# Proof of function property in group theory

The question states: Consider a function $$f: S\rightarrow S$$. Let $$f(A) = \{f(x)|x \in A\}$$ and $$f(B) = \{f(x)|x \in B\}$$. Prove that if $$A \subseteq B \subseteq S$$ then $$f(A) \subseteq f(B)$$.

The solution states:

$$B \subseteq S$$ and $$f : S\rightarrow S \Rightarrow f(B) \subseteq$$ range of $$B$$

Now if $$f(x) \in f(A)$$

$$\Rightarrow x \in A$$

$$\Rightarrow x \in B$$ {as $$A \subseteq B$$}

$$\Rightarrow f(x) \in f(B)$$

Thus $$f(x) \in f(B)$$ for all $$x \in A$$

Hence, $$f(x) \in f(A) \Rightarrow f(x) \in f(B)$$.

Thus $$f(A) \subseteq f(B)$$.

What is the point of the first line of this solution?

• This is definitely not a question related to group theory. – Dbchatto67 Apr 14 at 4:43
• @Dbchatto67 I am sorry this is from an introductory group theory textbook. Perhaps the basics aren't considered group theory yet? If so, please do not hesitate to change the title and tags – John Arg Apr 14 at 4:44
• what is the date of your book ? sometimes old books refer to range as the codomain, while nowadays there is no distinction between the image $f(B)$ and the range of $B$. Anyway, this line seems superfluous to me as well. – zwim Apr 14 at 4:45
• @zwim It is a new book. They say that the range is a subset of the codomain. – John Arg Apr 14 at 4:47

Let $$y \in f(A).$$ Then $$\exists$$ $$x \in A$$ such that $$f(x) = y.$$ Now $$A \subseteq B$$ and $$x \in A \implies x \in B.$$ Therefore $$y = f(x) \in f(B).$$ This shows that $$f(A) \subseteq f(B).$$
• Actually $f(B) =$ range of $B$ under the function $f \subseteq S.$ – Dbchatto67 Apr 14 at 4:47
• It is clear that the argument in your book is wrong because $f(x) \in f(A) \not\implies x \in A.$ – Dbchatto67 Apr 14 at 4:50
• Nope. Take the function $f(x) = 0,$ for all $x \in \Bbb R.$ Consider the set $A = \Bbb Z.$ Then it is clear that $f(x) \in f(A) = \{0 \}.$ Does it necessarily imply that $x \in A$? – Dbchatto67 Apr 14 at 4:58