In ℝ with the standard topology, what is the closure of the set of rational numbers ℚ?
Aℚ itself, since rationals form a closed subset of ℝ
Bℝ, because every real number is a limit point of ℚ
CThe algebraic numbers, since irrationals cannot be limits of rational sequences
DThe closure is undefined because ℚ is not bounded
ℚ is dense in ℝ — every real number can be approximated arbitrarily closely by rationals. Formally, every x ∈ ℝ is a limit point of ℚ: every open interval around x contains a rational. So cl(ℚ) = ℚ ∪ {all limit points} = ℝ. Option A is wrong because ℚ is not closed — its complement (the irrationals) is not open. This example shows that a 'small' countable set can have closure equal to the entire uncountable space.
Question 2 Multiple Choice
Let A = (0, 1) and B = (1, 2) be open intervals in ℝ. Which statement correctly describes their closures?
Acl(A ∩ B) = cl(A) ∩ cl(B) = {1}, showing closure distributes over intersections
Bcl(A ∪ B) = cl(A) ∪ cl(B) = [0, 2], but cl(A ∩ B) = ∅ while cl(A) ∩ cl(B) = {1}
Ccl(A ∩ B) = cl(A) ∪ cl(B) = [0, 1] ∪ [1, 2]
Dcl(A) ∩ cl(B) = ∅ because A and B do not overlap
A ∩ B = ∅, so cl(A ∩ B) = cl(∅) = ∅. But cl(A) = [0,1] and cl(B) = [1,2], so cl(A) ∩ cl(B) = {1} ≠ ∅. This shows cl(A ∩ B) ⊊ cl(A) ∩ cl(B) — containment is strict, not equality. The closure operator distributes exactly over unions: cl(A ∪ B) = cl(A) ∪ cl(B) = [0,2]. But for intersections, the closures of the pieces can 'see' a common boundary point ({1}) that the pieces themselves never reach.
Question 3 True / False
Applying the closure operator twice gives the same result as applying it once: cl(cl(A)) = cl(A) for any set A.
TTrue
FFalse
Answer: True
True — this is idempotence, the third Kuratowski axiom. cl(A) is already a closed set (it is an intersection of closed sets, and arbitrary intersections of closed sets are closed). The closure of a closed set is itself. Once you have captured all limit points of A to form cl(A), there are no new limit points to add — cl(A) already contains all its own limit points by definition of being closed.
Question 4 True / False
For any two sets A and B in a topological space, cl(A ∩ B) = cl(A) ∩ cl(B).
TTrue
FFalse
Answer: False
False. The correct statement is only the containment cl(A ∩ B) ⊆ cl(A) ∩ cl(B). The counterexample is A = (0,1) and B = (1,2): A ∩ B = ∅, so cl(A ∩ B) = ∅, but cl(A) ∩ cl(B) = [0,1] ∩ [1,2] = {1} ≠ ∅. The closures of the two pieces share the boundary point 1 even though the pieces themselves are disjoint. Equality holds for unions (cl(A ∪ B) = cl(A) ∪ cl(B)) but only inclusion holds for intersections.
Question 5 Short Answer
Why is cl(ℚ) = ℝ in the standard topology on ℝ, and what does this say about ℚ's relationship to ℝ?
Think about your answer, then reveal below.
Model answer: Every real number x is a limit point of ℚ: for any open interval (x − ε, x + ε), there exists a rational number inside it (the Archimedean property guarantees this). So every x ∈ ℝ belongs to ℚ ∪ {limit points of ℚ} = cl(ℚ). Since cl(ℚ) ⊆ ℝ trivially, we get cl(ℚ) = ℝ. This means ℚ is dense in ℝ — a set is dense in a space precisely when its closure equals the whole space.
Density is a fundamental concept: ℚ is 'everywhere' in ℝ in the sense that you can get arbitrarily close to any real number using rationals. This is why analysis works — theorems about real numbers are often proved by taking limits of rational approximations. The closure operation makes 'density' precise: S is dense in X if and only if cl(S) = X.