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From: Ostap Bender on 27 May 2010 05:28
On May 27, 2:26 am, Ostap Bender <ostap_bender_1...(a)hotmail.com>
wrote:
> On May 24, 4:21 am, William Elliot <ma...(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.
>
> PROOF: Suppose there exists a sequence S of elements {sj} from (0,1]
> for which sj -> 0 and the sequence f(sj) does not tend to 0. WLOG we
> can assume that all f(sj) >= 0. By definition of non-convergence,
> there must exist a b>0 and an infinite subsequence W of  {f(sj)}  s.t
> f(sj)>b. For each sj s.t. f(sj) is in W, let qj be a rational number

That's gj, not qj

> s.t. 0 <gj <sj and f(gj) > b/2. Such gj exists because f is left-
> continuous at sj. This sequence {gj} is in (0,1] /\ Q, gj -> 0; but
> f(gj) > b/2  for all j, and thus {f(gj)} does not converge to 0.
> Contradiction.
>
> QED
>
> This function f doesn't even have to be right-continuous. All you need
> is for f to be left-continuous.

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