Let X be a topological space and f,g continuous $X \rightarrow \mathbb{R}$
Is the set defined as A:={$x \in X : f(x) \neq g(x)$} open?
I tried to proof it but I couldn't. Would be nice if someone had a hint for me.
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The inverse image of the diagonal $\Delta = {(a,a) \in \mathbb{R} \times \mathbb{R} : a \in \mathbb{R}}$ by the continuous map $f\times g : X \rightarrow \mathbb{R} \times \mathbb{R} $ is $X-A$. Thus $A$ is open. – L Y Sep 24 '21 at 13:15
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1Hint: define $h(x)=f(x)-g(x)$. What is your set in terms of $h$? – lulu Sep 24 '21 at 13:16
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@lulu A={$x \in X: h(x) \neq 0$}, so i can write $A=h^{-1}(0)$. Now ${0} \subseteq \mathbb{R}$ is open, because Int({0})={0}. Is this correct? – John.W Sep 24 '21 at 13:32
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1No. ${0}$ is not open. And, $A\neq h^{-1}(0)$. – lulu Sep 24 '21 at 13:34
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@ lulu I think I got something wrong, it should be A= $h^{-1}$(X{0}) – John.W Sep 24 '21 at 13:43
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Now X{0} is open,thus A is open. Is this correct? – John.W Sep 24 '21 at 13:44
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Yes, that is correct. – lulu Sep 25 '21 at 10:14