Every ordered field has a countably infinite subset.
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
add a comment |
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
add a comment |
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
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
3,1422825
add a comment |
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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.
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
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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 |
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.
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 |
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.
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
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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