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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!

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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.

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  • $\begingroup$ 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? $\endgroup$ – Akira Oct 13 '18 at 2:27
  • $\begingroup$ Yes. If a set does not contain any interval, its complement is dense. $\endgroup$ – Asaf Karagila Oct 13 '18 at 13:19
  • $\begingroup$ Thank you so much for your response :) $\endgroup$ – Akira Oct 13 '18 at 13:21

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