Questions: Dense Sets and Separability

Question 1

Which of the following is equivalent to 'A is dense in X'?

• A. A intersects every non-empty open set in X
• B. X is a subset of A
• C. A contains every limit point of X
• D. cl(A) is a countable subset of X

Answer: (A)

A is dense in X iff cl(A) = X, which is equivalent to: for every non-empty open set U in X, U ∩ A ≠ ∅. If A missed some non-empty open set U, then U ⊆ Aᶜ (which is closed), so cl(A) ⊆ X \ U ≠ X — contradicting density. This open-set characterization is often easier to check in practice and makes density purely a statement about A's interaction with open sets, with no metric needed.

Question 2

Is the set of even integers dense in ℝ (with the standard topology)?

• A. No — the open interval (0.1, 0.9) contains no even integers, so the even integers miss a non-empty open set
• B. Yes — between any two real numbers there is an even integer
• C. Yes — the even integers are an infinite set, and infinite sets are always dense in ℝ
• D. No — a dense subset of ℝ must be uncountable

Answer: (A)

The open interval (0.1, 0.9) contains no even integers, so the even integers fail the open-set criterion for density. Option B mirrors the correct reasoning for ℚ (between any two reals there is a rational) but is false for even integers — there is no even integer between 0 and 1. Option C confuses infinite cardinality with density; the integers are infinite and not dense. Option D confuses density with cardinality in the opposite direction; ℚ is countable and dense.

Question 3

If A is dense in X, then A is expected to be uncountable.

• T. True
• F. False

Answer: False

The rationals ℚ are countable yet dense in ℝ. Density says nothing about cardinality — it says A approximates every point of X arbitrarily well (every point of X is in A or is a limit point of A). A separable space is precisely a space with a *countable* dense subset, so countable dense subsets are not just possible but definitionally important.

Question 4

If A is dense in X, then every point of X is either an element of A or a limit point of A.

• T. True
• F. False

Answer: True

This is exactly what cl(A) = X means. The closure cl(A) is A together with all its limit points, and density requires this to equal all of X. So for any x ∈ X, either x ∈ A or x is a limit point of A (meaning every neighborhood of x contains a point of A distinct from x). This is the precise sense in which A 'spreads throughout' X.

Question 5

Explain why ℚ is dense in ℝ and why this is surprising given that ℚ is countable while ℝ is uncountable.

Model answer: ℚ is dense in ℝ because between any two real numbers there is a rational number — equivalently, every open interval (a, b) contains rationals, so ℚ meets every non-empty open set. This means cl(ℚ) = ℝ: every real number is a limit of a sequence of rationals. The surprise is that density is about approximation, not cardinality. ℚ cannot 'fill up' ℝ (there are uncountably many irrationals not in ℚ), but it can come arbitrarily close to every real, which is all density requires.

This example is the prototype for understanding density. The key insight is that 'spreading throughout a space' (density) is a different property from 'being the whole space.' You don't need every point — you need every neighborhood of every point to be visited. ℚ does this despite being a countably infinite 'thin' set inside an uncountably infinite one, which explains why analysis can work so smoothly using rational approximations to real numbers.