# In a complete metric space $(X,\rho)$, show that if $E$ and $X\setminus E$ are dense, then at most one of them is a countable union of closed sets.

The problem statement is in the title.

I approached this proof using contradiction.

My attempt was:

Suppose that both $E$ and $X\setminus E$ are dense and that both are a countable union of closed sets. Then $E=\bigcup_{i=1}^\infty C_i$ where $C_i\subset E$ for every $i\in\mathbb{N}$. Thus, because $C_i$ is closed and $E$ dense, $C_i$ is nowhere dense. Now, $X\setminus E=\bigcap_{i=1}^\infty C_i^c=\bigcap_{i=1}^\infty O_i$ by DeMorgan's Laws and where $O_i=C_i^c$ is open for each $i\in\mathbb{N}$. Moreover, since each $C_i$ is nowhere dense, each $O_i$ is an open dense set. Since $E$ is dense, we have that $X\setminus E$ is hollow. Furthermore, because $O_i\subset X\setminus E$, this implies that $O_i$ is hollow, that is, $\text{int}(O_i)=\varnothing.$ Yet $O_i$ is open, that is, $\text{int}(O_i)\neq\varnothing$ and hence a contradiction. Thus, for $E$ and $X\setminus E$ dense, at most one of them is a countable union of closed sets.

I also wondering if I can just say that since both $E$ and $X\setminus E$ are dense, each of them is hollow and so $X=(X\setminus E)\cup E$ is also hollow. Yet I never use the fact that the sets $E$ and $X\setminus E$ are countable union of closed sets, which doesn't really sit right with me.

Thanks for any help or feedback!

• Did you try Baire? – Pedro Tamaroff Feb 23 '14 at 23:34
• I was thinking about using it, but I didn't see how it came into play, other than the fact that the if $E=\bigcup_{i=1}^\infty C_i$, where each $C_i$ is hollow and so the union is also hollow and that because $C_i$ is closed, it is nowhere dense. Also I got that $X\setminus E=\bigcap_{i=1}^\infty O_i$ is dense, since each $O_i$ is a dense open set. Where else can I use the Baire Category Theorem? – Shant Danielian Feb 23 '14 at 23:40
• $X$ is complete, so by the Baire category theorem the intersection of countably many dense open sets is dense. You’ve already shown that $X\setminus E$ is the intersection of countably many open sets. $E$ and $X\setminus E$ are interchangeable in the hypothesis, so the same argument shows that $E$ is the intersection of countably many open sets. But then $E\cap(X\setminus E)$ is the intersection of countably many open sets and therefore dense ... – Brian M. Scott Oct 9 '14 at 3:43
• By the way, there’s an error in your argument: $C_i\subseteq E$, so $O_i\supseteq X\setminus E$, not $O_i\subseteq X\setminus E$. – Brian M. Scott Oct 9 '14 at 3:44