0

I am trying to prove that the uniform metric over the set of bounded functions is complete. Let me define:

$ B(T,X) = \{f: T \to X: f(T)\; is\; bounded \} $

$ d_\infty: B(T,X) \times B(T,X) \to \mathbb{R} $

$ d_{\infty} = sup\{ ~d(f(t),g(t))~: t \in T \}$

$ Bd(x,\epsilon) = \{ y ~|~ d(x,y) < \epsilon \} $

They ask me to prove that:

Given $(X,d)$ a complete metric space. Show that $(B(T,X), d_\infty) $ is complete.

My try:

Let $(X,d)$ be a complete metric space. We need to show that $(B(T,X),d_\infty)$ is complete, that is,show that every Cauchy sequence of $(B(T,X),d_\infty)$ is convergent. To do that, let's take a Cauchy sequence $ (f_n)$ of points in $B(T,X)$ and let $Bd(f,\epsilon) \in \tau_{d_\infty} with \epsilon >0$. Since $(f_n)$ is Cauchy, then there exists $n_0 \in \mathbb{Z}_{+}$ such that $d_\infty(f_n, f_m) < \epsilon ~ \forall ~ n,m \geq n_0$. We must show that $ d_\infty(f_n, f) < \epsilon$.

Here is where I get stuck. I think I must use the fact that $ d_\infty(f_n, f) \leq d_\infty(f_n, f_m) + d_\infty(f_m, f)$ but I don't get any idea of how to do it. Any help is precious right now.

Henry Manson
  • 53
  • 1
    Once you have a candidate limit $f$ you can just let $m \to \infty$ in the Cauchy inequality pointwise to prove uniform convergence. The limit $f$ is the pointwise limit of $(f_n)$, which exists since $X$ is complete. – Mason Apr 05 '21 at 05:16
  • Yes, I get that, I must prove uniform convergence, I know already that $d_\infty(f_n,f_m) < \epsilon $ but I don't know how to get to $d_\infty(f_m,f) < \epsilon $ – Henry Manson Apr 05 '21 at 12:36

0 Answers0