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Will a “continuous†Cauchy sequence construction generate the real numbers?
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There’s two common ways to construct the real number system: Dedekind cuts and equivalence classes of Cauchy sequences. I’d like to try a different way.
Let $X$ be the set of all functions $f:Qrightarrow Q$ such that for any positive rational number $epsilon$ there exists a positive rational number $delta$ such that for all rational numbers $x,y>delta$, we have $|f(x)-f(y)|<epsilon$. (Intuitively, it means that its limit as $x$ goes $infty$ is a real number.) And let’s define an equivalence relation $sim$ on $X$ by saying that $fsim g$ if the limit of $f(x)-g(x)$ as $xrightarrowinfty$ is $0$.
My question is, is the set of equivalence classes of elements of $X$ under $sim$ isomorphic to the set of real numbers?
real-analysis real-numbers cauchy-sequences
edited Aug 2 at 2:12
Michael Hardy
204k23185460
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
add a comment |Â
up vote
2
down vote
favorite
There’s two common ways to construct the real number system: Dedekind cuts and equivalence classes of Cauchy sequences. I’d like to try a different way.
Let $X$ be the set of all functions $f:Qrightarrow Q$ such that for any positive rational number $epsilon$ there exists a positive rational number $delta$ such that for all rational numbers $x,y>delta$, we have $|f(x)-f(y)|<epsilon$. (Intuitively, it means that its limit as $x$ goes $infty$ is a real number.) And let’s define an equivalence relation $sim$ on $X$ by saying that $fsim g$ if the limit of $f(x)-g(x)$ as $xrightarrowinfty$ is $0$.
My question is, is the set of equivalence classes of elements of $X$ under $sim$ isomorphic to the set of real numbers?
real-analysis real-numbers cauchy-sequences
edited Aug 2 at 2:12
Michael Hardy
204k23185460
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
There’s two common ways to construct the real number system: Dedekind cuts and equivalence classes of Cauchy sequences. I’d like to try a different way.
Let $X$ be the set of all functions $f:Qrightarrow Q$ such that for any positive rational number $epsilon$ there exists a positive rational number $delta$ such that for all rational numbers $x,y>delta$, we have $|f(x)-f(y)|<epsilon$. (Intuitively, it means that its limit as $x$ goes $infty$ is a real number.) And let’s define an equivalence relation $sim$ on $X$ by saying that $fsim g$ if the limit of $f(x)-g(x)$ as $xrightarrowinfty$ is $0$.
My question is, is the set of equivalence classes of elements of $X$ under $sim$ isomorphic to the set of real numbers?
real-analysis real-numbers cauchy-sequences
edited Aug 2 at 2:12
Michael Hardy
204k23185460
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
There’s two common ways to construct the real number system: Dedekind cuts and equivalence classes of Cauchy sequences. I’d like to try a different way.
Let $X$ be the set of all functions $f:Qrightarrow Q$ such that for any positive rational number $epsilon$ there exists a positive rational number $delta$ such that for all rational numbers $x,y>delta$, we have $|f(x)-f(y)|<epsilon$. (Intuitively, it means that its limit as $x$ goes $infty$ is a real number.) And let’s define an equivalence relation $sim$ on $X$ by saying that $fsim g$ if the limit of $f(x)-g(x)$ as $xrightarrowinfty$ is $0$.
My question is, is the set of equivalence classes of elements of $X$ under $sim$ isomorphic to the set of real numbers?
real-analysis real-numbers cauchy-sequences
edited Aug 2 at 2:12
Michael Hardy
204k23185460
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
edited Aug 2 at 2:12
Michael Hardy
204k23185460
edited Aug 2 at 2:12
Michael Hardy
204k23185460
edited Aug 2 at 2:12
Michael Hardy
204k23185460
204k23185460
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
asked Aug 1 at 22:44
Keshav Srinivasan
1,74511338
1,74511338
One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13
add a comment |Â
One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13
One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13
One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes. Specifically, let $Y=X/sim$ and define a bijection $varphi:Yto mathbbR$ as follows. Given any $fin X$, define $varphi([f])$ to be the limit of $f(x)$ as $xto infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $xtoinfty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $fsim g$. To see that $varphi$ is a surjection, note that given any $rinmathbbR$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $varphi([f])=r$.
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes. Specifically, let $Y=X/sim$ and define a bijection $varphi:Yto mathbbR$ as follows. Given any $fin X$, define $varphi([f])$ to be the limit of $f(x)$ as $xto infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $xtoinfty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $fsim g$. To see that $varphi$ is a surjection, note that given any $rinmathbbR$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $varphi([f])=r$.
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
add a comment |Â
up vote
2
down vote
accepted
Yes. Specifically, let $Y=X/sim$ and define a bijection $varphi:Yto mathbbR$ as follows. Given any $fin X$, define $varphi([f])$ to be the limit of $f(x)$ as $xto infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $xtoinfty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $fsim g$. To see that $varphi$ is a surjection, note that given any $rinmathbbR$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $varphi([f])=r$.
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes. Specifically, let $Y=X/sim$ and define a bijection $varphi:Yto mathbbR$ as follows. Given any $fin X$, define $varphi([f])$ to be the limit of $f(x)$ as $xto infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $xtoinfty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $fsim g$. To see that $varphi$ is a surjection, note that given any $rinmathbbR$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $varphi([f])=r$.
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
Yes. Specifically, let $Y=X/sim$ and define a bijection $varphi:Yto mathbbR$ as follows. Given any $fin X$, define $varphi([f])$ to be the limit of $f(x)$ as $xto infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $xtoinfty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $fsim g$. To see that $varphi$ is a surjection, note that given any $rinmathbbR$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $varphi([f])=r$.
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
answered Aug 1 at 22:53
Eric Wofsey
161k12188297
161k12188297
add a comment |Â
add a comment |Â
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One should not write $f$~$g$. Proper usage is $fsim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly.
– Michael Hardy
Aug 2 at 2:13