Re: Tietze theorem and open sets

From: Valeriu Anisiu (vanisiu_at_personal.ro)
Date: 02/03/05 

Date: Thu, 3 Feb 2005 00:11:53 +0000 (UTC)

On 02 Feb 2005, lukasz wrote:
>I have the following version of Tietze theorem:
>
>Let A be a closed subset of X and f:A->R a continuous function.
>There exists a continuous function F:X->R such that F(x) = f(x) for x
>in A.
>Moreover, if |f(x)| <= u for x in A, then |F(x)| <= u and |F(x)| < u
>for x in X\A.
>
>As far as I know the condition for A to be closed is used in the
>theorem's proof to show certain function F is continuous. But there may
>be an other function F satisfying the theorem. So can anyone explain
>why the condition is important and/or provide a counterexample that
>shows the theorem does not apply when A is open?

Take X=R, A=(0,1), f(x)=1/x or f(x)=sin(1/x).

V. Anisiu

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