If R is a ring, and A has all the sets in R and it's complements, is A an algebra? (Halmos Measure Theory question) The question I have is related to problem 4.5 in chapter 1 of Halmos' text. Some definitions related to the question are the following. If $X$ is a set then a ring $\textbf{R}$ is a non-empty class of subsets such that if $E,F\in \textbf{R}$ then $E\cup F, E-F\in \textbf{R}$, where $E-F=E\cap F^c$.
Similarly, an algebra is a non-empty class of sets such that if $E,F\in \textbf{R}$ then $E\cup F\in \textbf{R}$ and $E\in\textbf{R}$ then $E^c\in\textbf{R}$.
The problem in the text is as follows. If $\textbf{R}$ is a ring, and $A=\{E\subseteq X|E\in \textbf{R}$ or $E^c\in\textbf{R}\}$, then show that $A$ is an algebra.
My trouble is in the following line of argument in showing so. If $E\in\textbf{R}$ and $F\in A$ is such that $F^c\in\textbf{R}$, why does it imply that $E\cup F\in A$?
I'm surely missing something. Here's what I've got till now. $E\cup F^c\in R\subseteq A\Rightarrow$ either $E\cup F^c\in R$ or $E^c \cap F\in R$. In the case of the latter, $E\cup F=(E-F)\cup (F-E)\cup (E\cap F)\in R$ by definition, but I'm stuck with the former and have tried other arguments like showing that it's in the intersection of all algebras that contain the ring and $X$, but can't wrap my head around why $E\cup F$ in this case must be in $A$, and am starting to doubt whether $A$ would even be an algebra.
Any pointers to this effect would help, thanks!

Note that $E$ and $F^c$ are in $\mathbf{R}$, hence so is $F^c-E=F^c\cap E^c = (F\cup E)^c = (E\cup F)^c$. But if $(E\cup F)^{c}\in\mathbf{R}$, then $E\cup F\in A$.
