Mixing
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is...
$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
Consider the spaces to be $Bbb R$
I took $f(x)=$ begin{cases} (x-1) &-1le xle 1\0 & x>1,x<-1end{cases}
and
$g(x)=$ begin{cases} (x-1) &x>1,x<-1\0 & -1le xle 1end{cases}
Though $gcirc f=0$ which is closed and $f$ is not closed since it takes $(-0.5,0.5)$ to $0$ but here $g$ is closed
I want examples where $g,f$ are not closed both. How to make $g$ not closed?
How to answer the second question?
real-analysis general-topology examples-counterexamples open-map closed-map
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
Consider the spaces to be $Bbb R$
I took $f(x)=$ begin{cases} (x-1) &-1le xle 1\0 & x>1,x<-1end{cases}
and
$g(x)=$ begin{cases} (x-1) &x>1,x<-1\0 & -1le xle 1end{cases}
Though $gcirc f=0$ which is closed and $f$ is not closed since it takes $(-0.5,0.5)$ to $0$ but here $g$ is closed
I want examples where $g,f$ are not closed both. How to make $g$ not closed?
How to answer the second question?
real-analysis general-topology examples-counterexamples open-map closed-map
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
$endgroup$
1
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
2
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21
add a comment |
$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
Consider the spaces to be $Bbb R$
I took $f(x)=$ begin{cases} (x-1) &-1le xle 1\0 & x>1,x<-1end{cases}
and
$g(x)=$ begin{cases} (x-1) &x>1,x<-1\0 & -1le xle 1end{cases}
Though $gcirc f=0$ which is closed and $f$ is not closed since it takes $(-0.5,0.5)$ to $0$ but here $g$ is closed
I want examples where $g,f$ are not closed both. How to make $g$ not closed?
How to answer the second question?
real-analysis general-topology examples-counterexamples open-map closed-map
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
$endgroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
Consider the spaces to be $Bbb R$
I took $f(x)=$ begin{cases} (x-1) &-1le xle 1\0 & x>1,x<-1end{cases}
and
$g(x)=$ begin{cases} (x-1) &x>1,x<-1\0 & -1le xle 1end{cases}
Though $gcirc f=0$ which is closed and $f$ is not closed since it takes $(-0.5,0.5)$ to $0$ but here $g$ is closed
I want examples where $g,f$ are not closed both. How to make $g$ not closed?
How to answer the second question?
real-analysis general-topology examples-counterexamples open-map closed-map
real-analysis general-topology examples-counterexamples open-map closed-map
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
edited Feb 2 at 21:21
Alex Ravsky
42.7k32383
42.7k32383
asked Jan 27 at 7:25
user596656
1
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
2
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21
add a comment |
1
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
2
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21
1
1
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
2
2
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21
add a comment |
1 Answer
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$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(Bbb R)=(0,infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $xge -1$ and $-1/x$, otherwise. Since $g(Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)={1}$ for each non-empty closed subset $A$ of $Bbb R$.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $gf:Bbb RtoBbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’inBbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=min gf([x,x’])$ and $M=max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))={y}$, a contradiction again. Thus $gf(x)ne gf(x’)$. Thefore the function $f:Bbb RtoBbb R$ in injective too. By invariance of domain, $f$ is an open map.
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
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$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(Bbb R)=(0,infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $xge -1$ and $-1/x$, otherwise. Since $g(Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)={1}$ for each non-empty closed subset $A$ of $Bbb R$.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $gf:Bbb RtoBbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’inBbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=min gf([x,x’])$ and $M=max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))={y}$, a contradiction again. Thus $gf(x)ne gf(x’)$. Thefore the function $f:Bbb RtoBbb R$ in injective too. By invariance of domain, $f$ is an open map.
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(Bbb R)=(0,infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $xge -1$ and $-1/x$, otherwise. Since $g(Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)={1}$ for each non-empty closed subset $A$ of $Bbb R$.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $gf:Bbb RtoBbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’inBbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=min gf([x,x’])$ and $M=max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))={y}$, a contradiction again. Thus $gf(x)ne gf(x’)$. Thefore the function $f:Bbb RtoBbb R$ in injective too. By invariance of domain, $f$ is an open map.
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(Bbb R)=(0,infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $xge -1$ and $-1/x$, otherwise. Since $g(Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)={1}$ for each non-empty closed subset $A$ of $Bbb R$.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $gf:Bbb RtoBbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’inBbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=min gf([x,x’])$ and $M=max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))={y}$, a contradiction again. Thus $gf(x)ne gf(x’)$. Thefore the function $f:Bbb RtoBbb R$ in injective too. By invariance of domain, $f$ is an open map.
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
Find continuous functions $f,g$ such that $gcirc f $ is closed and continuous but neither $g$ nor $f$ is closed map.
We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(Bbb R)=(0,infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $xge -1$ and $-1/x$, otherwise. Since $g(Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)={1}$ for each non-empty closed subset $A$ of $Bbb R$.
Find continuous functions $f,g$ such that $gcirc f $ is open and continuous but neither $g$ nor $f$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $gf:Bbb RtoBbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’inBbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=min gf([x,x’])$ and $M=max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))={y}$, a contradiction again. Thus $gf(x)ne gf(x’)$. Thefore the function $f:Bbb RtoBbb R$ in injective too. By invariance of domain, $f$ is an open map.
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
answered Feb 2 at 21:20
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
add a comment |
add a comment |
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1
$begingroup$
A little off topic question, but why the measure theory tag?
$endgroup$
– Keen-ameteur
Jan 27 at 7:36
$begingroup$
Does the topology have to be the usual one in $mathbb{R}$ ?
$endgroup$
– Jeffery
Jan 27 at 13:19
2
$begingroup$
Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed.
$endgroup$
– user87690
Jan 27 at 14:21
$begingroup$
Each of your functions $f$ and $g$ is discontinuous at $-1$.
$endgroup$
– Alex Ravsky
Feb 2 at 21:21