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A Darboux function is a function that has the intermediate value property. That is a function $f$ such that

$$ \forall a,b \in \mathbb{R} : f[a,b] \supseteq [f(a),f(b)] \cup[f(b),f(a)] $$

We define the sum of two functions as such

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

Now the question is:

If $f$ is a Darboux function and $g$ is a continuous function, must $f+g$ be a Darboux function as well?

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  • $\begingroup$ @WheatWizard My answer was incorrect, you are right. But the definition you are giving is incorrect too, isn't it? A Darboux function is such that $\forall y\in[f(a),f(b)]$ the exists $c\in[a,b]$ such that $f(c)=y$. But if I did not understand wrongly, you are saying that $f$ is Darboux if $\forall y\in[a,b]$, $f(y)\in[f(a),f(b)] \cup [f(b),f(a)]$, which is not the same. $\endgroup$ – AugSB Feb 20 '18 at 20:23
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    $\begingroup$ @AugSB Yes my definition was backwards. I've fixed it now. $\endgroup$ – Sriotchilism O'Zaic Feb 20 '18 at 20:30
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This is a very good question that leads into some fairly hard mathematics: the answer depends on what set theoretic assumptions you are prepared to make. If you assume the continuum hypothesis, there are what are called universally bad Darboux functions $f$ such that $f+g$ is not Darboux for any non-constant continuous $g$. See https://www.encyclopediaofmath.org/index.php/Darboux_property and the references it cites.

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