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I'm reading Eisworth's Elementary submodels and separable monotonically normal compacta and under the hypothesis that $X$ is monotonically normal, compact and separable, he mention that every cozero set is union of countably many zero sets. I have been thinking on it and can not see why it is true.

Just for the record: every monotonically normal space is normal and therefore completely regular.

Thank you very much!

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First, we know that, by definition, if $Z$ is a zero set, then there exist a continuous function $f:X\rightarrow [0,1]$ (originally, the function have $\mathbb{R}$ like codomain, but we can consider the absolute value of $g$, where $g$ is the function for the zero set, and then, define $\displaystyle\frac{|g|}{1+|g|}$, clearly, continuous with codomain $[0,1]$). A cozero set is the complement of a zero set, then, if we have $B$ a cozero set, then, $$B=f^{-1}[(0,1]]=\displaystyle\bigcup_{n\in\mathbb{N}}f^{-1}\left[\left[\frac{1}{n},1 \right] \right]$$and, for all $n\in\mathbb{N}$ we know that $f^{-1}\left[\left[\frac{1}{n},1 \right]\right]$ is a zero set.

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    $\begingroup$ Thank you very much Carlos! $\endgroup$ – Daniel Calderón Oct 31 '17 at 6:29
  • $\begingroup$ You're welcome. $\endgroup$ – Carlos Jiménez Oct 31 '17 at 6:33
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A zero set of $X$ is just a set of the form $f^{-1}[C]$ where $f:X \to \mathbb{R}$ is continuous and $C$ is closed in the reals.

So a cozero set is just the inverse continuous image of an open set of the reals. And in the reals open sets are $F_\sigma$ (as in any metric space). From these facts it follows right away..

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