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Let $X$ be an extremally disconnected compact Hausdorff space with no open points, and $f:X\to\mathbb{C}$ be a non-constant continuous function. Let $D_f$ be the linear span of the functions of the form $$u\to \gamma(u) f(\sigma(u))$$ where $\sigma:X\to X$ is a homeomorphism, and $\gamma:X\to\mathbb{C}$ is continuous with $|\gamma|=1$.

Question: What is an example of $X$ and $f$ so that $D_f$ isn't dense in $C(X)$?

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  • $\begingroup$ The condition that $|\gamma|=1$ is unnecessary since any continuous function from $X$ to $\mathbb{C}$ can be written as a linear combination of continuous functions $\gamma:X\rightarrow\mathbb{C}$ with $|\gamma|=1$. $\endgroup$
    – Joseph Van Name
    Commented Aug 15, 2023 at 21:42
  • $\begingroup$ @JosephVanName by the Stone-Weierstrass theorem? $\endgroup$
    – Onur Oktay
    Commented Aug 15, 2023 at 21:54
  • $\begingroup$ I was not thinking of the Stone-Weierstrass theorem, but you can use that too. Consider a continuous function $\iota:[0,1]\rightarrow S^1$ where $\text{Re}(\iota(x))=x$ for $x\in[0,1]$, so we can set $\iota(x)=x+i\sqrt{1-x^2}$. Then whenever $f:X\rightarrow[0,1]$ is continuous, we have $f=\frac{1}{2}(\iota\circ f+\overline{\iota}\circ f)$. And any continuous $f:X\rightarrow\mathbb{C}$ can be written as $a\cdot g+i\cdot b\cdot h+c$ where $a,b,c$ are real scalars and $g,h:X\rightarrow[0,1]$ are continuous. $\endgroup$
    – Joseph Van Name
    Commented Aug 16, 2023 at 10:20

1 Answer 1

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Take two extremally disconnected spaces $Y$ and $Z$ without open points and with different cardinality (say $Z$ is the one with larger cardinality). Define $X$ as the disjoint union of $Y$ and $Z$ and let $f$ be the indicator function of $Y$.

For every sequence $(g_n)$ in $D_f$ the union of the supports $\{g_n \not= 0\}$ is, for cardinality reasons, a proper subset of $X$. So $(g_n)$ cannot approximate the constant $1$-function on $X$.

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