Show that $\sigma(\xi)$ is a sigma field. I have an old exam question that reads:

Let $\xi:\Omega\rightarrow \mathbb{R}$ be an arbitrary mapping. Denote
by $\sigma(\xi)$ the following collection of subsets of
$\Omega:\{\xi^{-1}(B):B\in\mathcal{B}\},$ where $B$ runs through all
Borel subsets of $\mathcal{B}$.

*

*Show that $\sigma(\xi)$ is a $\sigma-$algebra.

*Find $\sigma(\xi)$ for a mapping $\xi$ taking just two distinct values.

*When does $\sigma(\xi)$ for such $\xi$ coincide with the set $2^\Omega$ for all subsets of $\Omega$?


For the first one, I need to show that the conditions in the definition of a $\sigma-$algebra are fulfilled. The definition in my book (Grimmett and Stirzaker, Probability and random processes, 3d edition, page 3) they give the following definition:

A collection of $\mathcal{F}$ of subsets of $\Omega$ is called a
$\sigma-$algebra if it satisfies the following conditions:
i) $\varnothing\in\mathcal{F}$
ii) if $A_1,A_2...\in\mathcal{F}$ then $\bigcup_{i=1}^\infty A_i\in\mathcal{F}.$
iii) if $A\in\mathcal{F}$ then $A^c\in\mathcal{F}$.

I find these abstract "show that" questions a bit tricky as I've not seen many solutions for these, and those I've seen are are extremely short with not much explanation between the lines. I'd appreciate if anyone could break down these questions.
Attempt to solve 1:
i)
Since $\xi$ is a mapping from $\Omega$ to $\mathbb{R}$ we know that $\Omega=\xi^{-1}(\mathbb{R})\in\sigma(\xi).$ Because of iii) we also know that $\Omega^c=\varnothing\in\sigma(\xi).$
ii)
Let $I$ be a countable set, then we have a family of elements of $\sigma(\xi)$ denoted by $\{\xi^{-1}(B_i)\}_{i\in I}$ and all their unions should be in $\sigma(\xi):$
$$\bigcup\limits_{i\in I}\xi^{-1}(B_i)=\xi^{-1}\left(\bigcup\limits_{i\in I}B_i\right),$$
since $\bigcup_{i\in I}B_i\in\mathcal{B}.$
iii)
For some $B\in\mathcal{B}$ we know that $\xi^{-1}(B)\in\sigma(\xi)$. We have that $(\xi^{-1}(B))^c=\xi^{-1}(B^c)\in\sigma(\xi)$ since $B\in\mathcal{B}$.
 A: Let's break down question 1.
Claim. $\varnothing \in \sigma(\xi)$
We have $\sigma(\xi) = \{\xi^{-1}(B): B \in \mathcal{B}\}$, where I assume $\mathcal{B}$ is some $\sigma$-algebra of subsets of $\Omega$.
As Burb commented, observe $\varnothing \in \mathcal{B}$, so $\xi^{-1}(\varnothing) = \varnothing \in \sigma(\xi)$.
Claim. $\sigma(\xi)$ is closed under countable unions.
Suppose we have $A_1, A_2, \dots \in \sigma(\xi)$. Our goal is to show that $\bigcup_{i=1}^{\infty}A_i \in \sigma(\xi)$.
To show that $\bigcup_{i=1}^{\infty}A_i \in \sigma(\xi)$, we must show that $\bigcup_{i=1}^{\infty}A_i = \xi^{-1}(B)$ for some $B \in \mathcal{B}$.
Going back to our assumptions, let's assess
$\bigcup_{i=1}^{\infty}A_i$. Since $A_1, A_2, \dots \in \sigma(\xi)$, there exist sets $B_1, B_2, \dots \in \mathcal{B}$ such that $\xi^{-1}(B_i) = A_i$ for each $i \geq 1$.
Thus
$$\bigcup_{i=1}^{\infty}\xi^{-1}(B_i) = \bigcup_{i=1}^{\infty}A_i$$
but there is a trick here: one must make use of the fact that
$$\bigcup_{i=1}^{\infty}\xi^{-1}(B_i) = \xi^{-1}\left(\bigcup_{i=1}^{\infty}B_i \right)$$
and since $B_1, B_2, \dots \in \mathcal{B}$, we know that $\bigcup_{i=1}^{\infty}B_i \in \mathcal{B}$. Thus the claim holds.
Claim. $\sigma(\xi)$ is closed under complements.
The idea is similar as above, but you must instead make use of the fact that $$A_i^c = \Omega \setminus A_i = \xi^{-1}(\mathbb{R}) \setminus \xi^{-1}(B_i) =  \xi^{-1}(\mathbb{R}\setminus B_i) = \xi^{-1}(B_i^c)\text{.}$$
I will leave this part to you.
