Statements about the preimages

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In summary, the conversation discusses the correctness of statements involving preimages under a function $f:M\rightarrow N$. The participants provide proofs for the statements and discuss their veracity. It is concluded that the implication $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$ is generally not true, but it can hold under certain conditions.
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
mathmari said:
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.
 
  • #3
Krylov said:
The empty set is a subset of any other set.
mathmari said:
5. $A\subseteq B \Rightarrow f^{-1}(B)\subseteq f^{-1}(A)$
So do you mean to take $A=\emptyset$ ? (Wondering)
 
  • #4
mathmari said:
So do you mean to take $A=\emptyset$ ? (Wondering)
Yes, and $B = N$.
What do you think about the veracity of statement #5?
 
  • #5
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)
 
  • #6
Krylov said:
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)
I like Serena said:
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)
 
  • #7
mathmari said:
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)
 
  • #8
I like Serena said:
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)
 
  • #9
mathmari said:
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)
 
  • #10
I like Serena said:
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)
 
  • #11
mathmari said:
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)
 
  • #12
Thank you very much! (Mmm)
 

What is a preimage?

A preimage is the original set of values or objects that undergo a transformation to create a new set of values or objects, known as the image.

Why are statements about preimages important in science?

Statements about preimages are important because they allow scientists to understand the relationship between the original set of values or objects and the transformed set. This can help with making predictions, analyzing patterns, and understanding the underlying mechanisms of a system.

What types of transformations can affect preimages?

There are many types of transformations that can affect preimages, including translation, rotation, reflection, dilation, and more complex transformations such as affine transformations and fractals. These transformations can occur in various fields of science, such as mathematics, physics, biology, and computer science.

Can preimages be used to prove a statement?

Yes, preimages can be used to prove a statement by showing that the original set of values or objects can be transformed into the desired outcome. This is often done through the use of mathematical proofs, where the preimage is manipulated to demonstrate that the resulting image supports the statement.

How can preimages be visualized or represented?

Preimages can be visualized through various means, such as diagrams, graphs, or physical models. They can also be represented mathematically using equations or matrices. In computer science, preimages are often represented as input values in a function that produces an output value or image.

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