Existence and uniqueness up to isomorphism of the real numbers from axioms
$begingroup$
Pretty much what the title says: how does one prove the existence and uniqueness of the real number system from the ordered field axioms together with the least-upper-bound property (or maybe some equivalent), instead of using some construction process?
Thanks in advance.
real-analysis elementary-set-theory real-numbers ordered-fields
$endgroup$
add a comment |
$begingroup$
Pretty much what the title says: how does one prove the existence and uniqueness of the real number system from the ordered field axioms together with the least-upper-bound property (or maybe some equivalent), instead of using some construction process?
Thanks in advance.
real-analysis elementary-set-theory real-numbers ordered-fields
$endgroup$
1
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
3
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34
add a comment |
$begingroup$
Pretty much what the title says: how does one prove the existence and uniqueness of the real number system from the ordered field axioms together with the least-upper-bound property (or maybe some equivalent), instead of using some construction process?
Thanks in advance.
real-analysis elementary-set-theory real-numbers ordered-fields
$endgroup$
Pretty much what the title says: how does one prove the existence and uniqueness of the real number system from the ordered field axioms together with the least-upper-bound property (or maybe some equivalent), instead of using some construction process?
Thanks in advance.
real-analysis elementary-set-theory real-numbers ordered-fields
real-analysis elementary-set-theory real-numbers ordered-fields
edited Jul 9 '14 at 9:57
Git Gud
28.9k1050101
28.9k1050101
asked Jul 9 '14 at 9:54
Bruno CantarelliBruno Cantarelli
800823
800823
1
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
3
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34
add a comment |
1
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
3
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34
1
1
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
3
3
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x bigcup {text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $forall x in mathbb{N}, x + 0 = x$
- $forall x in mathbb{N}forall y in mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $forall x in mathbb{N}, x times 0 = 0$
- $forall x in mathbb{N}forall y in mathbb{N}, x times S(y) = (x times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x + y) + z = x + (y + z)$
- $forall x in mathbb{N}forall y in mathbb{N}, x + y = y + x$
- $forall x in mathbb{N}forall y in mathbb{N}, x times y = y times x$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, x times (y + z) = (x times y) + (x times z)$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x times y) times z = x times (y times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $mathbb{Z}$. We can also invent an intuitive definition of + and times and show that it still has those properties and agrees with those operations on $mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $mathbb{Z}$. We can also create an intuitive definition of the relation $leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $times$ and $leq$ and show that $exists xexists y$ such that $(mathbb{R}, x, y, +, times, leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $times$, and $leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f861010%2fexistence-and-uniqueness-up-to-isomorphism-of-the-real-numbers-from-axioms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x bigcup {text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $forall x in mathbb{N}, x + 0 = x$
- $forall x in mathbb{N}forall y in mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $forall x in mathbb{N}, x times 0 = 0$
- $forall x in mathbb{N}forall y in mathbb{N}, x times S(y) = (x times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x + y) + z = x + (y + z)$
- $forall x in mathbb{N}forall y in mathbb{N}, x + y = y + x$
- $forall x in mathbb{N}forall y in mathbb{N}, x times y = y times x$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, x times (y + z) = (x times y) + (x times z)$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x times y) times z = x times (y times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $mathbb{Z}$. We can also invent an intuitive definition of + and times and show that it still has those properties and agrees with those operations on $mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $mathbb{Z}$. We can also create an intuitive definition of the relation $leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $times$ and $leq$ and show that $exists xexists y$ such that $(mathbb{R}, x, y, +, times, leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $times$, and $leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
$endgroup$
add a comment |
$begingroup$
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x bigcup {text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $forall x in mathbb{N}, x + 0 = x$
- $forall x in mathbb{N}forall y in mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $forall x in mathbb{N}, x times 0 = 0$
- $forall x in mathbb{N}forall y in mathbb{N}, x times S(y) = (x times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x + y) + z = x + (y + z)$
- $forall x in mathbb{N}forall y in mathbb{N}, x + y = y + x$
- $forall x in mathbb{N}forall y in mathbb{N}, x times y = y times x$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, x times (y + z) = (x times y) + (x times z)$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x times y) times z = x times (y times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $mathbb{Z}$. We can also invent an intuitive definition of + and times and show that it still has those properties and agrees with those operations on $mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $mathbb{Z}$. We can also create an intuitive definition of the relation $leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $times$ and $leq$ and show that $exists xexists y$ such that $(mathbb{R}, x, y, +, times, leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $times$, and $leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
$endgroup$
add a comment |
$begingroup$
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x bigcup {text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $forall x in mathbb{N}, x + 0 = x$
- $forall x in mathbb{N}forall y in mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $forall x in mathbb{N}, x times 0 = 0$
- $forall x in mathbb{N}forall y in mathbb{N}, x times S(y) = (x times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x + y) + z = x + (y + z)$
- $forall x in mathbb{N}forall y in mathbb{N}, x + y = y + x$
- $forall x in mathbb{N}forall y in mathbb{N}, x times y = y times x$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, x times (y + z) = (x times y) + (x times z)$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x times y) times z = x times (y times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $mathbb{Z}$. We can also invent an intuitive definition of + and times and show that it still has those properties and agrees with those operations on $mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $mathbb{Z}$. We can also create an intuitive definition of the relation $leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $times$ and $leq$ and show that $exists xexists y$ such that $(mathbb{R}, x, y, +, times, leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $times$, and $leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
$endgroup$
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x bigcup {text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $forall x in mathbb{N}, x + 0 = x$
- $forall x in mathbb{N}forall y in mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $forall x in mathbb{N}, x times 0 = 0$
- $forall x in mathbb{N}forall y in mathbb{N}, x times S(y) = (x times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x + y) + z = x + (y + z)$
- $forall x in mathbb{N}forall y in mathbb{N}, x + y = y + x$
- $forall x in mathbb{N}forall y in mathbb{N}, x times y = y times x$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, x times (y + z) = (x times y) + (x times z)$
- $forall x in mathbb{N}forall y in mathbb{N}forall z in mathbb{N}, (x times y) times z = x times (y times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $mathbb{Z}$. We can also invent an intuitive definition of + and times and show that it still has those properties and agrees with those operations on $mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $mathbb{Z}$. We can also create an intuitive definition of the relation $leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $times$ and $leq$ and show that $exists xexists y$ such that $(mathbb{R}, x, y, +, times, leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $times$, and $leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
edited Jan 30 at 0:45
answered Dec 13 '18 at 21:23
TimothyTimothy
326214
326214
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f861010%2fexistence-and-uniqueness-up-to-isomorphism-of-the-real-numbers-from-axioms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
See for example : DJH Garling, A Course in Mathematical Analysis Vol 1 (2013), page 91: para 3.3 The uniqueness of the real number system, for the proof of the theorem that a ordered field with the supremum property is unique up to isomorphism.
$endgroup$
– Mauro ALLEGRANZA
Jul 9 '14 at 11:22
3
$begingroup$
The uniqueness part of this answer may be what you're looking for. For existence you probably won't get away from noting that a particular construction does indeed satisfy the axioms.
$endgroup$
– Henning Makholm
Jul 9 '14 at 14:34