Exercise · ZW-CONTINUOUS-IMAGE-OF-A-COMPACT-SPACE-004 · ★☆☆ Basic
The continuous function \(f(x)=1/x\) on \((0,1]\) has image \([1,\infty)\), which is not compact. Why is this no contradiction?
Choose one option.
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- D. Because \((0,1]\) is not compact
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Explanation
The theorem requires a compact domain, and \((0,1]\) is not (it is not closed). Without that hypothesis the image need not be compact.
If \(f:X\to Y\) is continuous and the space \(X\) is compact, then the image \(f(X)\) is a compact subspace of \(Y\).
\[f(X)\ \text{compact}\]
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