I don't know how to prove this problem:
Given $D\neq \emptyset$ and the space $A=\{f:D\to \mathbb{R} : \exists k_f\in \mathbb{R}^+, |f(x)|\leq k_f, \forall x\in D \} $ with the metric $d(f,g)=\displaystyle \sup_{x\in D} \{|f(x)-g(x)|\} $, consider, $\forall a\in D$, the functions $f_a\in A$ such that $f_a(a)=1$ and $ f_a(x)=0, \forall x\neq a $. Prove that, if $a\neq b$, $B(f_a,1/2)\cap B(f_b,1/2)=\emptyset $.
I suppose that exists $h\in B(f_a,1/2)\cap B(f_b,1/2) $. Then, $\sup_{x\in D} \{|f_a(x)-h(x)|\} <1/2$ and $\sup_{x\in D} \{|f_b(x)-h(x)|\} <1/2$. Using that $|h(x)|\leq k_h $ I should get a contradiction, but I dont know where.
Thanks for your help!
$|f_a(x)-h(x)| <1/2$ for all $x$ so $|h(x)| <1/2$ for all $x\neq a$. Similarly $|h(x)| <1/2$ for all $x\neq b$. Hence $|h(x)| <1/2$ for all $x$. Now $|f_a(a)-h(a)| <1/2$, so $|h(a)|<1/2$ and $|1-h(a)| </1/2$. But there is no real number $h(a)$ such that $|1-h(a)| <1/2$ and $|h(a)| </1/2$ since the intervals $(-1/2,1/2)$ and $(1/2,3/2)$ are disjoint.