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Part CCSIR NET June 2025removing-the-rationals-leaves-a-set-that-is-still-dense-so-still-not-discrete

Removing the rationals leaves a set that is still dense so still not discrete

Consider R\mathbb{R} with the usual topology and S = {a+b2a + b\sqrt{2} | a,b∈Qa, b \in \mathbb{Q}} ⊂R\subset \mathbb{R} with the subspace topology. Which of the following statements are true?

  1. A.S is dense in R\mathbb{R}.✓ Correct
  2. B.S \ Q\mathbb{Q} is dense in R\mathbb{R}.✓ Correct
  3. C.S \ Q\mathbb{Q} is discrete with subspace topology on S.
  4. D.S is connected.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: An arbitrary intersection of open sets is open

More on this topic

The chapter behind this: Open and closed sets, closure, interior, boundary — free to read

From Metric Spaces › Open/closed sets, limit points, closure, interior

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