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It seems true that $f(\overline{X}) = \overline{f(X)}$ for $f:A\rightarrow B$ and $X$ is any subset of $A$ if and only if $f$ is bijective.But I couldn't write it as a formal way like epsilon argument.It makes sense to me but the trouble I have is with the formal prove.

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    $\begingroup$ Of course not ! Take $f(x)=x\boldsymbol 1_{\mathbb Q\cap [0,1]}$. Is surjective : $[0,1]\to \mathbb Q\cap [0,1]$ but $f\left(\overline{[0,1]\cap \mathbb Q}\right)\neq \overline{f([0,1]\cap \mathbb Q)}$ $\endgroup$ – Surb Dec 9 '18 at 13:02
  • $\begingroup$ What if it is bijective it is surely true. Let me change the question then $\endgroup$ – selman özlyn Dec 9 '18 at 13:12
  • $\begingroup$ Now it's not : $f(x)=x\boldsymbol 1_{\mathbb Q\cap [0,1]}-x\boldsymbol 1_{\mathbb R\setminus \mathbb Q\cap [0,1]}$. It's bijective $[0,1]\to (\mathbb Q\cap [0,1])\cup((\mathbb R\setminus \mathbb Q)\cap (0,-1])$ but $f(\overline{[0,1]\cap \mathbb Q})\neq \overline{f([0,1]\cap \mathbb Q)}$. What is exactely your exercise ? $\endgroup$ – Surb Dec 9 '18 at 13:33
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Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A \rightarrow B$. We have the following elementary propositions:

  1. $f$ is injective if and only if $$(\forall X)(X \subseteq A \implies f(\complement_{A}X) \subseteq \complement_{B}f(X))$$

Proof : Assuming first the injectivity of $f$, consider arbitrary $X \subseteq A$ and $y \in f(\complement_{A}X)$, such that $y=f(x)$ with $x \in A \setminus X$. If we were to assume by contradiction that $y \in f(X)$ it would entail that $y=f(t)$ for a certain $t \in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t \in X$ in contradiction to $x \notin X$. Hence $y \in B \setminus f(X)$ and the inclusion is established.

Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y \in A$ with $x \neq y$. This means that $y \in A \setminus \{x\}$ and thus by our assumption $$f(\{y\})=\{f(y)\} \subseteq f(A \setminus \{x\})\subseteq B \setminus f(\{x\})=B \setminus \{f(x)\}$$ from which we infer that $f(x) \neq f(y)$ and conclude $f$ is indeed injective. $\Box$

  1. $f$ is bijective if and only if: $$(\forall X)(X \subseteq A \implies f(\complement_{A}X)=\complement_{B}f(X))$$

Proof : Assume first $f$ is bijective and consider arbitrary $X \subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:

$$(\forall X)(X \subseteq A \implies f(A) \setminus f(X) \subseteq f(A \setminus X))$$

and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.

To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= \emptyset$, which yields: $$f(A \setminus \emptyset)=f(A)=B \setminus f(\emptyset)=B \setminus \emptyset =B$$ This concludes the proof. $\Box$

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