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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 for any $m \in M$ the sequence $\{f_k(m)\}_{k=1}^{\infty}$ is a Cachy sequence in $N$. Since N is complete, there exists a limit $f_m \in N$. This defines a function $f:M \ni m \to f_m \in N$ Why does "this" definie a function? Given $\epsilon > 0$

$$ d_N(f(m),f_k(m)) = lim_{j \to \infty} d_N(f_j(m),f_k(m)) \leq \epsilon $$ if $k\geq N(\epsilon)$. This shows that $f$ is a bounded function why? and that $$ \sup_{m\in M} d_N(f(m), f_k(m)) \to 0 \hspace{2cm} \text{when } k \to \infty $$

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

For a function to be bounded then their must $\exists m_0 \in M$ s.t $\sup_{m \in M} d_N(f(m), f(m_0)) < \infty$. But above I have $f(m)$ and $f_k(m)$. But $f(m) \not = f_k(m)$ ?

Olba12
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    To have a function $f:M\to N$, you have to be able to tell me what $f(m)$ is if I tell you what $m$ is (a point I specify in $M$), and your value of $f(m)$ must lie in $N$. Isn't that what is happening? Given an $m\in M$, there is a (unique) limit $f_m\in N$. The assignment is $f(m)=f_m$. That's the value. If the limit didn't exist, or wasn't in $N$, then the function would not be well-defined. – MPW Apr 04 '16 at 20:20
  • Thanks, that makes sence. But for a function to be bounded then their must $\exists m_0 \in M$ s.t $\sup_{m \in M} d_N(f(m), f(m_0)) < \infty$. But above I have $f(m)$ and $f_k(m)$. But $f(m) \not = f_k(m)$ ? @MPW – Olba12 Apr 04 '16 at 20:25
  • Check out this question, it seems to answer your question. The problem here is showing the candidate function is actually itself a bounded function. – MPW Apr 04 '16 at 21:02

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