Question

Let $f$ be a map. Prove that (a) $f[f^{-1}[A]]\subseteq A$ (b) $f[f^{-1}[A]]=A$ if $A\subseteq D_f^{\prime }$ (c) if $A\subseteq D_f$ and $f$ is one to one, $A=f^{-1}[f[A]]$. Is $f[A]\cap B\subseteq f[A\cap f^{-1}[B]]?$

Short answer

Yes, $f[A]\cap B\subseteq f[A\cap f^{-1}[B]]$.

Step-by-step solution

Understanding the Problem

We need to prove three properties involving a function $f$ and subsets $A$, focusing on preimages $f^{-1}[A]$ and images $f[A]$. For the additional question, we analyze relationships between a function's image under intersections.

Proving Part (a)

To prove $f[f^{-1}[A]]\subseteq A$, take an element $y\in f[f^{-1}[A]]$. Then there exists an $x$ such that $x\in f^{-1}[A]$ and $f(x)=y$. From $x\in f^{-1}[A]$, we know $f(x)\in A$. Therefore, $y\in A$, showing $f[f^{-1}[A]]\subseteq A$.

Proving Part (b)

Assume $A\subseteq D_f^{\prime }$, meaning $A$ contains only values in the range of $f$. Now, to prove $f[f^{-1}[A]]=A$, we combine the inclusion from part (a) and show $A\subseteq f[f^{-1}[A]]$. For any $a\in A$, since $a$ is in the range of $f$, $a=f(x)$ for some $x$. Thus $x\in f^{-1}[A]$, implying $a\in f[f^{-1}[A]]$. Therefore, $f[f^{-1}[A]]=A$.

Proving Part (c)

Assume $A\subseteq D_f$ and $f$ is one-to-one. We need to show $A=f^{-1}[f[A]]$. For any $x\in A$, $f(x)\in f[A]$, so $x\in f^{-1}[f[A]]$ giving one inclusion. Conversely, if $x\in f^{-1}[f[A]]$, then $f(x)=f(a)$ for some $a\in A$. Since $f$ is injective, $x=a$, thus $x\in A$. Both inclusions give us $A=f^{-1}[f[A]]$.

Analyzing the Question

For $f[A]\cap B\subseteq f[A\cap f^{-1}[B]]$, take any $y\in f[A]\cap B$. We know $y=f(x)$ for some $x\in A$ and $y\in B$. Hence, $x\in f^{-1}[B]$, meaning $x\in A\cap f^{-1}[B]$. Therefore, $y\in f[A\cap f^{-1}[B]]$, proving the subset inclusion.

Key concepts

Function Image

In mathematical terms, when we talk about a function's image, we're referring to the set of all output values it can produce. For a function $f:X\to Y$, the image of a subset $A\subseteq X$ is denoted as $f[A]$. This includes all elements $y\in Y$ such that $y=f(x)$ for some $x\in A$.
To understand it simply, consider $f[A]$ as a collection of results you get by applying the function to every member of $A$. This concept is handy in many areas of math because it helps us map relationships between different sets.
For instance, when dealing with part (a) and (b) of the exercise, knowing that the function image $f[f^{-1}[A]]$ being a subset of $A$ or equal to $A$ (under certain conditions), allows us to deeply understand how functions translate inputs to outputs.

One-to-One Function

A one-to-one function, also known as an injective function, is one where every element from the domain is mapped to a unique element in the codomain, meaning no two distinct elements in the domain point to the same element in the codomain.
Mathematically, a function $f$ is injective if, for every pair of distinct elements $x_1,x_2$ in the domain $D_f$, $f(x_1)eqf(x_2)$ holds true. This property is essential in proving part (c) of our exercise.
Here, we want to demonstrate that $A=f^{-1}[f[A]]$ if $f$ is one-to-one. An injective function ensures if $x$ and $y$ map to a common image, they must indeed be the same element, helping us prove our required equality.

Subset Inclusion

Subset inclusion is a principle where we say that one set $A$ is included in another set $B$ (written as $A\subseteq B$) if every element of $A$ is also an element of $B$. This is a foundational concept in proving mathematical statements involving sets.
In our exercise, subset inclusion plays a vital role in establishing our proofs, specifically in parts (a) and (b), where we show how images and pre-images relate and fit under inclusion properties. By demonstrating $f[f^{-1}[A]]\subseteq A$, we confirm that all elements of the image of the pre-image of $A$ are indeed within $A$ itself.
Understanding subset inclusion helps in forming a logical link between these different set-based operations and validates the integrity of mathematical arguments.

Mathematical Proof

Mathematical proof is a chain of logical statements and deductions that establish the truth of a given mathematical statement. It is crucial in fields like analysis, algebra, and geometry. A proof must be logically valid and comprehensive, starting from known truths or axioms.
In our exercise, several proofs exist to verify the statements provided. Each part—(a), (b), and (c)—offers a unique challenge requiring proper logical reasoning. For instance, proving $f[f^{-1}[A]]\subseteq A$ involves:

• Identifying elements in set definitions,
• Determining properties of function images and preimages,
• Utilizing subset inclusions.

Sadly, not everyone is naturally apt at proofs. However, practice and understanding these logical structures can dramatically improve one's ability to engage with mathematical exercises.