Part BCSIR NET December 2025a-constant-function-is-continuous-and-breaks-both-image-and-preimage-cardinality-claims-at-once

A constant function is continuous and breaks both image and preimage cardinality claims at once

Which of the following statements is necessarily true?

A.The set of all finite subsets of $\mathbb{Z}$ is uncountable.
B.Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous and one-one function. If $S \subseteq \mathbb{R}$ is a countably infinite set, then f(S) is a countably infinite set.✓
C.Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. Then there exists an uncountable subset $T \subseteq \mathbb{R}$ such that $f(T) \subseteq \mathbb{R}$ is uncountable.
D.Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. If R ⊆ image(f) is a countable set, then f⁻ $^{1}(R)$ is countable.

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.

50 are analysed free — try those first.

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