Exercise · ZW-CONTINUOUS-IMAGE-OF-A-COMPACT-SPACE-008 · ★★☆ Intermediate
Let \(f:X\to Y\) be continuous and surjective with \(X\) compact. What follows?
Choose one option.
Show answer
- D. \(Y\) is compact
Explanation
Surjective means \(f(X)=Y\), and \(f(X)\) is compact by the theorem. Nothing else follows: \(Y\) may be infinite and \(f\) not injective.
Review the concept: Continuous image of a compact space
If \(f:X\to Y\) is continuous and the space \(X\) is compact, then the image \(f(X)\) is a compact subspace of \(Y\).
\[f(X)\ \text{compact}\]
More practice problems: continuous image of a compact space
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