Boundedness in Function Spaces Let $(C(I),d_\infty)$ be a metric space, where $C(I)$ is the set of all continuous functions on a closed interval $I$ and $d_\infty=\sup_{x\in I}\{|f(x)-g(x)|\}$.
I am reading a book about real analysis and I am new to the subject. So I am not very familiar with function spaces. My question is: if there is a subset $K\subset C(I)$ and if $K$ is bounded can I say that the family of functions $f\in K$ is uniformly bounded (i.e. there is a constant N such that $\sup_x |f(x)|\leq N, \ \forall f\in K)$ as well? How about the converse?
 A: As you stated , if your closed interval is compact (which it is if $I$ is bounded) then you have (I will go ahead with the definition you gave in the comments)
Assuming your $K$ is nonempty, pick a fixed function $g\in K$ and observe that $\exists M$ such that $ |g(x)|\leq M$ for all $x\in I$.(since $I$ is compact), therefore by triangle inequality, we have for all $f\in K$,
$|f(x)|\leq |f(x)-g(x)|+ |g(x)|\leq r+M$. Take supremum to obtain the desired conclusion of uniform boundedness.
For the converse, assuming the uniform boundedness, just observe (again by triangle inequality)
$|f(x)-g(x)|\leq \sup_{x\in I}|f(x)|+\sup_{x\in I}|g(x)| \leq N $ for some $N$, so taking supremum would  give you, $d_{\infty}(f,g)\leq N $ for any two functions $f,g\in K\implies K$ is bounded.
Lastly, if $I$ is not bounded eg. $I=[0,+\infty )$ then such conclusions are not true. Just take the trivial example $K=\{ x,x+1\}$ to be two point set and observe none of the functions are bounded although $K$ is.
I hope this answers your question $\ldots $
