# How to Prove The Complement Of The Domain Is Complement Of The Image If f Is Bijective

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.

• 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 $f\left(\overline{[0,1]\cap \mathbb Q}\right)\neq \overline{f([0,1]\cap \mathbb Q)}$ – Surb Dec 9 '18 at 13:02
• What if it is bijective it is surely true. Let me change the question then – selman özlyn Dec 9 '18 at 13:12
• 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 ? – Surb Dec 9 '18 at 13:33

## 1 Answer

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. $$\Box$$