How to prove that a given collection is not sigma algebra?

Question

Let $\mathbf{F}$ be the collection of subsets of $\mathbb{R}$ and let $\tau$ be usual topology
Suppose that, $A\subseteq\mathbb{R}$ be open or closed with respect to $\tau$

How to prove that, $\mathbf{F}=\left\{ A\subseteq \mathbb{R}:A\,\,is\,\,open\,\,or\,\,closed\,\,\,w.r.t\,\,\tau \right\}$ is not a $\sigma$- algebra over $\mathbb{R}$

I start to think first to prove the singleton $x$ is not in $\mathbb{F}$ but this will leads me to show that also $\mathbb{Q}\not\in\mathbb{F}$

Answer 1

Singletons (being closed sets) lie in $\textbf{F}$, but $\Bbb{Q}$ (which is a countable union of singletons) does not lie in $\textbf{F}$, because the rationals are neither an open set nor a closed set.

Hence, the set $\textbf{F}$ is not closed under taking countable unions, hence not a $\sigma$-algebra.

Answer 2

We have $[0,1],(-1,1)\in\mathbf F$, but $[0,1]\cap(-1,1)=[0,1)$ is neither open nor closed.