MHB Are These Statements About Preimages Correct and Complete?

[mathmari]

Hey!

Let $f:M\rightarrow N$, $A,B\subset N$. I want to check if the following statements about the preimages are correct:

1. $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$
2. $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$
3. $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)$
4. $A\cap B=\emptyset \Rightarrow f^{-1}(A)\cap f^{-1}(B)=\emptyset$
5. $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$

I have done the following:

1. Let $x\in f^{-1}(A\cup B)$. Then it follows that $f(x)\in A\cup B$. This means that either $f(x)\in A$ or $f(x)\in B$, so $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$. It follows that $x\in f^{-1}(A)\cup f^{-1}(B)$.
   
   We have shown that $\displaystyle{f^{-1}(A\cup B)\subseteq f^{-1}(A)\cup f^{-1}(B)}$. Let $x\in f^{-1}(A)\cup f^{-1}(B)$. Then we either have that $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$, so either $f(x)\in A$ or $f(x)\in B$. This means that $f(x)\in A\cup B$. It follows that $x\in f^{-1}(A\cup B)$.
   
   We have shown that $\displaystyle{f^{-1}(A)\cup f^{-1}(B) \subseteq f^{-1}(A\cup B)}$. From these two inclusions it follows that $\displaystyle{f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)}$.
   
   $$$$
   
2. $$x\in f^{-1}(\overline{A}) \iff f(x) \in \overline{A} \iff f(x) \notin A \iff x\notin f^{-1}(A) \iff x\in \overline{f^{-1}(A)}$$
   Therefore, we get that $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$.
   
   $$$$
   
3. $$x\in f^{-1}(A\setminus B) \iff f(x)\in A\setminus B \iff f(x) \in A \text{ AND } f(x) \notin B \iff x\in f^{-1}(A) \text{ AND } x\notin f^{-1}(B) \\ \iff x\in f^{-1}(A)\setminus f^{-1}(B)$$
   Therefore, we get that $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)$.
   
   $$$$
   
4. We suppose that $f^{-1}(A)\cap f^{-1}(B)\neq\emptyset$. That means that $\exists x\in f^{-1}(A)\cap f^{-1}(B)$. That means that $x\in f^{-1}(A)$ AND $x\in f^{-1}(B)$. That implies that $f(x)\in A$ AND $f(x)\in B$, and so $f(x)\in A\cap B$. We get a contradiction since $A\cap B=\emptyset$.
   So, that what we supposed at the beginning was wrong, i.e., $f^{-1}(A)\cap f^{-1}(B)\neq\emptyset$ is not correct. So, it must be $f^{-1}(A)\cap f^{-1}(B)=\emptyset$.
   

Is everything correct and are the proofs complete? Or could I improve something?

Could you give me a hint about the last statement?

(Wondering)

 

[S.G. Janssens]

Could you give me a hint about the last statement?
(Wondering)

The empty set is a subset of any other set.

I'm sorry that I do not have the time to check your other work, but this is at least a hint about the last statement.

 

[mathmari]

The empty set is a subset of any other set.

5. $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$

So do you mean to take $A=\emptyset$ ? (Wondering)

 

[S.G. Janssens]

So do you mean to take $A=\emptyset$ ? (Wondering)

Yes, and $B = N$.
What do you think about the veracity of statement #5?

 

[I like Serena]

Hey mathmari! (Smile)

You wrote: '$f(x)∈A∪B$. This means that either $f(x)∈A$ or $f(x)∈B$.'

Using 'either ... or ...' means that they cannot both be true, but that's not the case is it?
Shouldn't it be: '$f(x)∈A∪B$. This means that $f(x)∈A$ or $f(x)∈B$.'? (Wondering)
I think all occurrences of 'either' should be removed.

Other than that I think everything is correct! (Nod)

 

[mathmari]

Yes, and $B = N$.
What do you think about the veracity of statement #5?

So, for $A=\emptyset$ and $B=N$ we have that $\emptyset \subseteq N$.
Let $x\in f^{-1}(N)$ then we have that $f(x)\in N$.
It cannot hold that $f(x)\in \emptyset$ and so it cannot hold that $x\in f^{-1}(\emptyset)$.
Therefore, we have that $f^{-1}(N)\nsubseteq f^{-1}(\emptyset)$.
And so generally the implication $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$ does not hold.

Is this correct? (Wondering)

You wrote: '$f(x)∈A∪B$. This means that either $f(x)∈A$ or $f(x)∈B$.'
Using 'either ... or ...' means that they cannot both be true, but that's not the case is it?
Shouldn't it be: '$f(x)∈A∪B$. This means that $f(x)∈A$ or $f(x)∈B$.'? (Wondering)
I think all occurrences of 'either' should be removed.
Other than that I think everything is correct! (Nod)

Ah ok! (Smile)

So, by '$f(x)∈A∪B$. This means that $f(x)∈A$ or $f(x)∈B$.' it is meant that both can be true, isn't it? Or do we have to mention it? (Wondering)

 

[I like Serena]

Is this correct?

So, by '$f(x)∈A∪B$. This means that $f(x)∈A$ or $f(x)∈B$.' it is meant that both can be true, isn't it? Or do we have to mention it?

Yep. All correct.
And there is no need to mention that both can be true - that is implied in the word 'or'.
If we want to be explicit about it, we can use 'and/or'. (Mmm)

 

[mathmari]

Yep. All correct.
And there is no need to mention that both can be true - that is implied in the word 'or'.
If we want to be explicit about it, we can use 'and/or'. (Mmm)

Great! (Mmm)

Is the implication $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$ true when a specific condition holds or is this never true? (Wondering)

 

[I like Serena]

Is the implication $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$ true when a specific condition holds or is this never true? (Wondering)

Can we find an example that it's true?
What happens if $A$ and $B$ are $\varnothing$, equal to each other, or $N$? (Wondering)

 

[mathmari]

Can we find an example that it's true?
What happens if $A$ and $B$ are $\varnothing$, equal to each other, or $N$? (Wondering)

Ah if $A=B$ we have that $A\subseteq B=A$ and that $f^{-1}(B)=f^{-1}(A)\subseteq f^{-1}(A)$. So, when the sets are equal to each other the implication is true, right? (Wondering)

 

[I like Serena]

Ah if $A=B$ we have that $A\subseteq B=A$ and that $f^{-1}(B)=f^{-1}(A)\subseteq f^{-1}(A)$. So, when the sets are equal to each other the implication is true, right?

Yep. (Nod)

 

[mathmari]

Thank you very much! (Mmm)