Questions: Compactness in Hausdorff Spaces

Question 1

In a topological space that is NOT Hausdorff, which of the following can occur?

• A. A compact subset fails to be closed — the Hausdorff property is essential for this conclusion
• B. A compact subset fails to have a finite subcover — compactness requires the Hausdorff property
• C. A continuous bijection from a compact space is automatically a homeomorphism
• D. Finite products of compact spaces fail to be compact

Answer: (A)

In a non-Hausdorff space, compact subsets need not be closed. A simple example: the Sierpiński space {0, 1} with topology {∅, {1}, {0,1}}. The set {1} is compact (it's finite) but not closed (its complement {0} is not open). The Hausdorff property — the ability to separate any two distinct points by disjoint open sets — is exactly what the proof that 'compact implies closed' uses. Without it, the proof fails and the conclusion can fail.

Question 2

Let f: X → Y be a continuous bijection where X is compact and Y is Hausdorff. Why is f automatically a homeomorphism?

• A. Because continuous bijections are always homeomorphisms when both spaces are connected
• B. Because the Tychonoff theorem guarantees the product space is compact
• C. Because the image of any closed set under f is compact (closed subsets of compacts are compact) and therefore closed in Y (compact subsets of Hausdorff spaces are closed), so f⁻¹ is continuous
• D. Because f being bijective implies its inverse exists and is bounded

Answer: (C)

The argument is: take any closed set C ⊆ X. Since C is a closed subset of the compact space X, C is compact. Since f is continuous, f(C) is compact. Since Y is Hausdorff and f(C) is compact, f(C) is closed in Y. But f(C) = (f⁻¹)⁻¹(C), so f⁻¹ pulls closed sets back to closed sets — which means f⁻¹ is continuous. This is striking because in general topology, a continuous bijection need not have a continuous inverse; both compactness of X and Hausdorff-ness of Y are needed.

Question 3

A compact subset K of a Hausdorff space and any point x ∉ K can always be separated by disjoint open sets — there exist open sets U ∋ x and V ⊇ K with U ∩ V = ∅.

• T. True
• F. False

Answer: True

This is the precise statement that compact subsets of Hausdorff spaces are closed — and more: K and any external point are separable by open sets. The proof goes: the Hausdorff property gives, for each k ∈ K, disjoint open sets Uₖ ∋ x and Vₖ ∋ k. The collection {Vₖ} covers K. By compactness, finitely many Vₖ₁, …, Vₖₙ suffice to cover K. Then V = Vₖ₁ ∪ … ∪ Vₖₙ covers K, and U = Uₖ₁ ∩ … ∩ Uₖₙ is an open neighborhood of x disjoint from V. This is the canonical pattern: Hausdorff gives local separation; compactness makes it global.

Question 4

In any topological space, compact subsets are closed.

• T. True
• F. False

Answer: False

This is false in general — it requires the Hausdorff property. Non-Hausdorff counterexamples are easy to construct: in the particular point topology on {0, 1, 2} with open sets ∅, {1}, {0,1}, {1,2}, {0,1,2}, the set {0} is compact (every cover has a finite subcover) but is not closed. The statement 'compact implies closed' is a theorem about Hausdorff spaces, not about topological spaces in general. This is why the Hausdorff condition is so useful in analysis — it brings compact sets into alignment with closed sets, as geometric intuition demands.

Question 5

Describe the proof strategy for showing that compact subsets of Hausdorff spaces are closed. Which role does compactness play, and which role does the Hausdorff property play?

Model answer: The Hausdorff property provides the local tool: for each point k in the compact set K and the external point x, disjoint open sets separate x from k individually. This gives infinitely many open sets, one per point of K. Compactness provides the global conversion: the collection of open sets covering K has a finite subcover, so finitely many of the separating open sets suffice to cover all of K. The intersection of the finitely many open sets around x (finite intersections of open sets are open) gives a single open neighborhood of x that is disjoint from an open cover of K. So x is in the interior of the complement of K, meaning K is closed. Compactness converts infinite local conditions into a finite combinable package; Hausdorff supplies those local conditions.

This proof pattern — Hausdorff gives local separation, compactness makes it global — recurs throughout topology. It is the template for why compactness and the Hausdorff property amplify each other, and why the Hausdorff assumption appears so frequently as a hypothesis in theorems about compact spaces.