MHB Set Theory for Beginners: How is A' ⊆ A and its Complement a Subset of A?

[Romono]

Could someone please explain how the image of a set A' ⊆ A is the set: f(A') = {b | b = f(a) for some a ∈ A'}. And how can the complement of A be a subset of A? Forgive my ignorance here, I'm a beginning student of set theory.

 

[Euge]

Hi Romono,

The answer to your first question is by definition. For your second question, the complement of $A$ (in $A$) is the empty set, and the empty set is a subset of $A$.

 

[Romono]

Hi Romono,

The answer to your first question is by definition. For your second question, the complement of $A$ (in $A$) is the empty set, and the empty set is a subset of $A$.

Hi Euge,

Thanks for your reply, but maybe I should rephrase my question: Could you explain what "the image of a set A' ⊆ A is the set: f(A') = {b | b = f(a) for some a ∈ A'}" actually means? Could you break it down? I don't understand what an image of a set is even after reading the definition here.

Also, thanks for answering my second question -- I think I understand that now.

 

[Euge]

Suppose you have function $f : A \to B$. This means that for every $a\in A$, there corresponds a unique $b\in B$ such that $f(a) = b$. Given $a \in A$, the element $f(a) \in B$ is called the image of $a$. Thus, given a subset $A'$ of $A$, $f(A')$ is the set of images $f(a')$, as $a'$ ranges over $A'$.

Let's consider an example. Define a function $f : \{1, 2, 3\} \to \{a, b, c\}$ by setting $f(1) = a$, $f(2) = b$, and $f(3) = c$. Since $1$ and $2$ are the only elements of $\{1, 2\}$, $f(\{1,2\}) = \{f(1), f(2)\} = \{a, b\}$. How about $f(\{2,3\})$? Since $2$ and $3$ are the only elements of $\{2, 3\}$, $f(\{2, 3\}) = \{f(2), f(3)\} = \{b, c\}$.

Here's another example. Let $\Bbb N$ denote the set of natural numbers. Define $g : \Bbb N \to \Bbb N$ by setting $g(n) = n + 1$ for all $n\in \Bbb N$. Let's find $g(2\Bbb N)$, where $2\Bbb N$ is the set of even natural numbers. Every element of $2\Bbb N$ is of the form $2n$ for some $n\in \Bbb N$. Now $g(2n) = 2n + 1$ for all $n \in \Bbb N$. Thus $g(2\Bbb N)$ consists of all natural numbers of the form $2n + 1$. In other words, $g(2\Bbb N)$ is the set of odd natural numbers.

 

[Romono]

Let's consider an example. Define a function $f : \{1, 2, 3\} \to \{a, b, c\}$ by setting $f(1) = a$, $f(2) = b$, and $f(3) = c$. Since $1$ and $2$ are the only elements of $\{1, 2\}$, $f(\{1,2\}) = \{f(1), f(2)\} = \{a, b\}$.

Just to be clear in this example, {a,b} would then be the image, wouldn't it? I think I'm understanding it now...

 

[Euge]

Just to be clear in this example, {a,b} would then be the image, wouldn't it? I think I'm understanding it now...

Yes, it is the image of the set $\{1,2\}$.