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Part CCSIR NET December 2025the-cofinite-topology-is-coarser-than-euclidean-yet-not-regular-any-two-nonempty-opens-there-intersect

The cofinite topology is coarser than euclidean yet not regular any two nonempty opens there intersect

Let τE\tau_E denote the euclidean topology on R\mathbb{R}. Which of the following statements are necessarily true?

  1. A.(R,τE)(\mathbb{R}, \tau_E) is a normal space.✓
  2. B.Let τ\tau be a topology on R\mathbb{R}. If the identity function of R\mathbb{R} is a continuous map from (R,τE)(\mathbb{R},\tau_E) to (R,τ)(\mathbb{R},\tau), then (R,τ)(\mathbb{R},\tau) is a regular space.
  3. C.Let τ\tau be a topology on R\mathbb{R}. If the identity function of R\mathbb{R} is a continuous map from (R,τ)(\mathbb{R},\tau) to (R,τE)(\mathbb{R},\tau_E), then (R,τ)(\mathbb{R},\tau) is a Hausdorff space.✓
  4. D.R\mathbb{R} is a regular space in the finite-complement topology.

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.

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50 are analysed free — try those first.

Related counterexample: A product of normal spaces is normal

More on this topic

The chapter behind this: Separation axioms and what survives which construction — free to read

From Topology › Continuity, homeomorphism, separation axioms

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