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 $R$ with the usual topology and S = { $a + b 2$ | $a, b \in Q$ } $\subset R$ with the subspace topology. Which of the following statements are true?

A.S is dense in $R$ .✓
B.S \ $Q$ is dense in $R$ .✓
C.S \ $Q$ is discrete with subspace topology on S.
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.

See pricing

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

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

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