One-to-one subset proof

Question

The question I am working on is:

Let $f:X \to Y$ be a function. Prove that:

If $f$ is one-to-one, then $A=f^{-1}(f(A))$ for every subset $A$ of $X$.

What I tried:

1. Looking up similar questions on here and in google:

Found this one: Prove that $A ⊆ f^{−1}(f(A))$. Wasn't able to get any further from looking around.

2. Tried writing down the definition of one-to-one, to see if that could somehow help make some sort of connection:

Proof:

Let $a, a' \in A \subset X$.

Then by the definition of $f$ being one-to-one, if $f(a)=f(a')$, then $a=a'$.

Answer 1

Suppose that $a\in A$. Then $f(a)\in f(A)$ and therefore $a\in f^{-1}\bigl(f(A)\bigr)$. This proves that $A\subset f^{-1}\bigl(f(A)\bigr)$

Now, suppose that $a\in f^{-1}\bigl(f(A)\bigr)$. That means that $f(a)\in f(A)$; in other words, $f(a)=f(a')$ for some $a'\in A$. But then, since $f$ is one-to-one, $a=a'\in A$. And this proves that $f^{-1}\bigl(f(A)\bigr)\subset A$.

Answer 2

Note by definition that $f^{-1}(f(A))=f^{-1}(\{y\in Y:y=f(a)\text{ for some }a\in A\})=\{x\in A:f(x)=f(a)\text{ for some }a\in A\}$.

If $a\in A$, clearly $a\in f^{-1}(f(A))$ by the last equality. On the other hand, if $x\in f^{-1}(f(A))$, then $f(x)=f(a)$ for some $a\in A$. But since $f$ is 1-1, $x=a$, so $x\in A$.