse fails. A connected set need not be arcwise connected, let alone polygon connected (see Problem 17). However, we have the following
theorem.

Theorem $$\PageIndex{4}$$

Every open connected set $$A$$ in $$E^{n}$$ (* or in another normed space) is also arcwise connected and even polygon connected.

Proof

If $$A=\emptyset,$$ this is "vacuously" true, so let $$A \neq \emptyset$$ and fix $$\overline{a} \in A$$.

Let $$P$$ be the set of all $$\overline{p} \in A$$ that can be joined with $$\overline{a}$$ by a polygon $$K \subseteq A$$ Let $$Q=A-P .$$ Clearly, $$\overline{a} \in P,$$ so $$P \neq \emptyset$$ . We shall show that $$P$$ is open, i.e., that each $$\overline{p} \in P$$ is in a globe $$G_{\overline{p}} \subseteq P .$$

Thus we fix any $$\overline{p} \in P .$$ As $$A$$ is open and $$\overline{p} \in A,$$ there certainly is a globe $$G_{\overline{p}}$$ contained in $$A$$ . Moreover, as $$G_{\overline{p}}$$ is convex, each point $$\overline{x} \in G_{\overline{p}}$$ is joined with $$\overline{p}$$ by the line segment $$L[\overline{x}, \overline{p}] \subseteq G_{\overline{p}} .$$ Also, as $$\overline{p} \in P,$$ some polygon $$K \subseteq A$$ joins $$\overline{p}$$ with $$\overline{a}$$ . Then

$K \cup L[\overline{x}, \overline{p}]$

is a polygon joining $$\overline{x}$$ and $$\overline{a},$$ and hence by definition $$\overline{x} \in P .$$ Thus each $$\overline{x} \in G_{\overline{p}}$$ is in $$P,$$ so that $$G_{\overline{p}} \subseteq P,$$ as required, and $$P$$ is open (also apen in $$A$$ as a subspace).

Next, we show that the set $$Q=A-P$$ is open as well. As before, if $$Q \neq \emptyset$$, fix any $$\overline{q} \in Q$$ and a globe $$G_{\overline{q}} \subseteq A,$$ and show that $$G_{\overline{q}} \subseteq Q .$$ Indeed, if some $$\overline{x} \in G_{\overline{q}}$$ were $$n o t$$ in $$Q,$$ it would be in $$P,$$ and thus it would be joined with $$\overline{a}$$ (fixed above) by a polygon $$K \subseteq A .$$ Then, however, $$\overline{q}$$ itself could be so joined by the polygon

$L[\overline{q}, \overline{x}] \cup K,$

implying that $$\overline{q} \in P,$$ not $$\overline{q} \in Q .$$ This shows that $$G_{\overline{q}} \subset Q$$ indeed, as claimed.

Thus $$A=P \cup Q$$ with $$P, Q$$ disjoint and open (hence clopen) in $$A .$$ The connectedness of $$A$$ then implies that $$Q=\emptyset . \quad(P \text { is not empty, as has been }$$ noted.) Hence $$A=P .$$ By the definition of $$P,$$ then, each point $$\overline{b} \in A$$ can be joined to $$\overline{a}$$ by a polygon. As $$\overline{a} \in A$$ was arbitrary, $$A$$ is polygon connected. $$\square$$

Finally, we obtain a stronger version of the intermediate value theorem.

Theorem $$\PageIndex{5}$$

If a function $$f : A \rightarrow E^{1}$$ is relatively continuous on a connected $$\operatorname{set} B \subseteq A \subseteq(S, \rho),$$ then $$f$$ has the Darboux property on $$B$$.

In fact, by Theorems 3 and $$2, f[B]$$ is a connected set in $$E^{1},$$ i.e., an interval. This, however, implies the Darboux property.