Mixing
Closed functions
Clash Royale CLAN TAG#URR8PPP
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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.
real-analysis general-topology proof-verification metric-spaces
asked 2 hours ago
duhdave
557
add a comment |Â
up vote
1
down vote
favorite
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.
real-analysis general-topology proof-verification metric-spaces
asked 2 hours ago
duhdave
557
1
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
1
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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.
real-analysis general-topology proof-verification metric-spaces
asked 2 hours ago
duhdave
557
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.
real-analysis general-topology proof-verification metric-spaces
real-analysis general-topology proof-verification metric-spaces
asked 2 hours ago
duhdave
557
asked 2 hours ago
duhdave
557
edited 18 mins ago
asked 2 hours ago
duhdave
557
asked 2 hours ago
duhdave
557
asked 2 hours ago
duhdave
557
557
1
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
1
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago
add a comment |Â
1
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
1
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago
1
1
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
1
1
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
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 $f^-1(V)subset C^c$.
answered 42 mins ago
Callus
4,318821
add a comment |Â
up vote
3
down vote
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.
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 $f^-1(V)subset C^c$.
answered 42 mins ago
Callus
4,318821
add a comment |Â
up vote
1
down vote
accepted
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 $f^-1(V)subset C^c$.
answered 42 mins ago
Callus
4,318821
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 $f^-1(V)subset C^c$.
answered 42 mins ago
Callus
4,318821
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 $f^-1(V)subset C^c$.
answered 42 mins ago
Callus
4,318821
answered 42 mins ago
Callus
4,318821
answered 42 mins ago
Callus
4,318821
answered 42 mins ago
Callus
4,318821
4,318821
add a comment |Â
add a comment |Â
up vote
3
down vote
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.
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
add a comment |Â
up vote
3
down vote
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.
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
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.
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
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.
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
answered 2 hours ago
Mohammad Riazi-Kermani
32.2k41853
32.2k41853
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
add a comment |Â
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
Thanks! I meant $f^-1(y) cap M-V=emptyset$ then $ycap f(M-V)=emptyset$!
– duhdave
2 hours ago
add a comment |Â
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1
There appear to be some typos. Could you double check? I can't figure out what the problem is supposed to be.
– Callus
2 hours ago
There was a typo in the problem! I meant $f^-1(y)subset Usubset V$! I cant find any other typo :/
– duhdave
2 hours ago
1
Are you sure this problem is stated correctly? For example, why not just take $U=V$? Maybe you mean "there exists a $Usubset N$ open such that $f^-1(y)subset f^-1(U) subset V$"?
– Callus
1 hour ago
Wow, yes! You're totally right! Thanks! I Edited again :)
– duhdave
58 mins ago