Extending a continuous function
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From: William Elliot on 24 May 2010 07:21 Assume f:(0,1] -> R is continuous and for all sequences {sj} of elements from (0,1] /\ Q for which sj -> 0, the sequence f(sj) -> 0. Show that for all sequences of elements {sj} from (0,1] for which sj -> 0, the sequence f(sj) -> 0 In otherwords, show f can be continuously extended to 0. Proposition. f in C((0,1],R), for all sequence s into (0,1] /\ Q, (s -> 0 ==> fs -> 0) ==> for all sequence s into (0,1], (s -> 0 ==> fs -> 0)
From: A N Niel on 24 May 2010 07:48 In article <20100524041147.B566(a)agora.rdrop.com>, William Elliot <marsh(a)rdrop.remove.com> wrote: > Assume f:(0,1] -> R is continuous and > for all sequences {sj} of elements from (0,1] /\ Q > for which sj -> 0, the sequence f(sj) -> 0. > > Show that for all sequences of elements {sj} from (0,1] > for which sj -> 0, the sequence f(sj) -> 0 > > In otherwords, show f can be continuously extended to 0. > > Proposition. > f in C((0,1],R), for all sequence s into (0,1] /\ Q, (s -> 0 ==> fs -> 0) > ==> for all sequence s into (0,1], (s -> 0 ==> fs -> 0) > > Are you telling us this exercise, or are you asking for assistance? Perhaps the contrapositive... If f(s_j) does not converge to zero for some sequence s_j, then there is a sequence s_j of rationals such that f(s_j) still does not converge to zero.
From: cwldoc on 24 May 2010 08:51 > Assume f:(0,1] -> R is continuous and > for all sequences {sj} of elements from (0,1] /\ Q > for which sj -> 0, the sequence f(sj) -> 0. > > Show that for all sequences of elements {sj} from > (0,1] > for which sj -> 0, the sequence f(sj) -> 0 > > In otherwords, show f can be continuously extended to > 0. > > Proposition. > f in C((0,1],R), for all sequence s into (0,1] /\ Q, > (s -> 0 ==> fs -> 0) > ==> for all sequence s into (0,1], (s -> 0 ==> fs > fs -> 0) > > Let sj be a sequence of elements of (0,1] such that sj -> 0 Supose that f(sj) fails to converge to zero. Then there exists e > 0 such that for every positive integer, J, there exists a j > J such that |f(sj)| > e Thus (letting J = 1), there exists a positive integer, j1, such that |f(s(j1))| > e. And ((letting J = j1)) there exists a positive integer, j2, such that |f(s(j2))| > e. Continuing in this manner, we define a subsequence of sj, denoted by s(jk), such that |f(s(jk))| > e for every k. By continuity of f, for each k, we can choose a dk > 0 such that |s(jk) - x| < dk implies |f(s(jk) - f(x)| < |f(s(jk)| - e and such that dk < 2^(-k) For each k, choose a rational element, tk, of (0,1] such that |tk - s(jk)| < dk Then tk -> 0. Furthermore, |f(tk)| > e, for all k, so f(tk) does not converge to zero. This is a contradiction, since tk is comprised of rational elements of (0,1].
From: William Elliot on 25 May 2010 05:01 From: cwldoc <cwldoc(a)aol.com> >> Proposition. Proposition 1. >> f in C((0,1],R), for all sequence s into (0,1] /\ Q, >> (s -> 0 ==> fs -> 0) >> ==> for all sequence s into (0,1], (s -> 0 ==> fs -> 0) > Let sj be a sequence of elements of (0,1] such that sj -> 0 > Supose that f(sj) fails to converge to zero. > Then there exists e > 0 such that for every positive integer, > J, there exists a j > J such that |f(sj)| > e > Thus (letting J = 1), there exists a positive integer, j1, Do you mean jk instead of j1? > such that |f(s(jk))| > e for every k. Are you're creating a subsequence s_jk of s by induction? > By continuity of f, for each k, we can choose a dk > 0 such that > |s(jk) - x| < dk implies |f(s(jk) - f(x)| < |f(s(jk)| - e Do you mean "such that for all x in (0,1], (|s(jk) - x| < dk implies |f(s(jk) - f(x)| < |f(s(jk)| - e)"? > and such that dk < 2^(-k) > For each k, choose a rational element, tk, > of (0,1] such that |tk - s(jk)| < dk > Then tk -> 0. Furthermore, |f(tk)| > e, for all k, > so f(tk) does not converge to zero. |f(s_jk) - f(tk)| < |f(s_jk)| - e > This is a contradiction, since tk is > comprised of rational elements of (0,1]. Proposition 2. Let X and Y be metrics spaces. Assume D dense subset X, U open subset X, f in C(U,Y) For all a in cl U - U, assume for all sequences sj of elements of U /\ D which converge to a, that the sequence f(sj) converges to f(a). Show for every a in cl U - U that for all sequences sj of elements of U which converges to a, the sequence f(sj) converges to f(a). Can your proof be extended to a proof of the above generalization? In the part |f(s(jk) - f(x)| < |f(s(jk)| - e d(f(s_jk), f(x)) can be used for |f(s(jk) - f(x)| and d(f(s_jk)), f(a)) - e can be used for |f(s(jk)| - e Would you concure that your proof of proposition 1 is a template for a proof of proposition 2? ----
From: cwldoc on 26 May 2010 11:45
> From: cwldoc <cwldoc(a)aol.com> > > >> Proposition. > Proposition 1. > > >> f in C((0,1],R), for all sequence s into (0,1] /\ > Q, > >> (s -> 0 ==> fs -> 0) > >> ==> for all sequence s into (0,1], (s -> 0 ==> fs > -> 0) > > > Let sj be a sequence of elements of (0,1] such that > sj -> 0 > > > Supose that f(sj) fails to converge to zero. > > > Then there exists e > 0 such that for every > positive integer, > > J, there exists a j > J such that |f(sj)| > e > > > Thus (letting J = 1), there exists a positive > integer, j1, > > Do you mean jk instead of j1? No. > > > such that |f(s(jk))| > e for every k. > > Are you're creating a subsequence s_jk of s by > induction? Yes. > > > By continuity of f, for each k, we can choose a dk > > 0 such that > > > |s(jk) - x| < dk implies |f(s(jk) - f(x)| < > |f(s(jk)| - e > > Do you mean "such that for all x in (0,1], Yes. > > (|s(jk) - x| < dk implies |f(s(jk) - f(x)| < > |f(s(jk)| - e)"? > > > and such that dk < 2^(-k) > > > For each k, choose a rational element, tk, > > of (0,1] such that |tk - s(jk)| < dk > > > Then tk -> 0. Furthermore, |f(tk)| > e, for all k, > > so f(tk) does not converge to zero. > > |f(s_jk) - f(tk)| < |f(s_jk)| - e which is equivalent to f(sjk) - (|f(sjk)| - e) < f(tk) < f(sjk) + (|f(sjk)| - e) For |f(sjk)| > 0, this becomes e < f(tk) < 2 f(sjk) - e While for |f(sjk)| < 0, this becomes 2 f(sjk) + e < f(tk) < -e Either way, this implies |f(tk)| > e > > > This is a contradiction, since tk is > > comprised of rational elements of (0,1]. > > Proposition 2. Let X and Y be metrics spaces. > Assume D dense subset X, U open subset X, f in C(U,Y) > For all a in cl U - U, assume for all sequences sj of > elements of U /\ D which converge to a, that the > sequence > f(sj) converges to f(a). Proposition 2 would have to be modified because f(a) is not defined, since a is not contained on U and f is only defined on U. > > Show for every a in cl U - U that for all sequences > sj of > elements of U which converges to a, the sequence > f(sj) > converges to f(a). > > Can your proof be extended to a proof of the above > generalization? > In the part > |f(s(jk) - f(x)| < |f(s(jk)| - e > > d(f(s_jk), f(x)) can be used for |f(s(jk) - f(x)| > and d(f(s_jk)), f(a)) - e can be used for |f(s(jk)| - > e > > Would you concure that your proof of proposition 1 > is a template for a proof of proposition 2? > > ---- |