I got a serious doubt ahead the question

Be $f:X\longrightarrow Y$ a function. If $A,B\subset X$, show that $f(A \cap B)\subset f(A)\cap f(B)$

I did as follows

$$\forall\;y\in f(A\cap B)\Longrightarrow \exists x\in A\cap B, \text{ such that } f(x)=y\\ \Longrightarrow x \in A\text{ and }x\in B\Longrightarrow f(x)\in f(A)\text{ and }f(x)\in f(B)\\ \Longrightarrow f(x)\in f(A)\cap f(B)\Longrightarrow y\in f(A)\cap f(B)$$

This ensures that $\forall y \in f(A\cap B)$ then $y\in f(A)\cap f(B)$, therefore $f(A\cap B)\subset f(A)\cap f(B)$.

Okay, we have the full demonstration.

We know that for equality to be valid, then $ f $ must be injective. But my question is when should I see that equality is not worth, not by counter example, but finding an error in the following demonstration

$$\forall\;y\in f(A)\cap f(B)\Longrightarrow y\in f(A)\text{ and }y\in f(B) \Longrightarrow \\ \exists x\in A \text{ and } B, \text{ such that } f(x)=y\\ \Longrightarrow x \in A\cap B\ \Longrightarrow f(x)\in f(A\cap B)\Longrightarrow y\in f(A\cap B)$$

Where is the error in the statement? Which of these steps can not do and why?


you wrote :

$$\forall\;y\in f(A)\cap f(B)\Longrightarrow y\in f(A)\text{ and }y\in f(B) \Longrightarrow \\ \exists x\in A \text{ and } B, \text{ such that } f(x)=y\\ $$

The problem is in the last implication : from $y\in f(A)\text{ and }y\in f(B)$ you get that there exist $x_A\in A$ and $x_B\in B$ such that $f(x_A)=y=f(x_B)$, you cannot assume that $x_A=x=x_B$.

  • $\begingroup$ Being quite honest with you, already researched a lot about this issue, and you in a few seconds I answered very clearly say that now I understand the resolution. Many Thanks. $\endgroup$ – marcelolpjunior Mar 25 '15 at 11:48
  • 1
    $\begingroup$ You're welcome, let me add that the only reason I can answer your question so quickly is because I have done the same kind of mistake myself few years ago... $\endgroup$ – Clément Guérin Mar 25 '15 at 12:47

Suppose $y\in f(A)\cap f(B)$. Then $y\in f(A)$, so there is $x_1\in A$ with $f(x)=y$. Moreover $y\in f(B)$, so there is $x_2\in B$ such that $f(x_2)=y$.

There's no reason why we should have $x_1=x_2$, except in the case when $f$ is injective.

Counterexample: $X=\{1,2\}$, $Y=\{0\}$, $f(1)=f(2)=0$. With $A=\{1\}$ and $B=\{2\}$ we have $$ f(A\cap B)=f(\emptyset)=\emptyset\\ f(A)\cap f(B)=\{0\}\cap\{0\}=\{0\} $$

If you had written the proof in words, instead of piling up symbols, you'd probably have discovered the issue.

Of course we might take $x_1=x_2$ in special cases. even when $f$ is not injective. For particuler subsets $A$ and $B$ we could have $f(A\cap B)=f(A)\cap f(B)$ (for example, when $B=X$), but not in general, unless $f$ is injective.

Actually, it's easy to prove that $f$ is injective if and only if, for all $A,B\subset X$, $f(A\cap B)=f(A)\cap f(B)$.


$\forall\;y\in f(A)\cap f(B)\Longrightarrow y\in f(A)\text{ and }y\in f(B) \Longrightarrow \exists x\in A \text{ and } B, \text{ such that } f(x)=y$

You have problem here. $y\in f(A)\text{ and }y\in f(B) \Longrightarrow \exists x_1\in A \text{ and } x_2 \in B, \text{ such that } f(x_1)=y=f(x_2).$ These $x_1$ and $x_2$ need not be equal. There you need the injectivity of $f$ to conclude that $x_1=x_2.$


True, if $y\in f(A)\cap f(B),$ then (1) there exists $x\in A$ such that $f(x)=y,$ and (2) there exists $x\in B$ such that $f(x)=y.$ The error is in assuming that these are referring to a particular $x$ and that the $x$ is the same in both cases!

Rather, we should interpret (1) as telling us that $\{x\in A:f(x)=y\}\neq\emptyset,$ and interpret (2) similarly. However, without injectivity, we can't necessarily conclude that $$\{x\in A:f(x)=y\}\cap\{x\in B:f(x)=y\}\neq\emptyset,$$ which is equivalent to saying that there is some $x\in A\cap B$ such that $f(x)=y.$


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.