Results 1 to 2 of 2

Math Help - intotoanalysis

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    13

    intotoanalysis

    Let K be a bounded non-empty subset of R.Let f:K->R be a uniformly continous function.Show that f is bounded.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by sisa View Post
    Let K be a bounded non-empty subset of R.Let f:K->R be a uniformly continous function.Show that f is bounded.
    Ok, so lets just establish a few things, the metric d(p,p') in \mathbb{R}^n is |p-p'|. Since assuming that K\subset\mathbb{R} we have that \phi:K\to\mathbb{R} is a real function. So for \phi to be uniformly continous we must have that for every \varepsilon>0 there exists a \delta>0 such that |p-p'|<\delta\implies|\phi(p)-\phi(p')|<\varepsilon.

    1. Now assume the converse and state that \phi(p) is unbounded. By the fact that we are mapping to the reals let us fix p' and let \phi(p') be finite. Now since \phi(p) is unbounded in K for every M\in\mathbb{R} we may find a p\in K such that M<|\phi(p)|, so for every M_0\in\mathbb{R} we can find a p\in K such that M_0<|\phi(p)-\phi(p')| which violates |\phi(p)-\phi(p')|<\varepsilon
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum