4.10: Arcs and Curves. Connected Sets
This page is a draft and is under active development.
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.
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 \( E ^ { 1 } \).
Each arc is a compact (hence closed and bounded) set (by Theorem 1 of §8).
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 \( A . ) \)
(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.
\(A\) 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 \((A, \rho)\).
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 \((S, \rho) .\)
(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.
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.
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. \(\square\)
All arcs and curves are connected sets (by Definition 2 and Theorems 2 and 3).
\(A\) 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. \(\square\)
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.
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.
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.