Papa Rudin theorem 1.12 $(a)$ There is the definition which is usable in the proof of the  theorem:

There is the theorem:
Suppose $\mathfrak M$ is a $\sigma$-algebra in $X$, and $Y$ is a topological space.
Let $f$ map $X$ into $Y$.
$(a)$ If $\Omega$ is the collection of all sets $E$ $\subset$ $Y$ such that $f^{-1}(E)$ $\in$ $\mathfrak M$, then $\Omega$ is a $\sigma$-algebra in $Y$.
There is the proof: ( it's from papa rudin book).
$(a)$ follows from the relations
$f^{-1}(Y) = X$,
$f^{-1}(Y-A) = X - f^{-1}(A)$,
$f^{-1}$($A_1$ $\cup$ $A_2$ $\cup$ ...) $=$ $f^{-1}(A_1)$ $\cup$ $f^{-1}(A_2)$ $\cup$ ...
I  don't understand How does $(a)$ follows from these relations.
Any help would be appreciated.
 A: In order to prove that $\Omega$ is a $\sigma$-algebra, you have to prove three things:

*

*$Y\in\Omega$.

*If $A\in\Omega$, $A^\complement\in\Omega$:

*If $A=\bigcup_{n=1}^\infty A_n$ with each $A_n\in\Omega$, then $A\in\Omega$.

And we have:

*

*$Y\in\Omega$ because $f^{-1}(Y)=X$ and $X\in\mathfrak M$.

*If $A\in\Omega$, then $f^{-1}(A)\in\mathfrak M$. So, $\left(f^{-1}(A)\right)^\complement\in\mathfrak M$. But $\left(f^{-1}(A)\right)^\complement=f^{-1}\left(A^\complement\right)$, and therefore $A^\complement\in\Omega$.

*If $(A_n)_{n\in\Bbb N}$ is a sequence of elements of $\Omega$, then, for each $n\in\Bbb N$, $f^{-1}(A_n)\in\mathfrak M$. So, $\bigcup_{n=1}^\infty f^{-1}(A_n)\in\mathfrak M$. But then$$f^{-1}\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty f^{-1}(A_n)\in\mathfrak M,$$and therefore $\bigcup_{n=1}^\infty A_n\in\Omega$.

A: You want to verify that, for sets $A,A_1,A_2,...$ in $\Omega$, we have: $$Y\setminus A,\bigcup A_n\in\Omega,\bigcap A_n\in\Omega$$By definition of $\Omega$, you just need to check if $f^{-1}$ applied to these sets gives an element of $\mathfrak{M}$.
For elements $E_n=f^{-1}(A_n)\in\mathfrak{M}$, their (countable) union, intersection and complement are contained in $\mathfrak{M}$ by definition of a $\sigma$-algebra. Since the preimage of a union is a union of preimages (e.g.), you get what you need.
$$f^{-1}\left(\bigcup A_n\right)=\bigcup f^{-1}(A_n)=\bigcup E_n\in\mathfrak{M}$$
