Proof for an equivalence of inverse functions when range of one function contains entire image set of other
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Munkres in his Topology book says
Let $f : X to Y$ be continuous. If $f(X) subset Z subset Y$ , we show that the function $g : X to Z$ obtained from $f$ is continuous. Let $B$ be open in $Z$. Then $B = Z cap U$ for some open set $U$ of $Y$.
Because $Z$ contains the entire image set $f (X)$, $f^−1(U)$ = $g^−1(B)$, by elementary set theory.
How does the last equation follow from set theory?
elementary-set-theory
edited Aug 22 at 10:53
Nash J.
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asked Aug 22 at 10:14
STEMExchanger
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up vote
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down vote
favorite
Munkres in his Topology book says
Let $f : X to Y$ be continuous. If $f(X) subset Z subset Y$ , we show that the function $g : X to Z$ obtained from $f$ is continuous. Let $B$ be open in $Z$. Then $B = Z cap U$ for some open set $U$ of $Y$.
Because $Z$ contains the entire image set $f (X)$, $f^−1(U)$ = $g^−1(B)$, by elementary set theory.
How does the last equation follow from set theory?
elementary-set-theory
edited Aug 22 at 10:53
Nash J.
1,094315
asked Aug 22 at 10:14
STEMExchanger
353
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Munkres in his Topology book says
Let $f : X to Y$ be continuous. If $f(X) subset Z subset Y$ , we show that the function $g : X to Z$ obtained from $f$ is continuous. Let $B$ be open in $Z$. Then $B = Z cap U$ for some open set $U$ of $Y$.
Because $Z$ contains the entire image set $f (X)$, $f^−1(U)$ = $g^−1(B)$, by elementary set theory.
How does the last equation follow from set theory?
elementary-set-theory
edited Aug 22 at 10:53
Nash J.
1,094315
asked Aug 22 at 10:14
STEMExchanger
353
Munkres in his Topology book says
Let $f : X to Y$ be continuous. If $f(X) subset Z subset Y$ , we show that the function $g : X to Z$ obtained from $f$ is continuous. Let $B$ be open in $Z$. Then $B = Z cap U$ for some open set $U$ of $Y$.
Because $Z$ contains the entire image set $f (X)$, $f^−1(U)$ = $g^−1(B)$, by elementary set theory.
How does the last equation follow from set theory?
elementary-set-theory
edited Aug 22 at 10:53
Nash J.
1,094315
asked Aug 22 at 10:14
STEMExchanger
353
edited Aug 22 at 10:53
Nash J.
1,094315
edited Aug 22 at 10:53
Nash J.
1,094315
edited Aug 22 at 10:53
Nash J.
1,094315
1,094315
asked Aug 22 at 10:14
STEMExchanger
353
asked Aug 22 at 10:14
STEMExchanger
353
asked Aug 22 at 10:14
STEMExchanger
353
353
add a comment |Â
add a comment |Â
1 Answer
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1
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You can verify the equality $f^-1(U)=g^-1(B)$ by proving that $$xin f^-1(U)iff xin g^-1(B)$$for arbitrary $xin X$.
That can be interpreted as an application of elementary set theory.
Edit (to make things more clear).
For a fixed $xin X$ the following statements are equivalent:
- $xin f^-1(U)$
- $f(x)in U$
- $f(x)in B=Zcap U$ (this because $f(X)subseteq Z$ so that $f(x)in Z$
- $g(x)in B$ (this because $g(y)=f(y)$ for every $yin X$, also for the fixed $x$)
- $xin g^-1(B)$
answered Aug 22 at 10:18
drhab
88k541120
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You can verify the equality $f^-1(U)=g^-1(B)$ by proving that $$xin f^-1(U)iff xin g^-1(B)$$for arbitrary $xin X$.
That can be interpreted as an application of elementary set theory.
Edit (to make things more clear).
For a fixed $xin X$ the following statements are equivalent:
- $xin f^-1(U)$
- $f(x)in U$
- $f(x)in B=Zcap U$ (this because $f(X)subseteq Z$ so that $f(x)in Z$
- $g(x)in B$ (this because $g(y)=f(y)$ for every $yin X$, also for the fixed $x$)
- $xin g^-1(B)$
answered Aug 22 at 10:18
drhab
88k541120
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
 |Â
show 4 more comments
up vote
1
down vote
You can verify the equality $f^-1(U)=g^-1(B)$ by proving that $$xin f^-1(U)iff xin g^-1(B)$$for arbitrary $xin X$.
That can be interpreted as an application of elementary set theory.
Edit (to make things more clear).
For a fixed $xin X$ the following statements are equivalent:
- $xin f^-1(U)$
- $f(x)in U$
- $f(x)in B=Zcap U$ (this because $f(X)subseteq Z$ so that $f(x)in Z$
- $g(x)in B$ (this because $g(y)=f(y)$ for every $yin X$, also for the fixed $x$)
- $xin g^-1(B)$
answered Aug 22 at 10:18
drhab
88k541120
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
 |Â
show 4 more comments
up vote
1
down vote
up vote
1
down vote
You can verify the equality $f^-1(U)=g^-1(B)$ by proving that $$xin f^-1(U)iff xin g^-1(B)$$for arbitrary $xin X$.
That can be interpreted as an application of elementary set theory.
Edit (to make things more clear).
For a fixed $xin X$ the following statements are equivalent:
- $xin f^-1(U)$
- $f(x)in U$
- $f(x)in B=Zcap U$ (this because $f(X)subseteq Z$ so that $f(x)in Z$
- $g(x)in B$ (this because $g(y)=f(y)$ for every $yin X$, also for the fixed $x$)
- $xin g^-1(B)$
answered Aug 22 at 10:18
drhab
88k541120
You can verify the equality $f^-1(U)=g^-1(B)$ by proving that $$xin f^-1(U)iff xin g^-1(B)$$for arbitrary $xin X$.
That can be interpreted as an application of elementary set theory.
Edit (to make things more clear).
For a fixed $xin X$ the following statements are equivalent:
- $xin f^-1(U)$
- $f(x)in U$
- $f(x)in B=Zcap U$ (this because $f(X)subseteq Z$ so that $f(x)in Z$
- $g(x)in B$ (this because $g(y)=f(y)$ for every $yin X$, also for the fixed $x$)
- $xin g^-1(B)$
answered Aug 22 at 10:18
drhab
88k541120
edited Aug 22 at 11:37
answered Aug 22 at 10:18
drhab
88k541120
answered Aug 22 at 10:18
drhab
88k541120
answered Aug 22 at 10:18
drhab
88k541120
88k541120
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
 |Â
show 4 more comments
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I am trying to proceed as follows. Let x∈$f^−1&(U). Since the image of f is contained in Z and U is contained in image of f we can conclude B = U. Now, since g is an entirely different function i am not able to figure out a way to conclude that x∈$g^−1&(U). Same concern for going the other way around since f here is an entirely different function, i cannot conclude that if x∈$g^−1&(U) it implies x∈$f^−1&(U).
– STEMExchanger
Aug 22 at 11:17
I have added something to my answer.
– drhab
Aug 22 at 11:38
I have added something to my answer.
– drhab
Aug 22 at 11:38
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
The domain of both g and f is X. if g(y)=f(y) for every y∈X, how are they different. What is the impact then of having a different range for g and f?
– STEMExchanger
Aug 22 at 12:13
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
@STEMExchanger If functions are only looked at as sets of ordered pairs then there is no difference between $f$ and $g$. But often functions are looked at as triples $f=(X,G,Y)$ where $X$ is the domain, $Y$ the codomain and $G$ the graphs (i.e. the set $(x,f(x)mid xin X$). Then here $g=(X,G,Z)$ with $Zneq Y$ hence $gneq f$. The impact concerns for a great deal the possibility of composition. If e.g. $h$ is another function with domain $Z$ then the composition $hcirc g$ exists, but the composition $fcirc h$ does not exist.
– drhab
Aug 22 at 12:21
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
Consider the elements of the set Zf(x), is it correct to state then that this set could be nonempty, that for this set, there cannot be preimage in f. (Not onto) Since the range of g is called a restriction, in fact, there could be an even bigger set Yf(x) that does not have a preimage in f. If this is the case, why not just say the co domain is the set f(x) instead of having a set Y that is the co domain which is bigger than f(X)? What is the need/applications of such a definition?
– STEMExchanger
Aug 22 at 13:06
 |Â
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