Connected Sets

Matt McCutchen April 10, 2005

Contents

1 Kinds of connectedness 1 2 All unsplittable sets are capacitor-connected 1 3 All compact, capacitor-connected sets are unsplittable 2

Note: In this document, a juxtaposition of two points denotes the distance between them.

1 Kinds of connectedness

Unsplittable. Two point sets T_{1} and T_{2} split a point set S if the following conditions all hold:

• T_{1} and T_{2} are disjoint open sets.

• T_{1} and T_{2} both intersect S (T_{1} ∩ S ≠ ∅ and T_{2} ∩ S ≠ ∅).

• T_{1} and T_{2} together cover S (S ⊆ T_{1} ∪ T_{2}). A point set S is unsplittable if there do not exist sets T_{1} and T_{2} that split it. According to several sources, when mathematicians say “connected”, they usually mean this definition. Arcwise connected. A point set S is arcwise connected if, for all a, b ∈ S, there exists a continuous curve lying entirely in S that connects a and b. That is, there is a continuous function f : [0, 1] → S with f(0) = a and f(1) = b. Capacitor-connected. A point set S is capacitor-connected if, for all a, b ∈ S and all positive real numbers δ, there exists a sequence of points a_{0}, …, a_{n} ∈ S with a_{0} = a, a_{n} = b, and a_{i}a_{i+1} < δ for i = 0, …, n − 1.

2 All unsplittable sets are capacitor-connected

Let S be an unsplittable point set, let a ∈ S, and let δ be a positive real number. Generate a sequence of point sets A_{0}, A_{1}, … as follows: A_{0} = {a}, and for all i ≥ 0,

A_{i+1} = S ∩ ⋃_{x∈A_{i}} Ball_{δ}(x).

Let A^{∗} = ⋃_{i≥0} A_{i}.

Now, let

U = ⋃_{x∈A^{∗}} Ball_{δ/2}(x) and V = ⋃_{x∈S\A^{∗}} Ball_{δ/2}(x).

Some observations about the sets U and V:

• U and V are both open because they are both unions of open balls.

• Together they cover S because A^{∗} ⊆ U and S \ A^{∗} ⊆ V.

• They are disjoint. To prove this, suppose there were a point x in both sets. By the definition of U, there must be a point x_{U} ∈ A^{∗} with xx_{U} < δ/2, and since x_{U} ∈ A^{∗}, x_{U} ∈ A_{n} for some n. By the definition of V, there must also be a point x_{V} ∈ S \ A^{∗} with xx_{V} < δ/2. But

x_{U}x_{V} ≤ xx_{U} + xx_{V} < δ/2 + δ/2 = δ,

so x_{V} ∈ Ball_{δ}(x_{U}) ⊆ A_{n+1} by the definition of the A_{i}. But then x_{V} ∈ A^{∗}, contradicting the fact that x_{V} ∈ S \ A^{∗}.

• U ∩ S is nonempty (it contains a).

If V ∩ S were also nonempty, U and V would split S, but we know S is unsplittable, so V doesn’t intersect S. This means S \ A^{∗} = ∅, because if there existed x ∈ S \ A^{∗}, we’d have x ∈ V ∩ S. Thus A^{∗} = S.

By the definition of A^{∗}, any point b ∈ S must be in A_{n} for some n. Generate a sequence of points b_{n}, …, b_{0} as follows: b_{n} = b, and b_{i−1} ∈ A_{i−1} with b_{i}b_{i−1} < δ for i = n, …, 1. A b_{i} must exist at every step because of the way the A_{i} were constructed, and b_{0} = a because a is the only point in A_{0}.

The points a=b_{0}, b_{1}, …, b_{n} =b form a sequence of hops in S from a to b of lengths less than δ. The foregoing argument can be used to create such a sequence for all a, b ∈ S and all positive real numbers δ, so S is capacitor-connected. ■

3 All compact, capacitor-connected sets are unsplittable

Let S be a compact, capacitor-connected set split by sets T_{1} and T_{2}; we’ll prove the section title by reaching a contradiction.

Lemma. Let O and C be open and compact point sets, respectively. If O ⊇ C, there exists a positive real number δ such that for all points x ∈ C and y ∉ O, xy ≥ δ.

Proof: For i = 0, 1, …, let O_{i} be the set of all points x such that Ball_{2^{−i}}(x) ⊆ O. Clearly O_{i} ⊆ O_{j} when i ≤ j. If x is any point in O, since O is open, there exists a positive real number r such that Ball_{r}(x) ⊆ O, so x ∈ O_{i} where i = max(0, ⌈−log_{2}r⌉); thus ⋃_{i} O_{i} = O.

The O_{i} form an infinite set of open sets that together cover O and therefore C, so by the Heine-Borel Theorem, there also exists a finite set U of O_{i} whose members together cover C. Let n be the highest index of any of the sets O_{i} in U. Since O_{n} ⊇ O_{i} for all i ≤ n, O_{n} = ⋃ U and therefore O_{n} ⊇ C.

Let δ = 2^{−n}. Let x be any point in C. Since C ⊆ O_{n}, the definition of O_{n} tells us that Ball_{2^{−n}}(x) ⊆ O. Let y be any point not in O. Since Ball_{2^{−n}}(x) ⊆ O, y ∉ Ball_{2^{−n}}(x), so xy ≥ 2^{−n} = δ. □

Applying Lemma 1 with O ↦ T_{1} ∪ T_{2} and C ↦ S, we find that any point x in S is always at least a fixed distance δ from any point y not in either T_{i}. I claim now that any point x in S ∩ T_{1} is further than δ from any point y ∉ T_{1}; it is enough to prove this for y ∈ T_{2}.

Let f : xy → ℝ^{n} parametrize the line segment connecting x and y in the obvious fashion with f(0) = x and f(xy) = y. Clearly f(t_{1})f(t_{2}) = |t_{1} − t_{2}| for all t_{1}, t_{2} ∈ [0, xy]. Let P be the set of all t ∈ [0, xy] for which f(t) ∈ T_{1}.

P is nonempty (it contains 0) and has an upper bound (xy), so by the completeness axiom, it has a supremum s. That is, t ∉ P for all t > s, and there exists no s′ < s for which t ∉ P for all t > s′. The second part implies that, for all s′ < s, there exists t ∈ (s′, s] with t ∈ P. I now claim that f(s) is a point in neither T_{1} nor T_{2}.

If f(s) were in T_{1}, then, since T_{1} is open, there would exist a positive real number ϵ such that Ball_{ϵ}(f(s)) ⊆ T_{1}. In particular, f(s + ϵ/2) would be in T_{1}, so s + ϵ/2 ∈ P, contradicting the last paragraph. If f(s) were in T_{2}, there would again exist a positive real number ϵ such that Ball_{ϵ}(f(s)) ⊆ T_{2}. Setting s′ = s − ϵ, we find that there is a number t ∈ (s′, s] with t ∈ P, which means f(t) ∈ T_{1}. But f(s)f(t) < ϵ, so f(t) ∈ T_{2}, contradicting the fact that T_{1} and T_{2} are disjoint.

We know that x ∈ S and f(s) ∈ ℝ^{n} \ (T_{1} ∪ T_{2}), so Lemma 1 tells us that xf(s) ≥ δ. f(s) is strictly between x and y on a line segment, so xy > xf(s) ≥ δ, proving the claim that xy ≥ δ for all x ∈ S ∩ T_{1} and y ∉ T_{1}.

Since T_{1} and T_{2} split S, there exist points a ∈ S ∩ T_{1} and b ∈ S ∩ T_{2}. Since S is capacitor-connected, there must exist points a_{0}, …, a_{k} ∈ S with a_{0} = a, a_{k} = b, and a_{i}a_{i+1} < δ for i = 0, …, n − 1. Let j be the smallest index for which a_{j} ∈ T_{1} but a_{j+1} ∉ T_{1}; some such j must exist because a_{0} ∈ T_{1} and a_{k} ∉ T_{1}. Applying the claim, we find that a_{j}a_{j+1} > δ, contradicting the validity of the sequence of hops a_{i}. ■