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A subset $U$ of a space $X$ is said to be a zero-set if there exists a continuous real-valued function $f$ on $X$ such that $U=\{x\in X: f(x)=0\}$. and said to be a Cozero-set if here exists a continuous real-valued function $g$ on $X$ such that $U=\{x\in X: g(x)\not=0\}$.

Is it true that every closed set in $\mathbb{R}$ is a Cozero-set?

I guess since $\mathbb{R}$ is a completely regular this implies that every closed set is Cozero-set, but by the same argument use completely regular property on $\mathbb{R}$, every closed subset of $\mathbb{R}$ is a zero-set. This argument is correct?

How can we discussed the relation between open & closed subset of $\mathbb{R}$ and zero and cozero-sets?


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up vote 5 down vote accepted

No non-empty proper closed subset of $\Bbb R$ is a cozero set: a cozero set is necessarily open, since it is the inverse image of an open set under a continuous map. Every closed subset of $\Bbb R$ is a zero set, however. Similarly, every open subset of $\Bbb R$ is a cozero set, and none (except $\varnothing$ and $\Bbb R$) is a zero set.

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Why every closed subset of $\mathbb{R}$ is a zero-set? – TXC Feb 7 '13 at 5:28
@TXC: Because $\Bbb R$ is perfectly normal, as is every metric space. If $\langle X,d\rangle$ is any metric space, and $H\subseteq X$ is closed, the function $f:X\to\Bbb R:x\mapsto d(x,H)$ is a continuous function that is $0$ exactly on $H$. – Brian M. Scott Feb 7 '13 at 5:33

This is more to add a small bit to Brian's answer.

In normal spaces, the zero sets correspond exactly to the closed G$_\delta$ subsets (and so the co-zero sets correspond exactly to the open F$_\sigma$ subsets).

The real line is perfectly normal which means in addition to normality that all closed sets are G$_\delta$ (or, equivalently, all open sets are F$_\sigma$; it is likely you have seen this second equivalent form before). Together with the above characterisation of zero sets in normal spaces, this means that the zero sets in $\mathbb R$ correspond exactly with the closed sets (and the co-zero sets correspond exactly with the open sets); exactly as Brian has stated.

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