Clash Royale CLAN TAG#URR8PPP

Closed functions

DEFINITION: Let M,N be two metric spaces, $f:Mrightarrow N$ is a closed function when for all closed $Fsubset M$, its image, $f(F)$ is closed in N.

PROBLEM Let $f:Mrightarrow N$ is a closed function if and only if for all $yin N$ and all open $Vsubset M$ with $f^-1(y)subset V$ exist an open $Usubset M$ such that $f^-1(y)subset f^-1(U) subset V$.

PROOF ATTEMPT: ($rightarrow$) Let $f$ be a closed function, $yin N$ and $Vsubset M$ such that $f^-1(y)subset V$. Since V is open and $f$ is closed, $M-V$ is a closed and $f(M-V)$ is closed. Then, $N-f(M-V)$ is open, and since $f^-1(y)cap M-V=emptyset$, then $ycap f(M-V)=emptyset$ $implies$ $yin N-f(M-V)$ which is open, then there is some open $U$ such that $yin Usubset N-f(M-V)implies Ucap f(M-V) = emptyset implies U subset f(V) implies ysubset Usubset f(V)$ and if we apply $f^-1$, then $f^-1(y)subset f^-1(U) subset V$.

QUESTION: Is ($rightarrow$) correctly proven? How could I prove ($leftarrow$) ? Thanks so much for your answers.

2 Answers
2

You still need to fix the statement of the problem, but your proof is correct now. To prove the other direction, suppose that $f$ is not closed, so there is a closed $Csubset M$ such that $f(C)$ is not closed. Let $yin barf(C)setminus f(C)$. $f^-1(y) in C^c$ open, so there is a $Vsubset N$ open such that $f^-1(y) in f^-1(V) subset C^c$. However, because $y$ is in the closure of $f(C)$, by definition this means that $Vcap f(C) neq emptyset$, so let $z$ be in the intersection. Since $zin f(C)$, there must be $xin C$ such that $f(x)=z$. This means that $xin f^-1(V)$ since $f(x)in V$. But $xnotin C^c$, which contradicts $f^-1(V)subset C^c$.

The following statement $ f^-1(y)cap f(M-V)=emptyset $ does not make sense because $ f^-1(y)$ and $ f(M-V)$ are in different metric spaces. You need to redo this part.

Drawing a graph will help you staying focused in your spaces.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.