3
$\begingroup$

I have come across a question while solving practice papers on the topic 'Functions'.

The question is as follows -

If $f : X \to Y $, find $f (X)$, when $f $ is a surjective or onto mapping.

Here $X $ and $Y $ are non-empty sets

Here is my approach -

As $f $ is a surjective mapping of $X$ to $Y$, then for each $y \in Y $, there exists an $x \in X $, such that $f (x) = y $. Thus $f (X) = Y$.

Can I be provided with a more formal proof ?

Suggestions for correction in my answer and a detailed answer with explanation would be helpful.

$\endgroup$
2
  • $\begingroup$ "...mapping of $X$ to $\color{red}Y$..." $\endgroup$ May 27, 2019 at 7:03
  • 4
    $\begingroup$ I think you said what needed to be said! If you want to be really pedantic, notice that your argument proves that $f(X) \supseteq Y$; one could mention that $f(X) \subseteq Y$ is immediate from the definition of $f(X)$, but in most circumstances that's so obvious as to not be worth mentioning. $\endgroup$ May 27, 2019 at 7:13

2 Answers 2

5
$\begingroup$

Just to be really really pedantic, let us prove it by double inclusion. Let us set $f(X):=\left\{y\in Y\mid \exists\,x\in X \text{ such that } y=f(x)\right\}$. Thus $f(X)\subseteq Y$. On the other hand, surjectivity means that for every $y\in Y$ there exists $x\in X$ such that $y=f(x)$, that is to say, that $Y\subseteq f(X)$. Conclusion: $Y=f(X).$

$\endgroup$
6
  • $\begingroup$ "$f(X):=\left\{y\in Y\mid \exists\,x\in X \text{ such that } y=f(x)\right\}$" and " surjectivity means that for every $y\in Y$ there exists $x\in X$ such that $y=f(x)$". The two quoted statements have actually the same meaning and then how have you proven $f (X) \subseteq Y$ and $Y \subseteq f (X)$ by the statements having the same meaning ? Can you Please explain it in a bit detail or by using an example ? $\endgroup$ May 27, 2019 at 8:34
  • 1
    $\begingroup$ @AdityaRaj I didn't understand your objection: the two doesn't have the same meaning. Take $f:\left\{\bullet,*\right\}\to\left\{\bullet,*\right\}$ given by $f(\bullet)=\bullet=f(*)$. Then $f(\left\{\bullet,*\right\})=\left\{\bullet\right\}\subsetneq \left\{\bullet,*\right\}$. Does this clarify the difference? In fact, it is always true that $f(X)\subseteq Y$, while surjectivity is a restatement of the fact that in addition $f(X)$ covers $Y$. $\endgroup$ May 27, 2019 at 8:50
  • 2
    $\begingroup$ @AdityaRaj They don't have the same meaning. The first is the definition for the set $f(X)$. The second is the definition for surjectivity of $f$. Note that $f(X)$ is a set, and the statement "$f$ is surjective" is a true or false statement. You can see that $f(X) \subseteq Y$, as the definition begins $f(X) := \{ \color{red}{y \in Y} \mid \ldots$ It's pretty trivial, and it holds regardless of whether $f$ is surjective or not. $\endgroup$ May 27, 2019 at 8:52
  • $\begingroup$ I have understood half...i am still no able to comprehend that how we are proving $Y \subseteq f (X) $ by defining that $f$ is surjective. Can I please be provided an element wise proof for the same ? $\endgroup$ May 27, 2019 at 9:47
  • 1
    $\begingroup$ @AdityaRaj Caveat: you are not defining $f$ to be surjective, you are assuming $f$ surjective. Being surjective is a property of $f$: $f$ is surjective if and only if every element in the codomain is also in the image of $f$ (in symbols, if and only if $Y=f(X)$). For example, consider $f:\{*,\bullet\}\to\{\bullet\}$ given by $f(*)=\bullet=f(\bullet)$. Then obviously $\{\bullet\}\subseteq f(\{*,\bullet\})\subseteq \{\bullet\}$ and hence $f(\{*,\bullet\})=\{\bullet\}$ and therefore you can claim that $f$ is surjective $\endgroup$ May 27, 2019 at 10:31
2
$\begingroup$

Below an indirect proof.

Remark : ( after incorrect edits have been made, then fixed) : since set equality means reciprocal inclusion, the indirect proof consists in showing that the negation of this reciprocal inclusion is false. Set equality is not defined in terms of reciprocal membership.

That " $Y$ is not included in $f(X)$ " is contradictory follows immediately from surjectivity.

To show that " $f(X)$ is not included in $Y$ " is contradictory, I use the definition of a function $f : X \to Y$ as relation between $X$ and $Y$ and therefore as subset of the cartesian product " $X × Y$". Saying that "$f(X)$ is not included in $Y$" would mean that $f$ contains an ordered pair that does not belong to the cartesian product " $X \times Y$" , which contradicts the definition of $f$.

enter image description here

$\endgroup$
2
  • 2
    $\begingroup$ "$f$ is injective" shouldn't be here at all. It's not a premise of the question, and so we need to prove the claim even if $f$ is not injective. $\endgroup$
    – aschepler
    May 27, 2019 at 15:22
  • 1
    $\begingroup$ @aschepter. You're right. I misread the question. ( Did not make use of this useless premise though). $\endgroup$
    – user654868
    May 27, 2019 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.