a) Let set $M$ be in Banach space $X$, and any continuous function on $M$ is uniformly continuous. Is it true that $M$ is compact?
b) Let M be compact in Banach space $X$ and $f:M\rightarrow Y$ be continuous mapping into Banach space $Y$. Is set $f(M)\subset Y$ compact?
c) Let M be a uniformly bounded set of functions in $C[a,b]$. Prove that the set of functions $y(t) = \int_{a}^{t}x(s)ds, x \in M$, is precompact
For a, I think the answer is no,but I can't seem to find a a function that makes M not compact.
For b, I have: Let {$V_a$} be an open cover of $f(M)$. Then since f is continuous, $f(V_a)$ is open. Since $M$ is compact, there are finitely many $a_i$'s such that $M \subset f^{-1}(V_{a_1}) \cup ... \cup f^{-1}(V_{a_n})$. For any $E\subset Y$ we have $f(f^{-1}(E))\subset E$ so $f(M)\subset V_{a_1} \cup ... \cup V_{a_n}$ which means $f(M)$ is compact. I'm just concerned here because I didn't use that $X$ is Banach
For c: By Arzela-Ascoli, we just need to show equicontinuity, but I can't seem to figure this out. Consider t,u where $|t-u| < \delta$. Then $|\int_{a}^{t}x(s)ds - \int_{a}^{u}x(s)ds| = |x(t)-x(a)-(x(u)-x(a))| = |x(t)-x(u)|\leq 2N$ where $N$ is the bound for $|x|$. I don't know where to go from here.
Any help is appreciated!
For a) take $X=\mathbb R$, $M=\mathbb N$. Every function on M is uniformly continuous but M is not compact. b) does not need the fact that X is Banach. One of the basic facts about compactness in topological spaces says continuous images of compact spaces are compact. For c) note that $|\int_a ^{t} x(s)ds-\int_a^{u} x(s)ds|=|\int_u^{t} x(s)ds|$ if $u<t$. This gives $|\int_a ^{t} x(s)ds-\int_a^{u} x(s)ds| \leq M |t-u|$ where M is an upper bound for the functions in M. By the same argument the inequality holds for $t<u$ also. We now have equi-continuity and Arzela - Ascoli applies.