Suppose that $(A,<)$ is a toset and that $A_1$ is dense in $A$. What are the conditions for which $A_2=A\setminus A_1$ is dense in $A$?

Suppose that $$(A,<)$$ is a totally ordered set and that $$A_1\cap A_2=\emptyset$$ and $$A_1 \cup A_2=A$$, and that $$A_1$$ is dense in $$A$$. What are the necessary and sufficient conditions for which $$A_2$$ is dense in $$A$$?

This question arose when i did an exercise that $$\Bbb R\setminus\Bbb Q$$ is dense in $$\Bbb R$$.

I would like to ask if this special property holds for a more general setting.

Thank you so much!

Since $$A_1$$ is dense if and only if $$A_1$$ meets every open interval, clearly the same should be true for $$A_2$$. It is not hard to verify that this translates to "$$A_1$$ does not contain any open interval".
Indeed, if $$A_1$$ does not contain any interval, then $$A\setminus A_1$$ meets every open interval as well.
• Hi @Asaf! I rephrase your answer in my own words: Suppose that $(A,<)$ is a totally ordered set and that $A_1\cap A_2=\emptyset$ and $A_1 \cup A_2=A$. We have: $A_1$ does not contain any open interval $\iff$ $A_1$ does not contain any interval $\implies$ $A_2$ is dense in $A$. Is my understanding correct? – Akira Oct 13 '18 at 2:27