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Part BCSIR NET December 2024on-a-finite-set-every-function-is-continuous

On a finite set every function is continuous

Let A, B be two non-empty subsets of R\mathbb{R}. Let S = {f:A→Rf : A \to \mathbb{R} | f is continuous}, T = {f:B→Rf : B \to \mathbb{R} | f is continuous}, U = {f:A×A→Rf : A \times A \to \mathbb{R} | f is continuous}. Which of the following statements is true?

  1. A.If A is finite, then there exists a bijection between S and U.✓ Correct
  2. B.If A is finite and B = [0, 1], then there is no bijection between S and T.
  3. C.There is no bijection between S and U for any choice of A.
  4. D.If A ≠ B, then there is no bijection between T and U.

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: Every ordered field is Archimedean

More on this topic

The chapter behind this: The real line: supremum, Archimedes and density — free to read

From The Real Line › Completeness, sup/inf, Archimedean property

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