Every ordered field has a countably infinite subset.
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Is the following argument correct?
Theorem. Every ordered field has countably infinite subset.
Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
$$A = {1,1+1,1+1+1,dots}subseteq F$$
To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
$$cdots>1+1+1+1>1+1+1>1+1>1$$
appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.
$blacksquare$
Note:
Definition $textbf{1.1.5}$ is just the standard definition of a field.
Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that
$(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$
$(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$
Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that
$(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.
$(ii)$ If $x<y$ and $y<z$ then $x<z$.
real-analysis proof-verification
asked Nov 24 at 16:01
Atif Farooq
3,1422825
add a comment |
up vote
2
down vote
favorite
Is the following argument correct?
Theorem. Every ordered field has countably infinite subset.
Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
$$A = {1,1+1,1+1+1,dots}subseteq F$$
To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
$$cdots>1+1+1+1>1+1+1>1+1>1$$
appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.
$blacksquare$
Note:
Definition $textbf{1.1.5}$ is just the standard definition of a field.
Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that
$(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$
$(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$
Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that
$(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.
$(ii)$ If $x<y$ and $y<z$ then $x<z$.
real-analysis proof-verification
asked Nov 24 at 16:01
Atif Farooq
3,1422825
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is the following argument correct?
Theorem. Every ordered field has countably infinite subset.
Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
$$A = {1,1+1,1+1+1,dots}subseteq F$$
To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
$$cdots>1+1+1+1>1+1+1>1+1>1$$
appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.
$blacksquare$
Note:
Definition $textbf{1.1.5}$ is just the standard definition of a field.
Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that
$(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$
$(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$
Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that
$(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.
$(ii)$ If $x<y$ and $y<z$ then $x<z$.
real-analysis proof-verification
asked Nov 24 at 16:01
Atif Farooq
3,1422825
Is the following argument correct?
Theorem. Every ordered field has countably infinite subset.
Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
$$A = {1,1+1,1+1+1,dots}subseteq F$$
To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
$$cdots>1+1+1+1>1+1+1>1+1>1$$
appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.
$blacksquare$
Note:
Definition $textbf{1.1.5}$ is just the standard definition of a field.
Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that
$(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$
$(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$
Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that
$(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.
$(ii)$ If $x<y$ and $y<z$ then $x<z$.
real-analysis proof-verification
real-analysis proof-verification
asked Nov 24 at 16:01
Atif Farooq
3,1422825
asked Nov 24 at 16:01
Atif Farooq
3,1422825
asked Nov 24 at 16:01
Atif Farooq
3,1422825
asked Nov 24 at 16:01
Atif Farooq
3,1422825
asked Nov 24 at 16:01
Atif Farooq
3,1422825
3,1422825
add a comment |
add a comment |
1 Answer
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active
oldest
votes
up vote
1
down vote
accepted
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.
answered Nov 24 at 16:25
Isaac Browne
4,60731132
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.
answered Nov 24 at 16:25
Isaac Browne
4,60731132
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
add a comment |
up vote
1
down vote
accepted
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.
answered Nov 24 at 16:25
Isaac Browne
4,60731132
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.
answered Nov 24 at 16:25
Isaac Browne
4,60731132
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.
answered Nov 24 at 16:25
Isaac Browne
4,60731132
answered Nov 24 at 16:25
Isaac Browne
4,60731132
answered Nov 24 at 16:25
Isaac Browne
4,60731132
answered Nov 24 at 16:25
Isaac Browne
4,60731132
4,60731132
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
add a comment |
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
– Atif Farooq
Nov 24 at 16:32
add a comment |
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