Question: I am trying to show that the $\sigma$-algebra generated by $F = \{A \subseteq \mathbb{R} : 0 \in A^{\circ} \text{or } 0 \in (A^c)^\circ\}$ contains all singleton sets $\{x\} \in \mathbb{R}$.

I know that $F$ is an algebra. I believe $F$ itself is **not** a $\sigma$-algebra, since, if we take an infinite sequence of $A\_n \in F$ such that $0 \in ((A\_n)^c)^\circ$ for each $n \in \mathbb{N}$ then we do not necessarily have that $0 \in \cup\_{n=1}^{\infty}A\_n$ (since $0 \in \cap\_{n=1}^{\infty}(A\_n^c)^\circ \supseteq (\cap\_{n=1}^{\infty}((A\_n)^c)^\circ = (\cup\_{n=1}^{\infty}A\_n)^\circ$). 

**So, I am not exactly sure how to explicitly construct and/or describe the $\sigma$-algebra generated by $F$.** (From there, I am pretty sure I should be able to show that the singleton sets are in $\sigma(F)$. I just don't know how to get to that point in the first place.)

[This link](http://math.ecnu.edu.cn/~wsun/teaching/Real_N_Complex_Analysis_2015/An%20explicit%20construction%20of%20sigma%20algebra%20generated%20by%20certain%20sets.pdf) gave an explanation about constructing a $\sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.

Answer: 
It contains $\{0\}$ because that is the intersection of the sets $A\_n = (-\frac{1}{n},\frac{1}{n})$, which all verify $0\in A\_n^\circ$, hence $A\_n \in F$. On the other hand, if $x\in\mathbb{R}\setminus \{0\}$, then we have that $\{x\}$ is closed and $\{x\}^c$ is an open set containing $0$, hence $\{x\} \in F$.
Every $S\subset \Bbb R$ belongs to $\sigma(F).$

(I)... $S\setminus (-1/n,1/n)\in F$ for each $n\in \Bbb N$ so $S\setminus \{0\}=\cup\_{n\in \Bbb N}\,(\,S\setminus (-1/n,1/n)\,)\in \sigma (F).$

(II)... $(-1/n,1/n)\in F$ for each $n\in \Bbb N$ so $\{0\}=\cap\_{n\in \Bbb N}(-1/n,1/n)\in \sigma (F).$

(III)... So either $S=S\setminus \{0\}\in \sigma(F)$ or $S=(S\setminus \{0\})\cup \{0\}\in \sigma(F).$