3
$\begingroup$

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?

thanks.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Why every closed subset of $\mathbb{R}$ is a zero-set? $\endgroup$ – TXC Feb 7 '13 at 5:28
  • 2
    $\begingroup$ @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 $H$. $\endgroup$ – Brian M. Scott Feb 7 '13 at 5:33
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

I just waant to know how $\phi$ and $\mathbb{R}$ are not zero set? as if i take $f(x) = 0 \forall x$ and $g(x) = e^{x} + 1 \forall x$ both are cts. Then $\phi$ and $\mathbb{R}$ are zero set.

$\endgroup$
1
  • $\begingroup$ Do you think that this is an answer to a three and half years old question ? $\endgroup$ – Claude Leibovici Nov 2 '16 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.