Exercise · ZW-CONTINUOUS-IMAGE-OF-A-COMPACT-SPACE-005 · ★☆☆ Basic
Let \(\mathbb{N}\) have the discrete topology and \(f(n)=1/n\), which is continuous. Is \(f(\mathbb{N})=\{1/n:n\ge1\}\) compact in \(\mathbb{R}\)?
Choose one option.
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- B. No: discrete \(\mathbb{N}\) is not compact and the image is not closed
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Explanation
The theorem needs a compact domain and discrete \(\mathbb{N}\) is not. The image omits its limit point \(0\), so it is neither closed nor 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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