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Consider the ring $R$ of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. The operations are defined as follows:

1.$(f+g)(x)=f(x)+g(x)$

2.$(fg)(x)=f(x)*g(x)$

for all $f,g \in R$, for all $x \in \mathbb{R}$.

We define a zero divisor in $R$ to be a non zero element $f$ such that $fg=0$ for some non zero $g \in R$.

The statement I'm trying to prove is this: If a non-constant function $f$ in this ring is such that $f^{-1}(\{0\})$ contains a non-empty open set $\mathcal{O}$, then this function is a zero divisor. In other words, I need to find a function $h \in R$ such that $f(x)h(x)=0$ for all $x\in \mathbb{R}$.

Here's how I tried it: We know that if a continuous real valued function $h:\mathbb{R} \rightarrow \mathbb{R}$ is non zero at some point, say, $k \in \mathbb{R}$ then there is a open set $\mathcal{O_{k}}$ containing $k$ such that $h$ is non zero on the open set as well. But, given an open set $\mathcal{O}$ can I find a function $h$ such that it does not vanish there?

Am I on the right track? Or is there another approach?

LaxHat1376
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    Hint: You only have find a $g$ for each open interval. – Thomas Andrews Apr 07 '23 at 19:12
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    Specificially, if $(a,b)\subseteq \mathcal O,$ then define: $$g(x)=\begin{cases}\frac{b-a}2-\left|\frac{a+b}2-x\right|&x\in(a,b)\0&\text{otherwise}\end{cases}$$ – Thomas Andrews Apr 07 '23 at 19:17
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    @AnneBauval The statement I mentioned was not the logical converse, I have corrected my mistake. – LaxHat1376 Apr 07 '23 at 19:20
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    @ThomasAndrews This function is precisely what I needed. Thank you! It is also continuous at the endpoints as well. It is enough for me if the function is non-zero at some subset of the open set, not the whole open set itself. Now I have a non zero function in R such that f(x)g(x)=0 for all x in \mathbb{R}. – LaxHat1376 Apr 07 '23 at 19:20
  • You might also look at Urysohn's lemma: if you choose a point $x \in \mathcal{O}$, then since ${ x }$ and $\mathbb{R} \setminus \mathcal{O}$ are disjoint closed subsets of $\mathbb{R}$, there exists a function $g : \mathbb{R} \to \mathbb{R}$ which is equal to 1 at $x$ and equal to 0 on $\mathbb{R} \setminus \mathcal{O}$. (Of course, as others have noted, this case of Urysohn's lemma is much easier to prove than the general case.) – Daniel Schepler Apr 07 '23 at 19:25
  • Does this answer your question? What are the zero divisors of $C[0,1]$? I know you consider $C(\Bbb R),$ not $C([0,1]).$ But the solution is the same, isn't it? – Anne Bauval Apr 07 '23 at 19:26
  • This answer of the proposed duplicate proves not only your "if...then..." but actually, "iff". See also (again about $C([0,1])$ but again: never mind) that other "duplicate". – Anne Bauval Apr 07 '23 at 19:35
  • @DanielSchepler The open set I am referring to here is a subset of the zeroes of this function. So the function would take 0 on {x} and 1 on $\mathbb{R}$ \O instead. – LaxHat1376 Apr 07 '23 at 19:35
  • @AnneBauval Thank you for the link. Yes, the question mentioned does prove both directions. – LaxHat1376 Apr 07 '23 at 19:36
  • @LaxHat1376 It would have to be continuous at all points or it wouldn't work. But it is definitely worth checking. – Thomas Andrews Apr 07 '23 at 19:43

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