Sigma Algebra: Seeking Help on Closure & Countable Unions

[Arenholt]

Hi everyone, didn't know where to post question on sigma algebra so here it is:-
View attachment 5671

What I've tried till now:
Let \(\displaystyle C\in G\)
1) For \(\displaystyle C=X, f^{-1}(B)=X\) which will be true for \(\displaystyle B=Y\) (by definition)
2) For closure under complementation, to show \(\displaystyle C^{c}\in G. So, C^{c}=X\setminus C=X\setminus f^{-1}(B)=f^{-1}(Y)\setminus f^{-1}(B).\)

Can someone suggest how to proceed from here in order to show that the set G is closed under complementation? And how to prove the third property of Countable unions of subsets of X lying in G?
Thanks

 

[caffeinemachine]

Hi everyone, didn't know where to post question on sigma algebra so here it is:-What I've tried till now:
Let \(\displaystyle C\in G\)
1) For \(\displaystyle C=X, f^{-1}(B)=X\) which will be true for \(\displaystyle B=Y\) (by definition)
2) For closure under complementation, to show \(\displaystyle C^{c}\in G. So, C^{c}=X\setminus C=X\setminus f^{-1}(B)=f^{-1}(Y)\setminus f^{-1}(B).\)

Can someone suggest how to proceed from here in order to show that the set G is closed under complementation? And how to prove the third property of Countable unions of subsets of X lying in G?
Thanks

Note that $f^{-1}(Y)\setminus f^{-1}(B)=f^{-1}(Y\setminus B)$. Now since $B$ is in the sigma algebra, so is $Y\setminus B$.

 

[Arenholt]

Note that $f^{-1}(Y)\setminus f^{-1}(B)=f^{-1}(Y\setminus B)$. Now since $B$ is in the sigma algebra, so is $Y\setminus B$.

This is exactly what I am trying to wrap my head around. I thought of this but how is $f^{-1}(Y)\setminus f^{-1}(B)$ equal to $f^{-1}(Y\setminus B)$.

 

[caffeinemachine]

This is exactly what I am trying to wrap my head around. I thought of this but how is $f^{-1}(Y)\setminus f^{-1}(B)$ equal to $f^{-1}(Y\setminus B)$.

To show that a set $A$ is equal to a set $B$, we show that $A\subseteq B$ and $B\subseteq A$.

Now we want to show $f^{-1}(Y)-f^{-1}(B)=f^{-1}(Y-B)$.

Step 1: We show $f^{-1}(Y)-f^{-1}(B)\subseteq f^{-1}(Y-B)$. To do this, pick $x\in f^{-1}(Y)-f^{-1}(B)$ arbitrarily. All we need to do is show that $x\in f^{-1}(Y-B)$. Now saying that $x\in f^{-1}(Y)-f^{-1}(B)$ is same as saying that $x\notin f^{-1}(B)$ since $X=f^{-1}(Y)$. This in turn is equivalent to saying that $f(x)\notin B$. Thus $f(x)\in Y-B$. This means the same thing as $x\in f^{-1}(Y-B)$ and we are done with step 1.

Can you do step 2, that is, show the reverse containment?

It is useful to draw some pictures to get an intuition about these things. Make a blob denoting $Y$ and a subblob denoting B etc.