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Topic: Looking for a counterexample
Replies: 4   Last Post: Nov 29, 2010 2:30 PM

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 Dan Luecking Posts: 26 Registered: 11/12/08
Re: Looking for a counterexample
Posted: Nov 29, 2010 2:30 PM

On Sun, 28 Nov 2010 20:30:03 +0000 (UTC), Rodolfo Conde
<rcm@gmx.co.uk> wrote:

>
>
> Hi all,
>
> I am trying to find a counterexample for the following
>proposition (which is false):
>
> Suppose that we have a sequence (f_n: K -> R^q) (where K is a
>compact subset of some euclidean space R^n) of continuous functions and this
>sequence converges uniformly to a function f: K -> R^q such that f is an
>imbedding of K. Then for n large enough, f_n is an imbedding

I assume by imbedding you mean at least that f is a
homeomorphism to ite image. Here is a straightforward
counterexample in one dimension: Let K be the interval
[0,1] and define f_n : K -> R by
f_n(x) = 0, 0 <= x < 1/n;
f_n(x) = x - 1/n, 1/n <= x <= 1.
This converges uniformly to the identity function f(x) = x.

One could clearly adapt this to provide an example for
any dimension. And one could adapt it to give f_n any
number of derivatives.

Dan
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Date Subject Author
11/28/10 Rodolfo Conde
11/29/10 Lee Rudolph
11/29/10 Robert Israel
11/29/10 Julia Kuznetsova
11/29/10 Dan Luecking