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?

  • $\begingroup$ This is definitely not a question related to group theory. $\endgroup$ – Dbchatto67 Apr 14 at 4:43
  • $\begingroup$ @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 $\endgroup$ – John Arg Apr 14 at 4:44
  • $\begingroup$ 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. $\endgroup$ – zwim Apr 14 at 4:45
  • $\begingroup$ @zwim It is a new book. They say that the range is a subset of the codomain. $\endgroup$ – 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 $f(A) \subseteq f(B).$

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    $\begingroup$ maybe, but that is not the question asked. $\endgroup$ – zwim Apr 14 at 4:43
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    $\begingroup$ Actually $f(B) =$ range of $B$ under the function $f \subseteq S.$ $\endgroup$ – Dbchatto67 Apr 14 at 4:47
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    $\begingroup$ It is clear that the argument in your book is wrong because $f(x) \in f(A) \not\implies x \in A.$ $\endgroup$ – Dbchatto67 Apr 14 at 4:50
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    $\begingroup$ 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 $x \in A$? $\endgroup$ – Dbchatto67 Apr 14 at 4:58
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    $\begingroup$ If the function was injective then definitely the proof in your book works. But it is not the case then you can proceed along the lines of my answer which I think is a valid proof of your question. $\endgroup$ – Dbchatto67 Apr 14 at 5:08

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