MATH 4630 — Perfect Sets

In the previous section we characterized compactness as being closed and bounded. We now introduce another important property of sets in ℝ: perfectness.

Question: Are all closed sets intervals?

Two ways a closed set can "fail to be an interval."

Recall that a point x ∈ A is an isolated point if some ε-neighbourhood V_ε(x) contains no other point of A; equivalently, x is not a limit point of A.

Consider these four sets. For each one, mark whether each point is a limit point (L) or an isolated point (I) of the set.

Set A	Sample point	L or I?	Closed?
[0,1]	any x ∈ (0,1)
[0,1]	the endpoint x = 0
ℕ = {1,2,3,…}	any n ∈ ℕ
{1/n : n ∈ ℕ} ∪ {0}	x = 1/n for fixed n
{1/n : n ∈ ℕ} ∪ {0}	x = 0

Observation. Closed sets come in (at least) two flavours:

Interval-like: [0,1] is closed and every point is a limit point.
Discrete-like: ℕ is closed but every point is isolated.
Mixed: {1/n} ∪ {0} has isolated points (1/n) and one limit point (0).

Question: What if we insist that a closed set has no isolated points at all? Then every point must be a limit point. This is exactly what compactness cannot guarantee on its own, and it leads to a new concept.

Your turn: For each set, decide whether it is perfect. If not, identify precisely which condition fails: not closed, has an isolated point, or both. Provide a brief justification.

[0,1] — Perfect?
{0} ∪ {1/n : n ∈ ℕ} — Perfect?
ℕ — Perfect?
[−1,0) ∪ (0,1] — Perfect?

What does "No Isolated Points" really mean?

The definition says every point of P is a limit point of P. Let us unpack this with a concrete family of examples that shows how delicate this condition is.

Example. Let A = {1/n : n ∈ ℕ} and B = A ∪ {0}.

(a) Is A closed?

(b) Is B = A ∪ {0} closed?

(d) Is the point x = 1/n (for fixed n) an isolated point of B?

(e) Is B perfect? What single element would you need to add to (or remove from) B to move closer to a perfect set?

Observation: Even adding the "right" limit point 0 does not cure the isolated-point problem—the isolated points 1/n remain. The simplest perfect sets are closed intervals.

Proof sketch. We must verify two things: (i) [a,b] is closed, and (ii) every x ∈ [a,b] is a limit point of [a,b].

Part (i): Why is [a,b] closed?

Part (ii): Let x ∈ [a,b] and let ε > 0 be arbitrary.

Step 1. Write down what V_ε(x) ∩ [a,b] looks like explicitly: $$V_\varepsilon(x) \cap [a,b] = \bigl(\underline{\hspace{2cm}}, \underline{\hspace{2cm}}\bigr).$$

Step 2. Show that this intersection is a nonempty open interval.

Step 3. Explain why a nonempty open interval cannot be a singleton {x}, and therefore must contain a point y ≠ x.

Conclude: V_ε(x) ∩ [a,b] contains a point y ∈ [a,b] with y ≠ x. Since ε > 0 was arbitrary, x is a limit point of [a,b]. Therefore, the closed interval [a,b] is perfect. ∎

A Shocking Theorem: Perfect Sets are Big

Activity:

Exercise 1. Let P be a perfect set and K compact.

Is P ∩ K always compact? Justify your answer. (Hint: Is the intersection of a closed set and a compact set compact?)
Is P ∩ K always perfect? If not, give a counterexample.

Exercise 2. Prove that no finite nonempty set F ⊆ ℝ is perfect.

Follow-up. Use this and Theorem 2 to give another proof that every perfect set is infinite.