In the book "Principles of Real Analysis" by Aliprantis and Burkinshaw I found the following exercise:
Let $\mathcal{S}$ be a semiring of subsets of a nonempty set $X$. What additional requirements must be satisfied for $\mathcal{S}$ in order to be a base for a topology on $X$?
Prove that if such is the case, the each member of $\mathcal{S}$ is both open and closed in this topology?
Recall the definition of semiring $\mathcal{S}$, as the collection of subsets of a given space $X$ such that:
1. $\varnothing \in \mathcal{S}$;
2. for all $A, B \in \mathcal{S}$, $A \cap B \in \mathcal{S}$;
3. for all $A, B \in \mathcal{S}$, there exists $C_1, \dots, C_n \in \mathcal{S}$ such that $A \setminus B = \bigcup_{k=1}^{n} C_k$.
My answer is that it is enough to have that, for every $x \in X$, $\{ x \} \in \mathcal{S}$.
Interestingly, at least to me, I found in the companion book "Problems in Real Analysis", that the solutions is that $\mathcal{S}$ is a base if and only if $\bigcup_{A \in \mathcal{S}} A = X$.
Now the question is, are the two answers equivalent?
I have the feeling that the one in the book is apparently less demanding than my condition, but they should really be the same. Am I right?
As always, thank you for your time.
Any feedback is most welcome.