I have proved the following statement and I would like to know if I have made any mistakes, thank you.
"Prove that the collection of Borel subsets of $\mathbb{R}$, $\mathcal{B}$, is dilation invariant. More precisely, prove that if $B\subset\mathbb{R}$ is a Borel set and $t\in\mathbb{R}$, then $tB:=\{tb:b\in B\}$ is a Borel set."
My proof:
(EDIT: I now believe only the parts in bold are necessary for the proof)
If $t=0$ the claim is trivial ($0B=\{0\}$ which is a Borel set, since all countable sets are Borel). so in the following we assume $t\in\mathbb{R}\setminus \{0\}$.
Let $t\in\mathbb{R}$: we now prove that $t\mathcal{B}$ is a $\sigma$-algebra
$\emptyset\in\mathcal{B}$ and $\emptyset\overset{*}{=}t\emptyset\in t\mathcal{B}$
*suppose for sake of contradiction that $t\emptyset\neq\emptyset$: this means there exists an element $a\in t\emptyset$ i.e. $a=to$, where $o\in\emptyset$, contradiction, since $\emptyset$ does not contain elements by definition.
$E\in t\mathcal{B}\Rightarrow E^c=tB^c, B^c\in\mathcal{B}$ so $E^c\in t\mathcal{B}$;
$E_1, E_2,\dots\in t\mathcal{B}\Rightarrow E_1=tB_1, E_2=tB_2, \dots\Rightarrow\bigcup_{k=1}^{\infty}E_k=\bigcup_{k=1}^{\infty}tB_k=t\bigcup_{k=1}^{\infty}B_k\in t\mathcal{B}$, since $\bigcup_{k=1}^{\infty}B_k\in\mathcal{B}$.
Now, let $O$ be an open subset of $\mathbb{R}$: then it can be written as the countable union of open intervals in $\mathbb{R}, O=\bigcup_{k=1}^{\infty}(a_k,b_k)\in\mathcal{B}$ and if we set $t\in\mathbb{R}\setminus \{0\}$ we can also write $O=t\bigcup_{k=1}^{\infty}(\frac{1}{t}a_k,\frac{1}{t} b_k)\in t\mathcal{B}$. So, $t\mathcal{B}$ contains every open subset of $\mathbb{R}$ and since by definition $\mathcal{B}$ is the smallest $\sigma$-algebra containing all the open subsets of $\mathbb{R}$ it follows that $\mathcal{B}\subset t\mathcal{B}.$
Now, let $t\in\mathbb{R}\setminus\{0\}$ and consider the function $f:\mathbb{R}\to\mathbb{R}, f(x):=\frac{1}{t}x$: then $f$, being a continuous function is also a Borel-measurable function so $f^{-1}(B)=tB$ is a Borel set for every Borel set $B$ so we also have that $t\mathcal{B}\subset \mathcal{B}$ thus $t\mathcal{B}=\mathcal{B}$, as desired.