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.