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Part BCSIR NET December 2025the-tube-lemma-needs-compactness-for-the-finite-subcover-nothing-else-gives-it

The tube lemma needs compactness for the finite subcover nothing else gives it

Let X and Y be topological spaces. Consider the following statement: S: For every open subset U ⊆ X × Y and every x ∈ X such that {x} × Y ⊆ U, there is a neighbourhood W of x in X such that W × Y ⊆ U. Which of the following statements is true?

  1. A.If X × Y is Hausdorff, then the statement S is true.
  2. B.If Y is connected, then the statement S is true.
  3. C.If X × Y is regular, then the statement S is true.
  4. D.If Y is compact, then the statement S is true.✓

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: Countably compact implies compact

More on this topic

The chapter behind this: Compactness in general spaces — free to read

From Topology › Compactness and Tychonoff

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