Show that lim f (x) exists in a Banach space Hi I'm struggling with a proof.
Let E be a normalized vector space and F a Banach space. We know that A ⊂ E, x0 ∈ Acc (A) and
f: A → F is a uniformly continuous function. How would you show that lim f (x) exists?
I have some ideas to how to prove with, but I'm probably missing a lot.
I would like to prove (())is Cauchy, because, from there, I will be able to prove that (() converges because we are in a Banach space.
I also think that if x0 ∈ Acc (A), it exists a sequence () as lim () = x0.
From there I don't know what to do. I'm kinda lost as you can see!
Thank you for your help!
 A: Consider $(A,d),(F,e)$ as metric spaces, with $d(a,a')=\|a-a'\|_E$ and $e(g,g')=\|g-g''\|_F.$
Let $x_0\in E$ and let $(a_n)_n$ be a sequence in $A$ with $\lim_{n\to\infty}\|x_0-a_n\|=0.$ Then $(a_n)_n$ is a $d$-Cauchy sequence in the space  $(A,d).$
For metric spaces $(A,d),(F,e)$  and a uniformly continuous $f:A\to F$, if $(a_n)_n $ is a $d$-Cauchy sequence then $(f(a_n))_n$ is an $e$-Cauchy sequence.
Proof. By contradiction, suppose $r\in \Bbb R^+$ and $\lim_{n\to\infty}\sup_{n\le n'<n''}\,e(f(a_{n'}),f(a_{n''})\,)>2r.$ Then there exists an infinite $S\subset \Bbb N$ such that
$(1*)$ for each $n'\in S$ there exists $n''>n'$ such that $e(f(a_{n'}),f(a_{n''}))>r.$
Now $f$ is uniformly continuous so take $\delta >0$ such that
$(2^*)\;\forall x,y\in A\,(\,e(x,y)<\delta \implies e(f(a_{n'}),f(a_{n''}))<r).$
And $(a_n)_n$ is a $d$-Cauchy sequence so take $n_0\in \Bbb N$ such that
$(3^*)\;n_0\le n'<n''\implies d(a_{n'},a_{n''})<\delta.$
But take $n'\in S$ with $n'\ge n_0.$ By $(2^*)$ and $(3^*)$ we have $\forall n''>n'\,(\,e(f(a_{n'}),f(a_{n''})<r)\,),$ contradicting $(1*)$.
Now if $F$ is a Banach space then $e$ is a complete metric.
