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Let $f:mathbb^*Rto mathbb^*R$ be an external function with $(x,yinmathbb^*R,xapprox y , xle y ) implies f(x)le f(y)$. Is $f$ monotonic?
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My main problem is that there's two methods yielding two different results:
1.We can count through $mathbb^*R$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $hinmathbb^*N-N$ steps, creating the sequence $x_i := frac i h$ which fulfills $x_iapprox x_i+1$ for all $i$.
We therefore have the increasing chain $0approx x_0 approx x_1 approx ... approx x_h-1 approx x_h approx 1$, and as such, for every $xin^*[0,1]$ the assessment $xapprox x_i$ holds for some $i$. Therefore, any $xin^*[0,1]$ can be reached in a series of infinitesimal steps, and so $f$ is monotonic in $^*[0,1]$.
This procedure then can easily be extendend to $mathbb^*R$.
2. Given $x,yinmathbb^*R$ with $xapprox y , xle y $ the following is true:
$st(x) = st(y)$, and
monad(x) = monad(st(x)) = monad(y)
I.e. $x$ and $y$ both have the same standard part, and therefore their monads (i.e. the set of points infinitesimally close to them) are identical.
This is in stark contrast to 1. (which should be the correct one), as it says that the premise $(x,yinmathbb^*R,xapprox y , xle y ) implies f(x)le f(y)$ merely says that $f$ is monotonic on any monad, but as each two monads around different standard points are disjunct, we can't extend this argument to conclude that $f$ on all of $^*mathbbR$ is monotonic.
I'm hoping to gain some understanding as to why my argument 2. fails, and how it should be correctly used to form a proof of the question in the title.
nonstandard-analysis
asked Jul 19 at 14:47
Sudix
7911316
 |Â
show 5 more comments
up vote
1
down vote
favorite
My main problem is that there's two methods yielding two different results:
1.We can count through $mathbb^*R$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $hinmathbb^*N-N$ steps, creating the sequence $x_i := frac i h$ which fulfills $x_iapprox x_i+1$ for all $i$.
We therefore have the increasing chain $0approx x_0 approx x_1 approx ... approx x_h-1 approx x_h approx 1$, and as such, for every $xin^*[0,1]$ the assessment $xapprox x_i$ holds for some $i$. Therefore, any $xin^*[0,1]$ can be reached in a series of infinitesimal steps, and so $f$ is monotonic in $^*[0,1]$.
This procedure then can easily be extendend to $mathbb^*R$.
2. Given $x,yinmathbb^*R$ with $xapprox y , xle y $ the following is true:
$st(x) = st(y)$, and
monad(x) = monad(st(x)) = monad(y)
I.e. $x$ and $y$ both have the same standard part, and therefore their monads (i.e. the set of points infinitesimally close to them) are identical.
This is in stark contrast to 1. (which should be the correct one), as it says that the premise $(x,yinmathbb^*R,xapprox y , xle y ) implies f(x)le f(y)$ merely says that $f$ is monotonic on any monad, but as each two monads around different standard points are disjunct, we can't extend this argument to conclude that $f$ on all of $^*mathbbR$ is monotonic.
I'm hoping to gain some understanding as to why my argument 2. fails, and how it should be correctly used to form a proof of the question in the title.
nonstandard-analysis
asked Jul 19 at 14:47
Sudix
7911316
For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
1
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
1
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My main problem is that there's two methods yielding two different results:
1.We can count through $mathbb^*R$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $hinmathbb^*N-N$ steps, creating the sequence $x_i := frac i h$ which fulfills $x_iapprox x_i+1$ for all $i$.
We therefore have the increasing chain $0approx x_0 approx x_1 approx ... approx x_h-1 approx x_h approx 1$, and as such, for every $xin^*[0,1]$ the assessment $xapprox x_i$ holds for some $i$. Therefore, any $xin^*[0,1]$ can be reached in a series of infinitesimal steps, and so $f$ is monotonic in $^*[0,1]$.
This procedure then can easily be extendend to $mathbb^*R$.
2. Given $x,yinmathbb^*R$ with $xapprox y , xle y $ the following is true:
$st(x) = st(y)$, and
monad(x) = monad(st(x)) = monad(y)
I.e. $x$ and $y$ both have the same standard part, and therefore their monads (i.e. the set of points infinitesimally close to them) are identical.
This is in stark contrast to 1. (which should be the correct one), as it says that the premise $(x,yinmathbb^*R,xapprox y , xle y ) implies f(x)le f(y)$ merely says that $f$ is monotonic on any monad, but as each two monads around different standard points are disjunct, we can't extend this argument to conclude that $f$ on all of $^*mathbbR$ is monotonic.
I'm hoping to gain some understanding as to why my argument 2. fails, and how it should be correctly used to form a proof of the question in the title.
nonstandard-analysis
asked Jul 19 at 14:47
Sudix
7911316
My main problem is that there's two methods yielding two different results:
1.We can count through $mathbb^*R$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $hinmathbb^*N-N$ steps, creating the sequence $x_i := frac i h$ which fulfills $x_iapprox x_i+1$ for all $i$.
We therefore have the increasing chain $0approx x_0 approx x_1 approx ... approx x_h-1 approx x_h approx 1$, and as such, for every $xin^*[0,1]$ the assessment $xapprox x_i$ holds for some $i$. Therefore, any $xin^*[0,1]$ can be reached in a series of infinitesimal steps, and so $f$ is monotonic in $^*[0,1]$.
This procedure then can easily be extendend to $mathbb^*R$.
2. Given $x,yinmathbb^*R$ with $xapprox y , xle y $ the following is true:
$st(x) = st(y)$, and
monad(x) = monad(st(x)) = monad(y)
I.e. $x$ and $y$ both have the same standard part, and therefore their monads (i.e. the set of points infinitesimally close to them) are identical.
This is in stark contrast to 1. (which should be the correct one), as it says that the premise $(x,yinmathbb^*R,xapprox y , xle y ) implies f(x)le f(y)$ merely says that $f$ is monotonic on any monad, but as each two monads around different standard points are disjunct, we can't extend this argument to conclude that $f$ on all of $^*mathbbR$ is monotonic.
I'm hoping to gain some understanding as to why my argument 2. fails, and how it should be correctly used to form a proof of the question in the title.
nonstandard-analysis
asked Jul 19 at 14:47
Sudix
7911316
edited Jul 21 at 15:47
asked Jul 19 at 14:47
Sudix
7911316
asked Jul 19 at 14:47
Sudix
7911316
asked Jul 19 at 14:47
Sudix
7911316
7911316
For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
1
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
1
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27
 |Â
show 5 more comments
For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
1
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
1
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27
For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
1
1
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
1
1
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27
 |Â
show 5 more comments
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For 2., note that the set of $nin^*mathbb N$ such that $f(x)le f(y)$ for all $yin[0,1/n]$ has a least element.
– Andrés E. Caicedo
Jul 19 at 14:52
Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $ninmathbbN$ in your set, and therefore we can't prove that there's a least element in the set.
– Sudix
Jul 19 at 15:07
Ah, $f$ is external. Sorry, I had missed that.
– Andrés E. Caicedo
Jul 19 at 16:06
1
Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal).
– Andrés E. Caicedo
Jul 19 at 16:13
1
No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$".
– Andrés E. Caicedo
Jul 19 at 18:27