Let $$X=\{f:[0,1]\rightarrow[0,1], f\text{ is Lipschitz continuous}\} $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
Best Answer
If $f_n(x)=x^n$ ($x\in[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_{n\in\mathbb N}$ has no convergente subsequence. Therefore, your space is not compact.