Let $X$ be a linear space (over $\mathbb{R}$) with $\mathcal{N}$ a family of semi-norms on $X,$ the $\mathcal{N}$- topology on $X$ is the weakest topology that makes each $\rho\in X$ continuous. Write $U_{\rho}=\left\{x:\rho(x)\leq 1\right\}$, then a local base for $X$ will be $$\mathcal{U}=\left\{U=\lambda\bigcap_{i=1}^{n}U_{\rho_i}:\rho_i\in X,\lambda>0\right\}$$ A set $A\subset X$ is said to be totally bounded if for any $U\in\mathcal{U},$ there is some finite subset $B$ of $A$ so that $A\subset B+U.$ I want to prove that $A\subset X$ is totally bounded $\Longleftrightarrow \forall\varepsilon>0,\forall\rho\in \mathcal{N},A$ has finite $(\varepsilon,\rho)$-net.
The "$\Rightarrow$" part is easy: $\forall\varepsilon>0,\forall\rho\in\mathcal{N},$ take $U=\frac{\varepsilon}{2} U_{\rho}\in\mathcal{U},$ then total boundedness of $A$ implies that there is some finite subset $B$ of $A$ such that $A\subset B+U,$ that's: $\forall a\in A,\exists b\in B,u\in U,s.t. a=b+u,$ so $\rho(a-b)=\rho(u)\leq \frac{\varepsilon}{2}<\varepsilon,$ thus $B$ is a finite $(\epsilon,\rho)$-net for $A.$
The "$\Leftarrow$" part seems to be hard for me: For each $U\in\mathcal{U},$ we should find some finite subset $B$ of $A$ so that $A\subset B+U,$ write $U$ as $U=\lambda\bigcap_{i=1}^{n}U_{\rho_i},$ so $B$ should satisfies $\forall a\in A,\exists b\in B,u\in U,s.t. a=b+u,$ while $u\in U$ implies $\rho_i(u)\leq\lambda,\forall i.$ But I have no idea how to find such a finite set, can someone help me?