Questions: Open Sets on the Real Line

Question 1

Which of the following sets is open in ℝ?

• A. [0, 1) — it contains 0 but not 1, so it is half-open and qualifies as open
• B. (0, 1) ∪ (2, 3) — the union of two open intervals
• C. [0, 1] — the closed interval contains all its boundary points
• D. {x ∈ ℝ : x ≠ 0} is not open because it excludes exactly one point

Answer: (B)

A finite or infinite union of open sets is always open. (0,1) and (2,3) are open intervals, so their union is open. [0,1) fails because 0 has no ε-neighborhood contained in the set — any interval (−ε, ε) extends to negative numbers, which are outside [0,1). [0,1] fails for both endpoints. The set {x : x ≠ 0} = (−∞,0) ∪ (0,∞) is actually open — it is a union of two open rays. The key test is always: does every point have a full ε-neighborhood inside the set?

Question 2

Consider the family of open intervals Iₙ = (−1/n, 1/n) for n = 1, 2, 3, …. What is their infinite intersection ∩ₙ Iₙ, and is it open?

• A. The empty set ∅, which is open
• B. The single point {0}, which is not open
• C. The interval (−1, 1), which is open
• D. All of ℝ, which is open

Answer: (B)

Every interval (−1/n, 1/n) contains 0, so 0 ∈ ∩ₙ Iₙ. But for any other point x ≠ 0, choosing n large enough so that 1/n < |x| excludes x from Iₙ, so x ∉ ∩ₙ Iₙ. Therefore the intersection is exactly {0}. The single-point set {0} is not open: any ε-neighborhood (−ε, ε) contains points other than 0, which are not in {0}. This counterexample shows why the axiom requires *finite* intersections of open sets to be open — infinite intersections can collapse to a non-open set.

Question 3

The empty set ∅ is open in ℝ because the condition for openness is vacuously satisfied — there are no points in ∅ to violate it.

• T. True
• F. False

Answer: True

The definition says: for every x ∈ U, there exists ε > 0 such that (x−ε, x+ε) ⊆ U. If U = ∅, the universal quantifier 'for every x ∈ ∅' ranges over an empty domain — no x exists to check — so the statement is vacuously true. This may feel unsatisfying, but the vacuous case is logically sound and necessary: ∅ must be open to satisfy the topology axioms (unions and intersections of open sets must be open, and ∅ = ℝ ∩ ∅ must itself be one of the axiomatically declared open sets).

Question 4

An infinite union of open sets is typically open, and so is an infinite intersection of open sets.

• T. True
• F. False

Answer: False

Arbitrary (including infinite) *unions* of open sets are always open. But infinite *intersections* need not be open. The canonical counterexample: ∩ₙ(−1/n, 1/n) = {0}, a single point that is not open. The intuition: to prove a point has an ε-neighborhood inside a *finite* intersection, you can take the minimum of finitely many ε values. With infinitely many, the infimum of those ε values may be zero — giving you no room to maneuver.

Question 5

Explain in your own words why the endpoint a of the closed interval [a, b] prevents [a, b] from being open.

Model answer: For [a,b] to be open, every point in it must have some ε-neighborhood entirely contained in [a,b]. But for the endpoint a, any ε-neighborhood (a−ε, a+ε) extends to the left of a by ε. The points in (a−ε, a) are less than a and therefore not in [a,b]. So no matter how small ε is, the ε-neighborhood of a always leaks outside [a,b]. Since a ∈ [a,b] but has no valid ε-neighborhood inside [a,b], the set fails the openness condition.

This is exactly what 'interior point' means: a point is an interior point if it has some wiggle room in all directions while staying in the set. The endpoints of [a,b] are boundary points — they are 'at the edge' with half their ε-neighborhood falling outside the set. Open sets consist entirely of interior points; that is the definition restated in geometric terms.