Exercise · ZW-CONTINUOUS-IMAGE-OF-A-COMPACT-SPACE-009 · ★★★ Advanced
Let \(X\) be compact and \(f:X\to\mathbb{R}\) continuous with \(f(x)>0\) for all \(x\). Is there \(c>0\) with \(f(x)\ge c\) on all of \(X\)?
Choose one option.
A Yes: \(f\) attains its minimum and it is positive B No: \(f\) may approach \(0\) without reaching it C Only if \(X\) is finite D Only if \(f\) is injective Show answer A. Yes: \(f\) attains its minimum and it is positive← Back to the question
Explanation \(f(X)\) is compact in \(\mathbb{R}\), so it has a minimum \(c=f(x_0)\) with \(x_0\in X\), and \(c>0\) by hypothesis. Without compactness it fails: \(x\) on \((0,1]\) approaches \(0\).
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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