MATH 4630 — Connected Sets

We have now characterized compactness and perfectness. We turn to a third property: connectedness. Informally, a connected set is one that is "all in one piece." Making this precise is more subtle than it sounds.

Question: Consider these two sets: $$E_1 = (1,2) \cup (2,5) \quad \text{and} \quad E_2 = (1,5).$$ Intuitively E₁ has a "gap" at x = 2 while E₂ does not. But can we make this precise using only the language of limit points and closures?

Let A = (1,2) and B = (2,5). Work through the following:

Is 2 ∈ A? Is 2 ∈ B?
So A ∩ B =
Is 2 a limit point of A = (1,2)?
So Ā =
Does Ā ∩ B = [1,2] ∩ (2,5) equal ∅?
Does A ∩ B̄ = (1,2) ∩ [2,5] equal ∅?
Now try C = (1,2] and D = (2,5): does C̄ ∩ D = ∅? Is C ∩ D̄ = ∅?

Observation: For A = (1,2) and B = (2,5): neither set contains a limit point of the other, even though they share the limit point 2. For C = (1,2] and D = (2,5): the closure of C hits D, so they are not separated in the same way. This "mutual closure-disjointness" is the right notion of separation.

Definition

Two nonempty sets A, B ⊆ ℝ are separated if $$\overline{A} \cap B = \emptyset \quad\text{and}\quad A \cap \overline{B} = \emptyset.$$ A set E ⊆ ℝ is disconnected if it can be written as E = A ∪ B for some nonempty separated sets A and B. A set is connected if it is not disconnected.

Example 1: Let A = (1,2) and B = (2,5). Consider: E = (1,2) ∪ (2,5). Is the set E connected?

Example 2: Let C = (1,2] and D = (2,5), so that C ∪ D = (1,5). Does the pair C, D give a separation of (1,5)?

Characterisation of Connected Sets in ℝ

(⟹): Suppose E is connected. Let a, b ∈ E with a < c < b. Define: $$A = (-\infty, c) \cap E \quad \text{and} \quad B = (c, \infty) \cap E.$$

Step 1: Prove that a ∈ A and b ∈ B.

Step 2: Prove that A and B are separated.

Step 3: Use the connectedness of E to prove that c ∈ E.

Therefore this direction is proved. ∎

(⟸): Conversely, assume that E is an interval: whenever a, b ∈ E satisfy a < c < b for some c, then c ∈ E.

Let E = A ∪ B with A, B nonempty and disjoint. We show A and B are not separated.

Pick a₀ ∈ A, b₀ ∈ B, and assume WLOG a₀ < b₀.

Step 1: Why is I₀ = [a₀, b₀] ⊆ E?

Step 2: Bisect I₀ at its midpoint m. The midpoint lies in A or B. Explain why we can always choose a half I₁ = [a₁, b₁] ⊆ I₀ so that a₁ ∈ A and b₁ ∈ B.

Step 3: Continuing, we get nested intervals I₀ ⊇ I₁ ⊇ I₂ ⊇ ⋯ with:

a_n ∈ A, b_n ∈ B for each n, and
b_n − a_n = (b₀ − a₀)/2ⁿ → 0.

By the Nested Interval Property, there exists

Step 4: Show that a_n → x and b_n → x.

Step 5: Prove that A and B are not separated.

Therefore, E is connected. ∎