0

Theorem

Let $M$ and $N$ be metric spaces, then the metric space $F_{b}(M,N)$ $= \{f:M \to N : \text{f bounded} \}$ is complete if $N$ is complete

Proof

Let $\{f_k\}_{k=1}^{\infty}$ be a Cauchy sequence in $F_{b}(M,N)$ $= \{f:M \to N : \text{f bounded} \}$. Then (why?) for any $m \in M$ the sequence $\{f_k(m)\}_{k=1}^{\infty}$ is a Cachy sequence in $N$.

I do not understand why the second statement (second sentence) follows from the first.

The metric on the set $F_b(M,N)$ is $d(f,g) = \sup_{m \in M} d_N(f(m),g(m))$

Olba12
  • 2,579

1 Answers1

1

I assume you are working with the following distance: $d(f_j,f_k)=\sup\{d(f_j(m),f_k(m));m\in M\}$. If $\{f_k\}$ is Cauchy, then for every $\epsilon>0$ there exist $N_\epsilon$ such that $d(f_j,f_k)<\epsilon$ for any $j,k\ge N_\epsilon$. This also implies that for any $m\in M$, $d(f_j(m),f_k(m))\le d(f_j,f_k)<\epsilon$ for any $j,k\ge N_\epsilon$. Hence $\{f_k(m)\}$is Cauchy also.

Catalin
  • 33