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An injective (one-to-one) function is a function $f$ which satisfies the property that if $a_1 \neq a_2$, then $f(a_1) \neq f(a_2)$, or equivalently, the contrapositive $f(a_1) = f(a_2)$ implies $a_1 = a_2$.

My question is whether this property also holds for sets. Specifically, if we let $A_1$ and $A_2$ be sets, then does $f(A_1) = f(A_2)$ imply that $A_1 = A_2$?

Joe
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Mathstudent123
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    Yes, $f^{-1}[f(A)] = A$ if $f$ is injective. – Dominik Kutek Jul 27 '24 at 11:54
  • I don't understand the vote to close here. The question is clear and well-motivated by the first paragraph. – Joe Jul 27 '24 at 12:03
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    @Joe It's clear and well motivated, but it's missing any efforts. (It's not my vote) – jjagmath Jul 27 '24 at 12:05
  • @jjagmath: I've deleted my answer because I now think that this question is likely a duplicate. (Mathstudent123, if you still want to look at my answer, it is: If $x\in A_1$, then $f(x)\in f(A_1)$, so $f(x)\in f(A_2)$, hence there is an $a\in A_2$ such that $f(x)=f(a)$. By injectivity of $f$, it follows that $x=a$, so $x\in A_2$. This establishes that $A_1\subseteq A_2$. The proof that $A_2\subseteq A_1$ is essentially the same.) – Joe Jul 27 '24 at 12:12
  • @Joe You don't need to explain me why you deleted your answer. But remember that questions that show no efforts in solving the problems are discouraged in this forum, even if they are well written and motivated. – jjagmath Jul 27 '24 at 12:36
  • @jjagmath I apologize for not putting in enough effort to solve the problem. I will make sure to do better next time. – Mathstudent123 Jul 27 '24 at 15:12
  • @Joe, I really appreciate your comment. Thank you! – Mathstudent123 Jul 27 '24 at 15:13

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Yes, if $x \in A_1$, then $f(x) \in f(A_1) = f(A_2)$. Thus there exists $y \in A_2$ such that $f(x) = f(y)$. By injectivity of $f$, $x = y \in A_2$. It proves that $A_1 \subset A_2$. Symletrically, $A_2 \subset A_1$.

Cactus
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