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Lesson 18 of 28

Compact images, closedness and inverse maps

Prove continuous images and closed subspaces of compact spaces are compact.

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The interval [0,1] maps continuously onto the unit circle by t↦exp(2πit). Compactness passes to that image even though the map identifies endpoints. This is an example of a property preserved by continuous images, not only by homeomorphisms.

A stronger result repairs a gap left in the quotient lesson: a continuous bijection from a compact space to a Hausdorff space has a continuous inverse. Both compactness and the target’s separation property matter.

Predict

Predict whether a compact subset is automatically closed. In a two-point indiscrete space, a singleton is compact because it is finite, but it is not closed. Hausdorffness is the missing hypothesis in the usual theorem.

Predict how compactness might help separate an exterior point from an entire compact set. Separate it from each set point first, then use a finite subcover to keep the intersection of the exterior neighborhoods open.

Experiment

Track the proof dependencies.

ClaimRequired hypotheses
Continuous image is compactCompact source, continuous map
Closed subset is compactClosed in a compact space
Compact subset is closedHausdorff ambient space
Continuous bijection is homeomorphismCompact source, Hausdorff target

The first two statements do not require Hausdorffness. Adding it everywhere would obscure which assumptions actually drive each proof.

Understand

For the continuous-image theorem, pull an open cover of f(X) back to an open cover of X. Compactness selects finitely many preimages covering X; their corresponding target sets cover f(X). Every image point has a preimage, which is the only surjectivity needed.

For a closed subset F of compact X, write its relative cover as intersections with ambient opens and add X without F. Compactness gives a finite cover of X. Removing the added complement leaves finitely many original members covering F.

Now let K be compact in Hausdorff X and choose x outside K. For each y∈K, choose disjoint opens U_y containing x and V_y containing y. Finitely many V_y cover K. The intersection of the corresponding U_y is an open neighborhood of x disjoint from K. Thus the complement of K is open, proving K closed. Finiteness is crucial because an arbitrary intersection of open U_y need not be open.

Finally, let f:X→Y be a continuous bijection, with X compact and Y Hausdorff. A closed subset of X is compact; its image is compact and therefore closed in Y. Hence f is a closed map. The closed-bijection criterion makes f⁻¹ continuous, proving it is a homeomorphism.

Compact Hausdorff spaces are also normal under our T₁ convention. The preceding point-versus-compact-set argument first gives regularity. For disjoint closed A and B, choose for each point of A an open neighborhood whose closure misses B. Compactness of A selects finitely many. Their union U contains A and its closure still misses B; U and the complement of its closure are the required disjoint open neighborhoods. This supplies the separation fact used later in compactification arguments.

These arguments also explain why quotient spaces can be compact even when separation fails: compactness follows from the continuous quotient map, but Hausdorffness must be checked separately.

Apply

Return to the endpoint quotient of [0,1]. Its quotient map is continuous, so the quotient is compact. The induced bijection to the unit circle is continuous, and the circle is Hausdorff as a metric subspace. The theorem now proves the intended homeomorphism, completing the earlier construction without assuming inverse continuity.

Repair “A continuous injection from a compact space into any target is an embedding.” The usual conclusion requires a Hausdorff target, or another condition ensuring the inverse on the image is continuous.

Continue

Local compactness will allow compact neighborhoods without requiring the whole space to be compact. Optional depth: prove a compact Hausdorff space is regular by separating an exterior point from a closed compact subset using the finite-neighborhood argument above, then extend the same idea to separation of two disjoint closed subsets.

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Practice and reasoning

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A map on a four-point compact space has fibers of sizes 1 and 3. How many points are in its image?

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Every compact subset of an arbitrary topological space is closed:

Your response

Prove the continuous image of a compact space is compact.

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Prove a closed subset F of a compact space X is compact.

Structure your argument and preserve revisions

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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

Prove a compact subset K of a Hausdorff space is closed.

Structure your argument and preserve revisions

These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.

Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

Prove a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Structure your argument and preserve revisions

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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

Reflect and continue

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