Extension of order-preserving bijection from rationals to reals.
$begingroup$
If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.
Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.
Next we would need to prove such an extension is continuous and continuous inverse?
real-analysis order-theory
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
$endgroup$
add a comment |
$begingroup$
If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.
Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.
Next we would need to prove such an extension is continuous and continuous inverse?
real-analysis order-theory
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
$endgroup$
2
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07
add a comment |
$begingroup$
If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.
Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.
Next we would need to prove such an extension is continuous and continuous inverse?
real-analysis order-theory
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
$endgroup$
If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.
Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.
Next we would need to prove such an extension is continuous and continuous inverse?
real-analysis order-theory
real-analysis order-theory
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
edited Dec 19 '18 at 19:52
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Dec 19 '18 at 18:02
ershersh
406113
asked Dec 19 '18 at 18:02
ershersh
406113
asked Dec 19 '18 at 18:02
ershersh
406113
406113
2
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07
add a comment |
2
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07
2
2
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.
Here is another approach using Lord Shark the Unknown natural idea.
Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then
$Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.
$F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$
$F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.
Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.
As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.
Here is another approach using Lord Shark the Unknown natural idea.
Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then
$Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.
$F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$
$F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.
Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.
As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
$endgroup$
add a comment |
$begingroup$
My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.
Here is another approach using Lord Shark the Unknown natural idea.
Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then
$Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.
$F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$
$F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.
Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.
As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
$endgroup$
add a comment |
$begingroup$
My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.
Here is another approach using Lord Shark the Unknown natural idea.
Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then
$Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.
$F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$
$F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.
Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.
As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
$endgroup$
My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.
Here is another approach using Lord Shark the Unknown natural idea.
Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then
$Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.
$F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$
$F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.
Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.
As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
edited Dec 19 '18 at 19:33
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
answered Dec 19 '18 at 19:15
yamete kudasaiyamete kudasai
1,090818
1,090818
add a comment |
add a comment |
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2
$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07