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.

  • 1
    $\begingroup$ 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

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. $\Box$


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.