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From: Kaba on 17 Jan 2010 06:29
Arturo Magidin wrote:
> Now, for the OP:
>
> Since f must send positive numbers to positive numbers, it must
> respect order: let r and s be real numbers. Then r<s if and only if s-
> r>0 if and only if f(s-r) > 0 (the converse follows from the
> argument above by considering f^{-1}) if and only if f(s)-f(r)>0, if
> and only if f(r)<f(s).
>
> To show your f is the identity, let x be an arbitrary real number; try
> finding monotonic sequences over which you have control that converge
> to x, and see what happens to their images.

Let d be a metric in R. Pick x in R and find a convergent sequence (x_n)
in (R, d) such that

x_n in Q and
lim_{n->oo} x_n = x

A convergent sequence is also a Cauchy sequence. Thus:
for all epsilon > 0: exists N: for all n, m > N:
d(x_n, x_m) < epsilon

I already proved
for all q in Q: f(q) = q

Thus (f(x_n)) is also a Cauchy sequence in (R, d). By completeness of R,
(f(x_n)) converges to some value s. I would now have to see that
actually

s = lim_{n->oo} f(x_n) = f(x)

Find a non-decreasing sequence (x_n) that converges to x. Then:

for all epsilon > 0: exists x_eps in Q:
x - epsilon < x_eps < x

Because (x_n) is monotonically convergent to x:
exists N: for all n > N: x_n >= x_eps
=> (for all q in Q: f(q) = q)
exists N: for all n > N: f(x_n) >= f(x_eps)
=> (f preserves order)
lim_{n->oo} f(x_n) >= f(x_eps)
=> (epsilon arbitrary)
lim_{n->oo} f(x_n) >= f(x)

However, because of monotonicity of (x_n) we also know
x_n <= x
=> f is order preserving
f(x_n) <= f(x)
=>
lim_{n->oo} f(x_n) <= f(x)

Thus by totality
lim_{n->oo} f(x_n) = f(x)

Then
f(x) = lim_{n->oo} f(x_n) = lim_{n->oo} x_n = x

Is that ok?

--
http://kaba.hilvi.org
From: Arturo Magidin on 18 Jan 2010 00:05
On Jan 17, 5:29 am, Kaba <n...(a)here.com> wrote:
> Arturo Magidin wrote:
> > Now, for the OP:
>
> > Since f must send positive numbers to positive numbers, it must
> > respect order: let r and s be real numbers. Then r<s if and only if s-
> > r>0 if and only if f(s-r) > 0   (the converse follows from the
> > argument above by considering f^{-1}) if and only if f(s)-f(r)>0, if
> > and only if f(r)<f(s).
>
> > To show your f is the identity, let x be an arbitrary real number; try
> > finding monotonic sequences over which you have control that converge
> > to x, and see what happens to their images.
>
> Let d be a metric in R.

You don't get to pick a metric here. Use the standard distance metric.

> Pick x in R and find a convergent sequence (x_n)
> in (R, d) such that

Do *I* have to find one, or should you, the one who wants to prove the
theorem, find it? And, since your metric was just *some* metric, what
makes you think that you can find such a sequence? I can come up with
plenty of metrics on R for which such a thing does not exist (e.g.,
the discrete metric).

> x_n in Q and
> lim_{n->oo} x_n = x
>
> A convergent sequence is also a Cauchy sequence. Thus:
> for all epsilon > 0: exists N: for all n, m > N:
> d(x_n, x_m) < epsilon
>
> I already proved
> for all q in Q: f(q) = q
>
> Thus (f(x_n)) is also a Cauchy sequence in (R, d). By completeness of R,

Again: since you simply said that d was *a* metric on R, you have no
grounds for claiming that R is complete *with respect to d*. So this
step is a non-sequitur because of how you set things up.

You picked *a* metric on R.

> (f(x_n)) converges to some value s. I would now have to see that
> actually
>
> s = lim_{n->oo} f(x_n) = f(x)

You have no grounds for claiming that the limit of the images of the
x_i is the image of x. Nor did you use the fact that the sequence was
increasing. So why did you define it as increasing?

>
> Find a non-decreasing sequence (x_n) that converges to x. Then:
>
> for all epsilon > 0: exists x_eps in Q:
> x - epsilon < x_eps < x
>
> Because (x_n) is monotonically convergent to x:
> exists N: for all n > N: x_n >= x_eps

What is "x_eps"?

> => (for all q in Q: f(q) = q)
> exists N: for all n > N: f(x_n) >= f(x_eps)
> => (f preserves order)
> lim_{n->oo} f(x_n) >= f(x_eps)
> => (epsilon arbitrary)
> lim_{n->oo} f(x_n) >= f(x)
>
> However, because of monotonicity of (x_n) we also know
> x_n <= x
> => f is order preserving
> f(x_n) <= f(x)
> =>
> lim_{n->oo} f(x_n) <= f(x)
>
> Thus by totality
> lim_{n->oo} f(x_n) = f(x)
>
> Then
> f(x) = lim_{n->oo} f(x_n) = lim_{n->oo} x_n = x
>
> Is that ok?

It is utterly confused, beginning with your idea of picking "a" metric
for R, and overly complicated.

Take x, and let x_n and y_n be increasing and decreasing sequences of
rationals, respectively, that converge to x. Then x_n <= x <= y_n for
all n, hence since f respects order, f(x_n) = x_n <= f(x) <= f(y_n) =
y_n for all n (since f(q) = q for all rationals q). Since this holds
for all n, you have that x = lim(x_n) <= lim(f(x)) <= lim(y_n) = x.

--
Arturo Magidin
From: William Elliot on 18 Jan 2010 03:44
On Sat, 16 Jan 2010, Kaba wrote:

> (Question at the end where the proof stumps on)
>
> Let
> * R be the set of real numbers,
> * Q be the set of rational numbers,
> * N be the set of natural numbers,
> * Z be the set of integers, and
> * f be a field automorphism in R.
>
You may weaken f to
for all x,y, f(xy) = f(x).f(y)
some r with f(r) /= 0
for all x, f(x + r) = f(x).

After you've shown f|Q = id_Q, ie f restricted to Q is the identity
map, show f is an increasing surjection, hence continuous and being
the identity on a dense set, f is the identity for the whole set.

> Claim:
> f is the identity function: f(x) = x
>
> Proof:
> 1) for all x:
> f(x) = f(1 * x) = f(1) * f(x)
> => f(1) = 1
>
> 2) for all x:
> f(x) = f(x + 0) = f(x) + f(0)
> => f(0) = 0
>
> 3) f(2) = f(1 + 1) = f(1) + f(1) = 2
> By induction, for all n in N: f(n) = n
>
> 4) for all x:
> 0 = f(0) = f(x + (-x)) = f(x) + f(-x)
> => f(-x) = -f(x)
>
> 5) for all z in Z, z < 0:
> f(-z) = -f(z) = -z
>
> 6) for all x != 0:
> 1 = f(1) = f(x / x) = f(x) * f(1 / x)
> => f(1 / x) = 1 / f(x)
>
> 7) for all q = (n / m) in Q:
> (n, m in Z, m != 0)
> f(n / m) = f(n) * f(1 / m) = f(n) / f(m) = n / m
>
> How do I prove f(x) = x for real numbers?
>
> --
> http://kaba.hilvi.org
>
From: Kaba on 18 Jan 2010 04:50
Arturo Magidin wrote:
> > Let d be a metric in R.
>
> You don't get to pick a metric here. Use the standard distance metric.

Right, I thought I could generalize the metric but I can't. Let d be the
standard metric.

> > Pick x in R and find a convergent sequence (x_n)
> > in (R, d) such that
>
> Do *I* have to find one, or should you, the one who wants to prove the
> theorem, find it? And, since your metric was just *some* metric, what
> makes you think that you can find such a sequence? I can come up with
> plenty of metrics on R for which such a thing does not exist (e.g.,
> the discrete metric).

It means I pushed this as a sub-problem, assuming I can do that, and see
what follows. With d changed it should now hold.

> > x_n in Q and
> > lim_{n->oo} x_n = x
> >
> > A convergent sequence is also a Cauchy sequence. Thus:
> > for all epsilon > 0: exists N: for all n, m > N:
> > d(x_n, x_m) < epsilon
> >
> > I already proved
> > for all q in Q: f(q) = q
> >
> > Thus (f(x_n)) is also a Cauchy sequence in (R, d). By completeness of R,
>
> Again: since you simply said that d was *a* metric on R, you have no
> grounds for claiming that R is complete *with respect to d*. So this
> step is a non-sequitur because of how you set things up.

Yep, fixed with changing d.

> > (f(x_n)) converges to some value s. I would now have to see that
> > actually
> >
> > s = lim_{n->oo} f(x_n) = f(x)
>
> You have no grounds for claiming that the limit of the images of the
> x_i is the image of x. Nor did you use the fact that the sequence was
> increasing. So why did you define it as increasing?

I am not claiming that. I am saying that the next thing I need to prove
is that thing. I did not yet define it as increasing.

> > Find a non-decreasing sequence (x_n) that converges to x. Then:

Here I add the property of being non-decreasing to (x_n).

> > for all epsilon > 0: exists x_eps in Q:
> > x - epsilon < x_eps < x

Here I define "x_eps".

> > Because (x_n) is monotonically convergent to x:
> > exists N: for all n > N: x_n >= x_eps
>
> What is "x_eps"?

See above.

> > => (for all q in Q: f(q) = q)
> > exists N: for all n > N: f(x_n) >= f(x_eps)
> > => (f preserves order)
> > lim_{n->oo} f(x_n) >= f(x_eps)
> > => (epsilon arbitrary)
> > lim_{n->oo} f(x_n) >= f(x)
> >
> > However, because of monotonicity of (x_n) we also know
> > x_n <= x
> > => f is order preserving
> > f(x_n) <= f(x)
> > =>
> > lim_{n->oo} f(x_n) <= f(x)
> >
> > Thus by totality
> > lim_{n->oo} f(x_n) = f(x)
> >
> > Then
> > f(x) = lim_{n->oo} f(x_n) = lim_{n->oo} x_n = x

The above then follows. I.e. after changing d to the standard metric it
seems to me my proof is ok, even if unnecessary complex. Do you agree?

> It is utterly confused, beginning with your idea of picking "a" metric
> for R, and overly complicated.
>
> Take x, and let x_n and y_n be increasing and decreasing sequences of
> rationals, respectively, that converge to x. Then x_n <= x <= y_n for
> all n, hence since f respects order, f(x_n) = x_n <= f(x) <= f(y_n) =
> y_n for all n (since f(q) = q for all rationals q). Since this holds
> for all n, you have that x = lim(x_n) <= lim(f(x)) <= lim(y_n) = x.

That is much cleaner.

--
http://kaba.hilvi.org
From: achille on 18 Jan 2010 08:06
On Jan 18, 5:50 pm, Kaba <n...(a)here.com> wrote:
>
> The above then follows. I.e. after changing d to the standard metric it
> seems to me my proof is ok, even if unnecessary complex. Do you agree?
>

> --http://kaba.hilvi.org

How about using the fact f is order preserving [*],
If f(x) <> x at any point x, pick a rational number
r between x and f(x), then

case1) x < f(x) => x < r AND f(x) > f(r).
case2) f(x) < x => r < x AND f(r) > f(x).

both cases contradict with [*], so f(x) = x for all x.



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