Hoi, I'm trying to prove that in a banach space: If f is locally lipschitz, then it maps bounded sets onto bounded sets.

Sounds fair right?

I started an argument as follows, but I couldn't finish it...

Choose $\displaystyle x_1\in B$, and let $\displaystyle U_1:=U(x_1)$ be the largest neighborhood in $\displaystyle B$ containing $\displaystyle x_1$, such that $\displaystyle f|_{U_1}$ is Lipschitz with constant $\displaystyle L_1$. For arbitrary $\displaystyle n$, if $\displaystyle B_n:=B\setminus \bigcup_{i<n} U_{i}$ is nonempty, choose $\displaystyle x_n\in B_n$ and define $\displaystyle U_n$ similarly. Repeat indefinately. If we find a finite sequence $\displaystyle U_1,\cdots, U_n$ we're done since $\displaystyle f(B) = \bigcup_{i\leq n} f(U_i)$ is clearly bounded since all $\displaystyle f(U_i)$ are bounded. If we find an infinite sequence $\displaystyle U_1,U_2,\cdots$ we have that $\displaystyle L_n\to \infty$, and we must show that $\displaystyle f(B)$ remains bounded. Clearly, since $\displaystyle B$ is bounded we have diam($\displaystyle U_n)\to 0$...

, and I wanted to use this. But, now I'm thinking that my argument wouldn't work because $\displaystyle f(B)$ could still be unbounded, if the diam($\displaystyle f(B_i)$) doesn't go to 0 fast enough...

what can I use here?? I dont see how this would especially work in a banach space. Maybe, the fact that we are working in a banach space is not even necessary?