Part CCSIR NET June 2025continuity-controls-images-of-connected-sets-not-preimages

Continuity controls images of connected sets not preimages

Let $p : R \to R$ be a nonconstant polynomial. Which of the following statements are true?

A.The preimage of a compact set under p is a compact set.✓
B.The preimage of a connected set under p is a connected set.
C.Every point $x \in R$ has an open neighbourhood U_x such that the restriction $p ∣_{U_{x}}$ is a homeomorphism onto an open set in $R$ .
D.The image of a bounded set under p is a bounded set.✓

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 converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Related counterexample: Bounded ⇒ totally bounded

The chapter behind this: Compactness — the implication map — free to read

From Metric Spaces › Compactness: open covers, sequential, Heine–Borel

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