Showing that $F = \{A \subseteq \mathbb{R} : 0 \in A^{\circ} \text{or } 0 \in (A^c)^\circ\}$ is an algebra. I am trying to show that $F = \{A \subseteq \mathbb{R} : 0 \in A^{\circ} \text{or } 0 \in (A^c)^\circ\}$ is an algebra.
Here, $A^{\circ}$ denotes the interior of A. 
I am having trouble showing, in particular, that: $A, B \in F$ with $0 \in (A^c)^\circ, 0 \in (B^c)^\circ$ implies that $A \cup B \in F$.
I can't figure out how to put things together to show that $0$ is in either $(A \cup B)^{\circ}$ or $((A \cup B)^c)^{\circ}$. 
A few (potentially) useful facts I know:


*

*$((A \cup B)^c)^{\circ} \subseteq (((A \cup B)^{\circ})^c)^{\circ}$

*$A^{\circ} \cup B^{\circ} \subseteq (A \cup B)^{\circ}$
 A: As $0 \in (A^c)^\mathrm{o}$ and $0 \in (B^c)^\mathrm{o}$, then $0 \in (A^c)^\mathrm{o} \cap (B^c)^\mathrm{o}$. But $(A^c)^\mathrm{o} \cap (B^c)^\mathrm{o} = ((A^c)\cap (B^c))^\mathrm{o}=(A \cup B)^\mathrm{o}$, since the set interior of a finite intersection is the finite intersection of interiors. Hence $0 \in (A \cup B)^\mathrm{o}$.
A: To show $A,B\in F\implies A\cup B\in F$.


*

*$0\in A^{\circ}$ and $0\in B^{\circ}$ then $0\in A^{\circ}\cup B^{\circ}\subseteq (A\cup B)^{\circ}$

*$0\in A^{\circ} $ and $0\in (B^c)^{\circ}$ then since $0\in A^{\circ}\subset  (A\cup B)^{\circ} $ since $A\subset A\cup B$. The  same is true for $0\in (A^c)^{\circ} $ and $0\in B^{\circ}$.

*$0\in (A^c)^{\circ} $ and $0\in (B^c)^{\circ}$ then $0\in (A^c)^{\circ} \cap (B^c)^{\circ}=(A^c\cap B^c)^{\circ}\subset ((A\cup B)^c)^{\circ}$
A: $A,B \in F$ then we have, wlog, 3 possibilities:
1) $0\in A^{\circ}$ and $0\in B^{\circ}$
2) $0\in (A^c)^{\circ}$ and $0\in (B^c)^{\circ}$
3) $0\in A^{\circ}$ and $0\in (B^c)^{\circ}$
For 1) and 3), we can conclude that $0\in A^{\circ}\cup B^{\circ} \implies 0\in (A\cup B)^{\circ}$ as said your fact 2. Hence $A\cup B\in F$
2) We have that there is some open ball $B_0$ centered at zero with radius $\varepsilon$ such that $B_0\subseteq A^c$ and some $B'_0$ with radius $\varepsilon'$such that $B'_0\subseteq B^c$. Then we can construct a ball $B$ centered at zero with radius $\min\{\varepsilon , \varepsilon'\}$ sucht that $B\subseteq A^c\cap B^c$. Hence $0\in (A^c\cap B^c)^{\circ}$, but $A^c \cap B^c = (A\cup B)^c$.
