Mixing
The complete ordered field is unique up to isomorphism
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The complete ordered field is unique up to isomorphism.
I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{A}=langle A,<',oplus,odot,0',1' rangle$ and $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. It suffices to prove that $mathfrak{R}$ is isomorphic to $mathfrak{A}$. Let $langle C,<',oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$. It is well-known that there exists a unique isomorphism $f:Bbb Q to C$ between $langle Bbb Q,<,+,cdot,0,1 rangle$ and $langle C,<',oplus,odot,0',1' rangle$ and that $C$ is dense in $A$. Let $sup, sup'$ be supremums w.r.t $<$ and $<'$ respectively.
We define a mapping $F:Bbb R to A$ by $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ for all $xinBbb R$.
$F$ is well-defined
$Bbb R$ has no endpoints and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_0,q_0in Bbb Q$ such that $q_0<x<p_0$ $implies { pin Bbb Q mid p<x}$ is nonempty and bounded from above by $p_0inBbb Q$. Then ${f(p) mid pin Bbb Q text{ and } p<x}$ is nonempty and bounded from above by $f(p_0)$. Hence $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ exists.
$forall xin Bbb Q:F(x)=f(x)$
$begin{align}forall xin Bbb Q:F(x)&= sup' {f(p) mid pin Bbb Q text{ and } p<x}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } f(p)<'f(x)}\ &= sup' {c mid cin C text{ and } c<'c'},space c:=f(p) text{ and } c':=f(x)\ &= c'\ &=f(x)end{align}$
- $forall x,y in Bbb R: x<y iff F(x)<' F(y)$
Since $<$ and $<'$ are linear orders, it suffices to prove that $forall x,y in Bbb R: x<y implies F(x)<' F(y)$.
$x,y in Bbb R$ , $x<y$ , and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_1,p_2in Bbb Q$ such that $x<p_1<p_2<y$.
First, $p_1<p_2$ and $p_1,p_2in Bbb Q implies f(p_1) <' f(p_2) implies F(p_1) <' F(p_2)$. Second, $x<p_1 implies {f(p) mid pin Bbb Q text{ and } p<x} subsetneq {f(p) mid pin Bbb Q text{ and } p<p_1} implies$ $sup'{f(p) mid pin Bbb Q text{ and } p<x} le' sup'{f(p) mid pin Bbb Q text{ and } p<p_1} implies F(x)le' F(p_1)$.
Similarly, $F(p_2)le' F(y)$. Hence $F(x)le' F(p_1)<'F(p_2)le' F(y)$ and thus $F(x)<' F(y)$.
$F$ is surjective
For $ain A$, let $X'={cin C mid c<'a} subseteq C$. Since $C$ is dense in $A$, $sup'X'=a$. Let $X=f^{-1}[X']$. Then $Xsubseteq Bbb Q$ and $f[X]=X'$.
$X'$ is nonempty bounded from above and $f$ is an isomorphism between $Bbb Q$ and $C$ $implies$ $X'$ is nonempty bounded from above. Let $x=sup X in Bbb R$. Next we prove $F(x)=a$.
$begin{align} F(x)&=sup' {f(p) mid pin Bbb Q text{ and } p<x} \ &= sup' {f(p) mid pin Bbb Q text{ and } p<sup X}\ &= sup' {f(p) mid pin Bbb Q text{ and } exists p'in X:p<p'}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } exists f(p')in f[X]:f(p)<'f(p')}\ &= sup' {f(p) mid f(p)in C text{ and } exists f(p')in X':f(p)<'f(p')}\ &= sup' {c mid cin C text{ and } exists c'in X':c<'c'},space c:=f(p) text{ and }c':=f(p')\ &= sup' {c mid cin C text{ and } c<'sup' X'}\ &= sup' {c mid cin C text{ and } c <' a},space sup' X'=a\ &= aend{align}$
- $forall x,y in Bbb R: F(x+y) = F(x) oplus F(y)$
By definition, $x+y = sup {p+q mid p,qinBbb Q text{ and } p<x text{ and } q<y}$. Furthermore, $rin Bbb Q$ and $r < x+y iff$ $rin Bbb Q$ and $r<p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y iff$ $r=p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' B$
$begin{align}F(x+y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x+y} \&= sup' {f(p+q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}\&=sup' {f(p) oplus f(q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}end{align}$
It follows that $F(x+y)=sup ' (A+B)$ and thus $F(x+y)=F(x) oplus F(y)$.
- $forall x,y in Bbb R: F(x cdot y) = F(x) odot F(y)$
WLOG, we assume $x>0$ and $y>0$.
By definition, $x cdot y = sup {p cdot q mid p,qinBbb Q text{ and } 0<p<x text{ and } 0<q<y}$. Furthermore, $rin Bbb Q$ and $0 < r < x cdot y iff$ $rin Bbb Q$ and $0 < r <p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y iff$ $r=p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' {f(p) mid pin Bbb Q text{ and } 0<p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' {f(q) mid qin Bbb Q text{ and } 0<q<y}=sup' B$
$begin{align}F(x cdot y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x cdot y} \&=sup' {f(r) mid rin Bbb Q text{ and } 0<r<x cdot y} \&= sup' {f(p cdot q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}\&=sup' {f(p) odot f(q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}end{align}$
It follows that $F(x cdot y)=sup ' (A cdot B)$ and thus $F(x cdot y)=F(x) odot F(y)$.
proof-verification ordered-fields
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
add a comment |
$begingroup$
The complete ordered field is unique up to isomorphism.
I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{A}=langle A,<',oplus,odot,0',1' rangle$ and $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. It suffices to prove that $mathfrak{R}$ is isomorphic to $mathfrak{A}$. Let $langle C,<',oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$. It is well-known that there exists a unique isomorphism $f:Bbb Q to C$ between $langle Bbb Q,<,+,cdot,0,1 rangle$ and $langle C,<',oplus,odot,0',1' rangle$ and that $C$ is dense in $A$. Let $sup, sup'$ be supremums w.r.t $<$ and $<'$ respectively.
We define a mapping $F:Bbb R to A$ by $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ for all $xinBbb R$.
$F$ is well-defined
$Bbb R$ has no endpoints and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_0,q_0in Bbb Q$ such that $q_0<x<p_0$ $implies { pin Bbb Q mid p<x}$ is nonempty and bounded from above by $p_0inBbb Q$. Then ${f(p) mid pin Bbb Q text{ and } p<x}$ is nonempty and bounded from above by $f(p_0)$. Hence $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ exists.
$forall xin Bbb Q:F(x)=f(x)$
$begin{align}forall xin Bbb Q:F(x)&= sup' {f(p) mid pin Bbb Q text{ and } p<x}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } f(p)<'f(x)}\ &= sup' {c mid cin C text{ and } c<'c'},space c:=f(p) text{ and } c':=f(x)\ &= c'\ &=f(x)end{align}$
- $forall x,y in Bbb R: x<y iff F(x)<' F(y)$
Since $<$ and $<'$ are linear orders, it suffices to prove that $forall x,y in Bbb R: x<y implies F(x)<' F(y)$.
$x,y in Bbb R$ , $x<y$ , and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_1,p_2in Bbb Q$ such that $x<p_1<p_2<y$.
First, $p_1<p_2$ and $p_1,p_2in Bbb Q implies f(p_1) <' f(p_2) implies F(p_1) <' F(p_2)$. Second, $x<p_1 implies {f(p) mid pin Bbb Q text{ and } p<x} subsetneq {f(p) mid pin Bbb Q text{ and } p<p_1} implies$ $sup'{f(p) mid pin Bbb Q text{ and } p<x} le' sup'{f(p) mid pin Bbb Q text{ and } p<p_1} implies F(x)le' F(p_1)$.
Similarly, $F(p_2)le' F(y)$. Hence $F(x)le' F(p_1)<'F(p_2)le' F(y)$ and thus $F(x)<' F(y)$.
$F$ is surjective
For $ain A$, let $X'={cin C mid c<'a} subseteq C$. Since $C$ is dense in $A$, $sup'X'=a$. Let $X=f^{-1}[X']$. Then $Xsubseteq Bbb Q$ and $f[X]=X'$.
$X'$ is nonempty bounded from above and $f$ is an isomorphism between $Bbb Q$ and $C$ $implies$ $X'$ is nonempty bounded from above. Let $x=sup X in Bbb R$. Next we prove $F(x)=a$.
$begin{align} F(x)&=sup' {f(p) mid pin Bbb Q text{ and } p<x} \ &= sup' {f(p) mid pin Bbb Q text{ and } p<sup X}\ &= sup' {f(p) mid pin Bbb Q text{ and } exists p'in X:p<p'}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } exists f(p')in f[X]:f(p)<'f(p')}\ &= sup' {f(p) mid f(p)in C text{ and } exists f(p')in X':f(p)<'f(p')}\ &= sup' {c mid cin C text{ and } exists c'in X':c<'c'},space c:=f(p) text{ and }c':=f(p')\ &= sup' {c mid cin C text{ and } c<'sup' X'}\ &= sup' {c mid cin C text{ and } c <' a},space sup' X'=a\ &= aend{align}$
- $forall x,y in Bbb R: F(x+y) = F(x) oplus F(y)$
By definition, $x+y = sup {p+q mid p,qinBbb Q text{ and } p<x text{ and } q<y}$. Furthermore, $rin Bbb Q$ and $r < x+y iff$ $rin Bbb Q$ and $r<p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y iff$ $r=p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' B$
$begin{align}F(x+y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x+y} \&= sup' {f(p+q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}\&=sup' {f(p) oplus f(q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}end{align}$
It follows that $F(x+y)=sup ' (A+B)$ and thus $F(x+y)=F(x) oplus F(y)$.
- $forall x,y in Bbb R: F(x cdot y) = F(x) odot F(y)$
WLOG, we assume $x>0$ and $y>0$.
By definition, $x cdot y = sup {p cdot q mid p,qinBbb Q text{ and } 0<p<x text{ and } 0<q<y}$. Furthermore, $rin Bbb Q$ and $0 < r < x cdot y iff$ $rin Bbb Q$ and $0 < r <p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y iff$ $r=p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' {f(p) mid pin Bbb Q text{ and } 0<p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' {f(q) mid qin Bbb Q text{ and } 0<q<y}=sup' B$
$begin{align}F(x cdot y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x cdot y} \&=sup' {f(r) mid rin Bbb Q text{ and } 0<r<x cdot y} \&= sup' {f(p cdot q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}\&=sup' {f(p) odot f(q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}end{align}$
It follows that $F(x cdot y)=sup ' (A cdot B)$ and thus $F(x cdot y)=F(x) odot F(y)$.
proof-verification ordered-fields
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
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– reuns
Jan 22 at 14:00
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Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
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– Le Anh Dung
Jan 22 at 14:04
1
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@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
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– Max
Jan 28 at 10:10
add a comment |
$begingroup$
The complete ordered field is unique up to isomorphism.
I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{A}=langle A,<',oplus,odot,0',1' rangle$ and $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. It suffices to prove that $mathfrak{R}$ is isomorphic to $mathfrak{A}$. Let $langle C,<',oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$. It is well-known that there exists a unique isomorphism $f:Bbb Q to C$ between $langle Bbb Q,<,+,cdot,0,1 rangle$ and $langle C,<',oplus,odot,0',1' rangle$ and that $C$ is dense in $A$. Let $sup, sup'$ be supremums w.r.t $<$ and $<'$ respectively.
We define a mapping $F:Bbb R to A$ by $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ for all $xinBbb R$.
$F$ is well-defined
$Bbb R$ has no endpoints and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_0,q_0in Bbb Q$ such that $q_0<x<p_0$ $implies { pin Bbb Q mid p<x}$ is nonempty and bounded from above by $p_0inBbb Q$. Then ${f(p) mid pin Bbb Q text{ and } p<x}$ is nonempty and bounded from above by $f(p_0)$. Hence $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ exists.
$forall xin Bbb Q:F(x)=f(x)$
$begin{align}forall xin Bbb Q:F(x)&= sup' {f(p) mid pin Bbb Q text{ and } p<x}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } f(p)<'f(x)}\ &= sup' {c mid cin C text{ and } c<'c'},space c:=f(p) text{ and } c':=f(x)\ &= c'\ &=f(x)end{align}$
- $forall x,y in Bbb R: x<y iff F(x)<' F(y)$
Since $<$ and $<'$ are linear orders, it suffices to prove that $forall x,y in Bbb R: x<y implies F(x)<' F(y)$.
$x,y in Bbb R$ , $x<y$ , and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_1,p_2in Bbb Q$ such that $x<p_1<p_2<y$.
First, $p_1<p_2$ and $p_1,p_2in Bbb Q implies f(p_1) <' f(p_2) implies F(p_1) <' F(p_2)$. Second, $x<p_1 implies {f(p) mid pin Bbb Q text{ and } p<x} subsetneq {f(p) mid pin Bbb Q text{ and } p<p_1} implies$ $sup'{f(p) mid pin Bbb Q text{ and } p<x} le' sup'{f(p) mid pin Bbb Q text{ and } p<p_1} implies F(x)le' F(p_1)$.
Similarly, $F(p_2)le' F(y)$. Hence $F(x)le' F(p_1)<'F(p_2)le' F(y)$ and thus $F(x)<' F(y)$.
$F$ is surjective
For $ain A$, let $X'={cin C mid c<'a} subseteq C$. Since $C$ is dense in $A$, $sup'X'=a$. Let $X=f^{-1}[X']$. Then $Xsubseteq Bbb Q$ and $f[X]=X'$.
$X'$ is nonempty bounded from above and $f$ is an isomorphism between $Bbb Q$ and $C$ $implies$ $X'$ is nonempty bounded from above. Let $x=sup X in Bbb R$. Next we prove $F(x)=a$.
$begin{align} F(x)&=sup' {f(p) mid pin Bbb Q text{ and } p<x} \ &= sup' {f(p) mid pin Bbb Q text{ and } p<sup X}\ &= sup' {f(p) mid pin Bbb Q text{ and } exists p'in X:p<p'}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } exists f(p')in f[X]:f(p)<'f(p')}\ &= sup' {f(p) mid f(p)in C text{ and } exists f(p')in X':f(p)<'f(p')}\ &= sup' {c mid cin C text{ and } exists c'in X':c<'c'},space c:=f(p) text{ and }c':=f(p')\ &= sup' {c mid cin C text{ and } c<'sup' X'}\ &= sup' {c mid cin C text{ and } c <' a},space sup' X'=a\ &= aend{align}$
- $forall x,y in Bbb R: F(x+y) = F(x) oplus F(y)$
By definition, $x+y = sup {p+q mid p,qinBbb Q text{ and } p<x text{ and } q<y}$. Furthermore, $rin Bbb Q$ and $r < x+y iff$ $rin Bbb Q$ and $r<p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y iff$ $r=p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' B$
$begin{align}F(x+y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x+y} \&= sup' {f(p+q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}\&=sup' {f(p) oplus f(q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}end{align}$
It follows that $F(x+y)=sup ' (A+B)$ and thus $F(x+y)=F(x) oplus F(y)$.
- $forall x,y in Bbb R: F(x cdot y) = F(x) odot F(y)$
WLOG, we assume $x>0$ and $y>0$.
By definition, $x cdot y = sup {p cdot q mid p,qinBbb Q text{ and } 0<p<x text{ and } 0<q<y}$. Furthermore, $rin Bbb Q$ and $0 < r < x cdot y iff$ $rin Bbb Q$ and $0 < r <p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y iff$ $r=p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' {f(p) mid pin Bbb Q text{ and } 0<p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' {f(q) mid qin Bbb Q text{ and } 0<q<y}=sup' B$
$begin{align}F(x cdot y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x cdot y} \&=sup' {f(r) mid rin Bbb Q text{ and } 0<r<x cdot y} \&= sup' {f(p cdot q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}\&=sup' {f(p) odot f(q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}end{align}$
It follows that $F(x cdot y)=sup ' (A cdot B)$ and thus $F(x cdot y)=F(x) odot F(y)$.
proof-verification ordered-fields
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
The complete ordered field is unique up to isomorphism.
I am quite sure that Parts 1,2,3 are correct. Could you please help me verify Parts 4,5,6 at the end of the proof. Do Parts 4,5,6 look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{A}=langle A,<',oplus,odot,0',1' rangle$ and $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. It suffices to prove that $mathfrak{R}$ is isomorphic to $mathfrak{A}$. Let $langle C,<',oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$. It is well-known that there exists a unique isomorphism $f:Bbb Q to C$ between $langle Bbb Q,<,+,cdot,0,1 rangle$ and $langle C,<',oplus,odot,0',1' rangle$ and that $C$ is dense in $A$. Let $sup, sup'$ be supremums w.r.t $<$ and $<'$ respectively.
We define a mapping $F:Bbb R to A$ by $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ for all $xinBbb R$.
$F$ is well-defined
$Bbb R$ has no endpoints and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_0,q_0in Bbb Q$ such that $q_0<x<p_0$ $implies { pin Bbb Q mid p<x}$ is nonempty and bounded from above by $p_0inBbb Q$. Then ${f(p) mid pin Bbb Q text{ and } p<x}$ is nonempty and bounded from above by $f(p_0)$. Hence $F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}$ exists.
$forall xin Bbb Q:F(x)=f(x)$
$begin{align}forall xin Bbb Q:F(x)&= sup' {f(p) mid pin Bbb Q text{ and } p<x}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } f(p)<'f(x)}\ &= sup' {c mid cin C text{ and } c<'c'},space c:=f(p) text{ and } c':=f(x)\ &= c'\ &=f(x)end{align}$
- $forall x,y in Bbb R: x<y iff F(x)<' F(y)$
Since $<$ and $<'$ are linear orders, it suffices to prove that $forall x,y in Bbb R: x<y implies F(x)<' F(y)$.
$x,y in Bbb R$ , $x<y$ , and $Bbb Q$ is dense in $Bbb R$ $implies$ there exist $p_1,p_2in Bbb Q$ such that $x<p_1<p_2<y$.
First, $p_1<p_2$ and $p_1,p_2in Bbb Q implies f(p_1) <' f(p_2) implies F(p_1) <' F(p_2)$. Second, $x<p_1 implies {f(p) mid pin Bbb Q text{ and } p<x} subsetneq {f(p) mid pin Bbb Q text{ and } p<p_1} implies$ $sup'{f(p) mid pin Bbb Q text{ and } p<x} le' sup'{f(p) mid pin Bbb Q text{ and } p<p_1} implies F(x)le' F(p_1)$.
Similarly, $F(p_2)le' F(y)$. Hence $F(x)le' F(p_1)<'F(p_2)le' F(y)$ and thus $F(x)<' F(y)$.
$F$ is surjective
For $ain A$, let $X'={cin C mid c<'a} subseteq C$. Since $C$ is dense in $A$, $sup'X'=a$. Let $X=f^{-1}[X']$. Then $Xsubseteq Bbb Q$ and $f[X]=X'$.
$X'$ is nonempty bounded from above and $f$ is an isomorphism between $Bbb Q$ and $C$ $implies$ $X'$ is nonempty bounded from above. Let $x=sup X in Bbb R$. Next we prove $F(x)=a$.
$begin{align} F(x)&=sup' {f(p) mid pin Bbb Q text{ and } p<x} \ &= sup' {f(p) mid pin Bbb Q text{ and } p<sup X}\ &= sup' {f(p) mid pin Bbb Q text{ and } exists p'in X:p<p'}\ &= sup' {f(p) mid f(p)in f[Bbb Q] text{ and } exists f(p')in f[X]:f(p)<'f(p')}\ &= sup' {f(p) mid f(p)in C text{ and } exists f(p')in X':f(p)<'f(p')}\ &= sup' {c mid cin C text{ and } exists c'in X':c<'c'},space c:=f(p) text{ and }c':=f(p')\ &= sup' {c mid cin C text{ and } c<'sup' X'}\ &= sup' {c mid cin C text{ and } c <' a},space sup' X'=a\ &= aend{align}$
- $forall x,y in Bbb R: F(x+y) = F(x) oplus F(y)$
By definition, $x+y = sup {p+q mid p,qinBbb Q text{ and } p<x text{ and } q<y}$. Furthermore, $rin Bbb Q$ and $r < x+y iff$ $rin Bbb Q$ and $r<p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y iff$ $r=p+q$ for some $p,qinBbb Q$ such that $p<x$ and $q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' B$
$begin{align}F(x+y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x+y} \&= sup' {f(p+q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}\&=sup' {f(p) oplus f(q) mid p,qin Bbb Q text{ and } p<x text{ and } q<y}end{align}$
It follows that $F(x+y)=sup ' (A+B)$ and thus $F(x+y)=F(x) oplus F(y)$.
- $forall x,y in Bbb R: F(x cdot y) = F(x) odot F(y)$
WLOG, we assume $x>0$ and $y>0$.
By definition, $x cdot y = sup {p cdot q mid p,qinBbb Q text{ and } 0<p<x text{ and } 0<q<y}$. Furthermore, $rin Bbb Q$ and $0 < r < x cdot y iff$ $rin Bbb Q$ and $0 < r <p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y iff$ $r=p cdot q$ for some $p,qinBbb Q$ such that $0<p<x$ and $0<q<y$.
$F(x)=sup' {f(p) mid pin Bbb Q text{ and } p<x}=sup' {f(p) mid pin Bbb Q text{ and } 0<p<x}=sup' A$
$F(y)=sup' {f(q) mid qin Bbb Q text{ and } q<y}=sup' {f(q) mid qin Bbb Q text{ and } 0<q<y}=sup' B$
$begin{align}F(x cdot y)&=sup' {f(r) mid rin Bbb Q text{ and } r<x cdot y} \&=sup' {f(r) mid rin Bbb Q text{ and } 0<r<x cdot y} \&= sup' {f(p cdot q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}\&=sup' {f(p) odot f(q) mid p,qin Bbb Q text{ and } 0<p<x text{ and } 0<q<y}end{align}$
It follows that $F(x cdot y)=sup ' (A cdot B)$ and thus $F(x cdot y)=F(x) odot F(y)$.
proof-verification ordered-fields
proof-verification ordered-fields
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
asked Jan 15 at 3:55
Le Anh DungLe Anh Dung
1,1921621
1,1921621
$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
$endgroup$
– reuns
Jan 22 at 14:00
$begingroup$
Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
$endgroup$
– Le Anh Dung
Jan 22 at 14:04
1
$begingroup$
@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
$endgroup$
– Max
Jan 28 at 10:10
add a comment |
$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
$endgroup$
– reuns
Jan 22 at 14:00
$begingroup$
Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
$endgroup$
– Le Anh Dung
Jan 22 at 14:04
1
$begingroup$
@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
$endgroup$
– Max
Jan 28 at 10:10
$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
$endgroup$
– reuns
Jan 22 at 14:00
$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
$endgroup$
– reuns
Jan 22 at 14:00
$begingroup$
Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
$endgroup$
– Le Anh Dung
Jan 22 at 14:04
$begingroup$
Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
$endgroup$
– Le Anh Dung
Jan 22 at 14:04
1
1
$begingroup$
@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
$endgroup$
– Max
Jan 28 at 10:10
$begingroup$
@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
$endgroup$
– Max
Jan 28 at 10:10
add a comment |
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$begingroup$
How do you define complete, how do you exclude your field being $mathbb{R}(x),<$ or the surreal numbers ? If $mathbb{Q},<$ is dense in $K,<$ and the latter has the Least-upper-bound property then $K,<$ is isomorphic to $mathbb{R},<$.
$endgroup$
– reuns
Jan 22 at 14:00
$begingroup$
Hi @reuns, from my textbook: Let $(P, <)$ be a dense linearly ordered set. $P$ is complete if every nonempty $S subseteq P$ bounded from above has a supremum. Since I am just exposed to introductory set theory, I have not learned surreal numbers yet.
$endgroup$
– Le Anh Dung
Jan 22 at 14:04
1
$begingroup$
@reuns : $mathbb{R}(x)$ is not complete : $mathbb{R}$ is bounded in it (by $x$ for instance) but has no least upper bound (such a l.u.b. must be smaller than $x$ so have degree at most $1$, can't be real, so have degree exactly $1$, but $ax+b-1<ax+b$ is an upper bound whenever $ax+b$ is). So it suffices that $K$ is complete, density of $mathbb{Q}$ will follow
$endgroup$
– Max
Jan 28 at 10:10