Calculus of Real and Complex Variables

2.7 Connected Sets

Stated informally, connected sets are those which are in one piece. In order to define what is meant by this, I will first consider what it means for a set to not be in one piece. This is called separated. Connected sets are defined in terms of not being separated. This is why theorems about connected sets sometimes seem a little tricky.

Proof: First of all, denote by C the set of closed sets which contain A. Then

-- A = ∩C

and this will be closed if its complement is open. However,

--C { C } A = ∪ H : H ∈ C .

Each H^C is open and so the union of all these open sets must also be open. This is because if x is in this union, then it is in at least one of them. Hence it is an interior point of that one. But this implies it is an interior point of the union of them all which is an even larger set. Thus A is closed.

The interesting part is the next claim. First note that from the definition, A ⊆A so if x ∈ A, then x ∈A. Now consider y ∈ A^′ but y

A. If yA, a closed set, then there exists B ⊆A^C. Thus y cannot be a limit point of A, a contradiction. Therefore,

′ -- A ∪ A ⊆ A

Next suppose x ∈A and suppose x

A. Then if B contains no points of A different than x, since x itself is not in A, it would follow that B ∩A = ∅ and so recalling that open balls are open, B^C is a closed set containing A so from the definition, it also contains A which is contrary to the assertion that x ∈A. Hence if xA, then x ∈ A^′ and so

-- A ∪ A′ ⊇ A ■

Now is a definition about what it means to not be connected. This is called separated.

Note that the concept of connected sets is defined in terms of what it is not. This makes it somewhat difficult to understand. One of the most important theorems about connected sets is the following.

Proof: Suppose

K = A ∪B

where A ∩ B = B ∩ A = ∅,A≠∅,B≠∅. Let U ∈U. Then

U = (U ∩ A)∪ (U ∩B )

and this would separate U if both sets in the union are nonempty since the limit points of U ∩ B are contained in the limit points of B. It follows that every set of U is contained in one of A or B. Suppose then that some U ⊆ A. Then all U ∈U must be contained in A because if one is contained in B, this would violate the assumption that they all have a point p in common. Thus K is connected after all because this requires B = ∅. Alternatively, p is in one of these sets. Say p ∈ A. Then by the above argument every U must be in A because if not, the above would be a separation of U. Thus B = ∅. ■

The intersection of connected sets is not necessarily connected as is shown by the following picture.

Proof: To do this you show f

is not separated. Suppose to the contrary that f = A∪B where A and B separate f. Then consider the sets f⁻¹ and f⁻¹. If z ∈ f⁻¹, then f ∈ B and so f is not a limit point of A. Therefore, there exists an open set, U containing f such that U ∩ A = ∅. But then, the continuity of f implies that f⁻¹ is an open set containing z such that f⁻¹ ∩ f⁻¹ = ∅. Therefore, f⁻¹ contains no limit points of f⁻¹. Similar reasoning implies f⁻¹ contains no limit points of f⁻¹. It follows that X is separated by f⁻¹ and f⁻¹, contradicting the assumption that X was connected. ■

An arbitrary set can be written as a union of maximal connected sets called connected components. This is the concept of the next definition.

Proof: Let C denote the connected subsets of S which contain p. By Theorem 2.7.4, ∪C = C_p is connected. If x ∈ C_p ∩C_q, then from Theorem 2.7.4, C_p ⊇ C_p ∪C_q and so C_p ⊇ C_q. The inclusion goes the other way by the same reason. ■

This shows the connected components of a set are equivalence classes and partition the set.

A set, I is an interval in ℝ if and only if whenever x,y ∈ I then

⊆ I. The following theorem is about the connected sets in ℝ.

Proof: Let C be connected. If C consists of a single point, p, there is nothing to prove. The interval is just

. Suppose p < q and p,q ∈ C. You need to show ⊆ C. If

x ∈ (p,q) ∖C

let C ∩

≡ A, and C ∩ ≡ B. Then C = A∪B and the sets A and B separate C contrary to the assumption that C is connected.

Conversely, let I be an interval. Suppose I is separated by A and B. Pick x ∈ A and y ∈ B. Suppose without loss of generality that x < y. Now define the set,

S ≡ {t ∈ [x,y] : [x,t] ⊆ A}

and let l be the least upper bound of S. Then l ∈A so l

B which implies l ∈ A. But if lB, then for some δ > 0,

(l,l+ δ)∩ B = ∅

contradicting the definition of l as an upper bound for S. Therefore, l ∈B which implies l

A after all, a contradiction. It follows I must be connected. ■

This yields a generalization of the intermediate value theorem from one variable calculus.

Proof: From Theorem 2.7.5, f

is a connected subset of ℝ. By Theorem 2.7.8 f must be an interval. In particular, it must contain y. This proves the corollary. ■

The following theorem is a very useful description of the open sets in ℝ.

Proof: Let p ∈ U and let z ∈ C_p, the connected component determined by p. Since U is open, there exists, δ > 0 such that

⊆ U. It follows from Theorem 2.7.4 that

(z − δ,z + δ) ⊆ Cp.

This shows C_p is open. By Theorem 2.7.8, this shows C_p is an open interval,

where a,b ∈. There are therefore at most countably many of these connected components because each must contain a rational number and the rational numbers are countable. Denote by _i=1^∞ the set of these connected components. ■

An example of an arcwise connected space would be any subset of ℝⁿ which is the continuous image of an interval. Arcwise connected is not the same as connected. A well known example is the following.

{ ( ) } x,sin 1- : x ∈ (0,1] ∪ {(0,y) : y ∈ [− 1,1]} (2.18) x

(2.18)

You can verify that this set of points in ℝ² is not arcwise connected but is connected.

Proof: This is easy from the convexity of the set. If x,y ∈ B

, then let γ = x + t for t ∈.

showing γ stays in B.■

Proof:   Let p ∈ X. Then by assumption, for any x ∈ X, there is an arc joining p and x. This arc is connected because it is the continuous image of an interval which is connected. Since x is arbitrary, every x is in a connected subset of X which contains p. Hence C_p = X and so X is connected. ■

Proof: By Proposition 2.7.13 it is only necessary to verify that if U is connected and open in the context of this theorem, then U is arcwise connected. Pick p ∈ U. Say x ∈ U satisfies P if there exists a continuous function, γ :

→ U such that γ = p and γ = x.

A ≡ {x ∈ U such that x satisfies P.}

If x ∈ A, then Lemma 2.7.12 implies B

⊆ U is arcwise connected for small enough r. Thus letting y ∈ B, there exist intervals, and and continuous functions having values in U, γ,η such that γ = p,γ = x,η = x, and η = y. Then let γ₁ : → U be defined as

{ γ (t) if t ∈ [a,b] γ1(t) ≡ η (t+ c− b) if t ∈ [b,b+ d − c]

Then it is clear that γ₁ is a continuous function mapping p to y and showing that B

⊆ A. Therefore, A is open. A≠∅ because since U is open there is an open set, B containing p which is contained in U and is arcwise connected.

Now consider B ≡ U ∖A. I claim this is also open. If B is not open, there exists a point z ∈ B such that every open set containing z is not contained in B. Therefore, letting B

be such that z ∈ B ⊆ U, there exist points of A contained in B. But then, a repeat of the above argument shows z ∈ A also. Hence B is open and so if B≠∅, then U = B ∪ A and so U is separated by the two sets B and A  contradicting the assumption that U is connected.

It remains to verify the connected components are open. Let z ∈ C_p where C_p is the connected component determined by p. Then picking B

⊆ U, C_p ∪B is connected and contained in U and so it must also be contained in C_p. Thus z is an interior point of C_p. ■

As an application, consider the following corollary.

Proof: Suppose not. Then it achieves two different values, k and l≠k. Then Ω = f⁻¹

∪f⁻¹ and these are disjoint nonempty open sets which separate Ω. To see they are open, note

( ( ) ) f−1({m ∈ ℤ : m ⁄= l}) = f−1 ∪m⁄=l m − 1,m + 1 6 6

which is the inverse image of an open set while f⁻¹

= f⁻¹ also an open set. ■