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Let $X$ be a compact Hausdorff space and let $A \subseteq X$ be closed. By Tietze extension theorem for any function $f\in C(A, \mathbb{C})$ we find a map $F\in C(X, \mathbb{C})$ such that $F|_A=f$. Is it possible to choose $F$ with $F(X)=f(A)$?

worldreporter
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  • It's certainly possible when you consider real-valued functions. Can you see how to prove that? In the complex case, you might end up in the convex hull of $f(A)$. – Ted Shifrin May 09 '18 at 20:48

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Let $X=[0,1],A=\{0,1\}, f(0)=0,f(1)=1$. If $F$ is a continuous extension then $F(X)$ is connected subset of $\mathbb C$ containing 0 and 1, so the range of $F$ is different from that of $f$. [ Note that the assertion if false even for real valued functions.

Kavi Rama Murthy
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