Continuity problem over the reals
José Carlos Santos
Let _f_ be a function from R into itself with this property: whenever a
real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
Best regards,
Jose Carlos Santos
master1729
no.
David C. Ullrich
<jcsa...@fc.up.pt> wrote:
>Hi all,
>
>Let _f_ be a function from R into itself with this property: whenever a
>real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
>then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
Yes. This is actually easy.
Say f is discontinuous at 0. Let's say f(0) = 0 but there
is a sequence of reals r_n -> 0 with f(r_n) > 1 for all n.
For each n, consider a sequence of rationals approaching
r_n, and you see that there exists a rational q_n with
f(q_n) > 1/2 and |r_m - q_n| < |r_n|/2. Hence q_n -> 0,
contradicting the hypothesis.
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
David C. Ullrich
wrote:
Giggle. How do you prove that?
(giggle. I suppose you figured you had a 50-50 chance
of guessing right. Too bad about that.)
José Carlos Santos
>> Let _f_ be a function from R into itself with this property: whenever a
>> real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
>> then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
>
> Yes. This is actually easy.
I bet you only wrote that to make me feel bad. :-)
> Say f is discontinuous at 0. Let's say f(0) = 0 but there
> is a sequence of reals r_n -> 0 with f(r_n) > 1 for all n.
> For each n, consider a sequence of rationals approaching
> r_n, and you see that there exists a rational q_n with
> f(q_n) > 1/2 and |r_m - q_n| < |r_n|/2.
You meant r_n here, not r_m.
> Hence q_n -> 0, contradicting the hypothesis.
Got it. Thanks a lot.
David C. Ullrich
Jos� Carlos Santos <jcsa...@fc.up.pt> wrote:
> On 10-11-2009 15:20, David C. Ullrich wrote:
>
> >> Let _f_ be a function from R into itself with this property: whenever a
> >> real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
> >> then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
> >
> > Yes. This is actually easy.
>
> I bet you only wrote that to make me feel bad. :-)
Sorry. If it makes you feel any better, although it _is_
easy it took me a minute to decide.
(Ok, I actually wrote that to make someone else feel bad,
I'll let you guess who.)
>
> > Say f is discontinuous at 0. Let's say f(0) = 0 but there
> > is a sequence of reals r_n -> 0 with f(r_n) > 1 for all n.
> > For each n, consider a sequence of rationals approaching
> > r_n, and you see that there exists a rational q_n with
> > f(q_n) > 1/2 and |r_m - q_n| < |r_n|/2.
>
> You meant r_n here, not r_m.
Just making sure you were paying attention, haha.
> > Hence q_n -> 0, contradicting the hypothesis.
>
> Got it. Thanks a lot.
>
> Best regards,
>
> Jose Carlos Santos
--
David C. Ullrich
Gc
> Hi all,
>
> Let _f_ be a function from R into itself with this property: whenever a
> real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
> then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
Follows directly from that every real number is always a (equivalence
class of) limit(s) of a cauchy sequence of rationals. You didn`t
remember this?
Gc
I mean "quite" directly. I was thinking a limit changing procedure, to
which, now I think, is little more complicated because I think you
need to prove a equicontinuity. But it is sometimes nice to have also
a direct proof.
David C. Ullrich
wrote:
Remmeber what, exactly? Saying every real number is an equivalence
class of limits of a cauchy sequence of rationals is nonsense. A
cauchy sequence has one limit, and a real _is_ the limit of a
cauchy sequence, not an equivalence class of such limits.
Every real _is_ the limit of a cauchy sequence of rationals.
(And one construction of the reals says a real is an equivalence
class of cauchy sequences of rationals.)
Gc
> On Tue, 10 Nov 2009 09:48:03 -0800 (PST),Gc<gcut...@hotmail.com>
> wrote:
>
> >On 10 marras, 13:19, José Carlos Santos <jcsan...@fc.up.pt> wrote:
> >> Hi all,
>
> >> Let _f_ be a function from R into itself with this property: whenever a
> >> real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
> >> then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
>
> >Follows directly from that every real number is always a (equivalence
> >class of) limit(s) of a cauchy sequence of rationals. You didn`t
> >remember this?
>
> Remmeber what, exactly? Saying every real number is an equivalence
> class of limits of a cauchy sequence of rationals is nonsense. A
> cauchy sequence has one limit, and a real _is_ the limit of a
> cauchy sequence, not an equivalence class of such limits.
This is what I meant...
> Every real _is_ the limit of a cauchy sequence of rationals.
> (And one construction of the reals says a real is an equivalence
> class of cauchy sequences of rationals.)
...and somewhere when writing about it confused with this.
Gc
> wrote:
>
> >On 10 marras, 13:19, José Carlos Santos <jcsan...@fc.up.pt> wrote:
> >> Hi all,
>
> >> Let _f_ be a function from R into itself with this property: whenever a
> >> real number _x_ is the limit of a sequence (q_n)_n of rational numbers,
> >> then lim_n f(q_n) = f(x). Does it follow that _f_ is continuous?
>
> >Follows directly from that every real number is always a (equivalence
> >class of) limit(s) of a cauchy sequence of rationals. You didn`t
> >remember this?
>
> Remmeber what, exactly? Saying every real number is an equivalence
> class of limits of a cauchy sequence of rationals is nonsense. A
> cauchy sequence has one limit, and a real _is_ the limit of a
> cauchy sequence, not an equivalence class of such limits.
Well, let the equivalence relation be "=" :)