"Introduction to Real Analysis" (Robert G. Bartle) Chapter 1, Question 13: Show that if $f:A\to B$ and $G,H$ are subsets of $B$, then $f^{-1}(G\cup H)=f^{-1}(G) \cup f(H)$ and $f^{-1}(G\cap H)= f^{-1}(G) \cap f^{-1}(H)$
EDIT: Despite @ArtudoMagdin's comments, I still cannot form a complete, correct solution. I require a full answer. I crossed out the solution verification tag.
Attempt:
- Proving $f^{-1}(G\cup H)=f^{-1}(G)\cup f^{-1}(H)$
We know $\require{enclose} \enclose{horizontalstrike}{f(x)\in G}$ and $\require{enclose} \enclose{horizontalstrike}{G\subseteq G\cup H}$ so $\require{enclose} \enclose{horizontalstrike}{f(x)\in G\cup H}$ meaning $\require{enclose} \enclose{horizontalstrike}{f^{-1}(G)\subseteq f^{-1}(G\cup H)}$ and $\require{enclose} \enclose{horizontalstrike}{f(x)\in H}$ and $\require{enclose} \enclose{horizontalstrike}{H \subseteq G\cup H}$ so $\require{enclose} \enclose{horizontalstrike}{f(x)\in G\cup H}$ meaning $\require{enclose} \enclose{horizontalstrike}{f^{-1}(H)\subseteq f^{-1}(G\cup H)}$. Since $\require{enclose} \enclose{horizontalstrike}{f(x)}$ is independetly in $\require{enclose} \enclose{horizontalstrike}{G}$ or $\require{enclose} \enclose{horizontalstrike}{H}$, $\require{enclose} \enclose{horizontalstrike}{f^{-1}(G)\cup f^{-1}(H) = f^{-1}(G\cup H)}$
Edit: Here is my new attempt:
Second Edit: I made additional changes by the request of @ArturoMagidin
- If $x\in f^{-1}(G)$ then $f(x) \in G \subseteq G\cup H$, hence $f^{-1}(G)\subseteq f^{-1}(G\cup H)$
- If $x\in f^{-1}(H)$ then $f(x)\in H\subseteq G \cup H$, hence $f^{-1}(H)\subseteq f^{-1}(G\cup H)$
- If $x\in f^{-1}(G)\cup f^{-1}(H)$, it follows 1. and 2. that $f(x)\in G \cup H$ which also means that $x\in f^{-1}(G\cup H)$.
- It follows from 3. that $f^{-1}(G)\cup f^{-1}(H)\subseteq f^{-1}(G\cup H)$
- If $x\in f^{-1}(G\cup H)$ then $f(x)\in G\cup H$, hence $f(x)\in G$ or $f(x)\in H$ which means $x\in f^{-1}(G)$ or $x\in f^{-1}(H)$. Hence, $f^{-1}(G\cup H)=f^{-1}(G)\cup f^{-1}(H)$
Is my proof correct, it seems this part is right?
- Proving $f^{-1}(G\cap H)\subseteq f(G) \cap f(H)$
We know $\require{enclose} \enclose{horizontalstrike}{f(x)\in G}$ and $\require{enclose} \enclose{horizontalstrike}{G\supseteq G\cap H}$ so if $\require{enclose} \enclose{horizontalstrike}{f(x)\in G\cap H}$ meaning $\require{enclose} \enclose{horizontalstrike}{f^{-1}(G)\supseteq f^{-1}(G\cap H)}$ and $\require{enclose} \enclose{horizontalstrike}{f(x)\in H}$ and $\require{enclose} \enclose{horizontalstrike}{H \supseteq G\cup H}$ so $\require{enclose} \enclose{horizontalstrike}{f(x)\in G\cup H}$ meaning $\require{enclose} \enclose{horizontalstrike}{f^{-1}(H)\subseteq f^{-1}(G\cup H)}$. I am not sure how to proceed from here. Is my approach correct? If so, how would this imply that $\require{enclose} \enclose{horizontalstrike}{f^{-1}(G\cap H)=f^{-1}(G)\cap f^{-1}(H)}$
Edit: Here is my new attempt for 2.
Second Edit: I made additional changes by the request of @ArturoMagidin
Third Edit: I shortened my steps per @ArturoMagidin's comments. I figure I could have done the same for the first proof.
- If $x\in f^{-1}(G)\cap f^{-1}(H)$ then $f(x)\in G$ and $f(x)\in H$ therefore $f(x)\in G\cup H$ so $f^{-1}(G\cap H)\subseteq f^{-1}(G)\cap f^{-1}(H)$
- If $x\in f^{-1}(G\cap H)$ then $f(x)\in G\cap H$, hence $f(x)\in G$ and $f(x)\in H$ which means $x\in f^{-1}(G)$ and $x\in f^{-1}(H)$. Hence, $f^{-1}(G\cap H) \supseteq f^{-1}(G)\cap f^{-1}(H)$
- From 1. and 2. we state that $f^{-1}(G\cap H)\subseteq f^{-1}(G)\cap f^{-1}(H)$
I am not satisfied with step 5.? How do we show $\require{enclose} \enclose{horizontalstrike}{f^{-1}(G\cap H)\supseteq f^{-1}(G)\cap f^{-1}(H)}$?
Is this correct? EDIT: According to @ArtudoMagdin, it is still wrong. What is the full solution for this proof?
Your proof is not complete, and it is also rather confusing. You are trying to prove $f^{-1}(G\cup H)=f^{-1}(G)\cup f^{-1}(H)$, and you start by saying “We know that $f(x)\in G$.”
Wait a second! What is $x$? Where did it come from? What are you trying to prove here? I don’t know who $x$ is, so I certainly do not know what happens to $x$ under $f$.
Same thing with the second proof.
You really need to be more intentional. It appears you are trying to prove the equality by double inclusion. Fine. So then start with an element of one side. So, “We want to show that $f^{-1}(G)\cup f^{-1}(H)\subseteq f^{-1}(G\cup H)$ first. To that end, let $x\in f^{-1}(G)$. Then $f(x)\in G\subseteq G\cup H$, so $x\in f^{-1}(G\cup H)$. Therefore, we have $f^{-1}(G)\subseteq f^{-1}(G\cup H)$.” Then you do the same thing with $x\in f^{-1}(H)$. Then you want to argue this shows $f^{-1}(G)\cup f^{-1}(H)\subseteq f^{-1}(G\cup H)$.
But at this point you have not proven equality. You have only proven one inclusion. So you are not done. Now you must show that if $x\in f^{-1}(G\cup H)$, then it must be the case that $x\in f^{-1}(G)\cup f^{-1}(H)$. And you have not done so.
Similar problems occur with your argument so far for the second part. Not only do you apparently start in the middle of the argument, you are comparing the wrong things.
For the second part, let us first show that $f^{-1}(G\cap H)\subseteq f^{-1}(G)\cap f^{-1}(H)$. Let $x\in f^{-1}(G\cap H)$. Then $f(x)\in G\cap H\subseteq G$, so $x\in f^{-1}(G)$.
Now you should be able to show that we also have $x\in f^{-1}(H)$. Then you want to argue this shows $x\in f^{-1}(G)\cap f^{-1}(H)$. This will show one inclusion. Not equality (yet).
So then you want to take $x\in f^{-1}(G)\cap f^{-1}(H)$, and show it lies in $f^{-1}(G\cap H)$. Well, $f(x)\in G$ because $x\in f^{-1}(G)$; and $f(x)\in H$ because....
... and finish it off.
Added (May 13, 15:35 CDT)
A proof of the second part proceeds along the same lines.
To prove that $f^{-1}(G\cap H)\subseteq f^{-1}(G)\cap f^{-1}(H)$, let $x\in f^{-1}(G\cap H)$. Then $f(x)\in G\cap H$, hence $f(x)\in G$ and $f(x)\in H$. This means that $x\in f^{-1}(G)$, and $x\in f^{-1}(H)$, and therefore that $x\in f^{-1}(G)\cap f^{-1}(H)$, proving the inclusion.
To prove that $f^{-1}(G)\cap f^{-1}(H)\subseteq f^{-1}(G\cap H)$, let $x\in f^{-1}(G)\cap f^{-1}(H)$. Then $x\in f^{-1}(G)$, so $f(x)\in G$; and $x\in f^{-1}(H)$, so $f(x)\in H$. Therefore, $f(x)\in G\cap H$, and thus by definition we have $x\in f^{-1}(G\cap H)$. This proves the second inclusion, and hence the equality.